A graph with at most one vertex of degree less than 3 has a subdivision of $K_4$

This is about a problem in the Bondy and Murty's book on Graph Theory.

Let $G$ be a nontrivial simple graph with $\delta(G)\geq 3$ or a vertex $v\in V(G)$ such that $\delta(G-v)\geq 3$. Then $G$ has a subdivision of $K_4$.

This is also about a question made by me in math.stackexchange about a case when the graph was $2-$connected but not $3-$connected. My original reasoning was as follows:

There are some remarks to do: First, we can reduce the cases to the first one: $\delta(G)\geq 3$. If $d(v)<3$ we can take an edge $e$ such that $v\in e$ and by induction on $|G|$, $G/e$ has a subdivision of $K_4$, so $G$ also has one. We can also suppose $G$ is connected since we can also use induction on its components. We can also suppose $G$ is planar since both $K_5$ and $K_{3,3}$ have subdivisions of $K_4$.

The first thing I tried to do was check (as stated above) that we could suppose $G$ was planar, but my last attempt was checking the connectedness of the graph as follows:

If $G$ is connected, has $\delta(G)\geq 3$ and is not $2-$connected, then take a cut-vertex $v$ and a component $C$ of $G-v$. Then the induced subgraph of $C\cup v$ has at most one vertex of degree $<3$ (the cut-vertex), and by induction we're done. So we can suppose $G$ is $2-$connected. If we can reduce the case to one on which $G$ is $3-$connected then any $3$ vertices are in a common cycle, and then we can apply the Fan Lemma or Menger Theorem appropiately to get the subdivision of $K_4$ (It will be a cycle of length at least $3$ not covering all the vertices and a $3-$fan from one of the remaining vertices to that cycle).

But as Misha Lavrov comments there, there is a flaw in the first paragraph of the attempt, which is shown in the following graph.

When contracting an edge incident to the red vertex it becomes the following graph:

which has two vertices of degree two. This can still be solved and the reasoning to do it will be shown next.

Using the Misha's answer and my own remarks, and ignoring the fan-lemma we have the following proof:

Proof. Suppose first that $G$ is not $3-$connected. For such a graph, we'll prove the result by induction on $n=|G|$. For $n=4$ the result is trivial ($G$ must be $K_4$), so we can suppose that $n>1$ and that any graph with less than $n$ vertices and at most one vertex of degree $<3$ contains a subdivision of $K_4$. Now, if, in fact, $G$ is not connected, then its connected components all have at most a vertex of degree $<3$. By induction hypothesis, each of them, besides a possibly trivial one (on which case the rest of them would have minimum degree $\geq 3$) contains a subdivision of $K_4$ so $G$ too. Let's now suppose $G$ is connected, but has a cut-vertex $v$. Let $C$ be a connected component of $G-v$ with no vertex of $G$ with degree $\leq 2$ (this is how we solve the problem mentioned above). Then $G[C\cup v]$ has at most one vertex of degree $2$ ($v$ is the only candidate). By induction hypothesis, $G[C\cup v]$ contains a subdivision of $K_4$; so does $G$, then. 

Now suppose $G$ is $2-$connected but contains a separating set of minimum size $S$ with two vertices. Take $S=\{v,w\}$ in such a way that a component $C$, with no vertices of $G$ with degree $\leq 2$, of minimum size of $G-S$ is as smallest as possible.  

This means that both $v$ and $w$ have two adjacent vertices in $C$. Indeed, if $v$ (resp. $w$) has only one adjacent vertex $v'$ in $C$, then $\{v',w\}$ is another separating set in $G$, with a component being $C'=C-v'$. Also, $C'\neq \varnothing$, in such a case it would happen that $C=\{v'\}$ and $d(v')=2$, a contradiction.

Furthermore, since $v,w$ both have adjacent vertices in any other component of $G-S$, then there exists a $v-w-$path, say, $\gamma$ not meeting $C$.

Take a new graph $G'$ defined as follows: 

$$V(G')=V(C)\cup S,$$

$$E(G')=E(C)\cup E(S,C)\cup \{vw\}.$$

By the choice of $C$, again, $v,w$ have each one at least two edges to $C$, and the edge $vw$, adding $3$ to both degrees, so $d(v),d(w)\geq 3$. Then $\delta(G')\geq 3$ (Since the only changing degrees are the ones of $v,w$). By induction hypothesis, $G'$ contains a subdivision of $K_4$. If $vw\in E(G)$ we're done. If this subdivision uses $vw$ and $vw\notin E(G)$, then, in $G$, replace $vw$ by $\gamma$. It'll still be a subdivision of $K_4$. Therefore, every non-$3-$connected graph with at most one vertex of degree $\leq 2$ has a subdivision of $K_4$.

Now suppose $G$ is $3-$connected. We know, by Tutte's wheel theorem that $K_4$ is a minor of $G$, but we're looking for $K_4$ as a topological minor which is not the same. We'll work as follows: Take a cycle $\zeta$ of $G$ of minimum length, in such a way it's an induced cycle. Since $G$ is $3-$connected, then $G\neq \zeta$ and since $\zeta$ is induced, there is $v\in V(G)-V(\zeta)$. Also $\delta(G)\geq 3$, thus $d(v)\geq 3$. Since $N(v)$ separates $v$ fron $\zeta$, then by Menger's theorem there exist at least three disjoint $v-\zeta-$paths. Three of them, along with $\zeta$, make a subdivision of $K_4$.

Highly connected subgraph of a finite graph

In the article Forcing highly connected subgraphs from Maya Jakobine Stein there is a statement about a theorem of Mader which guarantees the existence of highly connected subgraphs of a given graph with a high minimum degree. Then a well known theorem of Mader which guarantees the existence given a high average degree is stated.

Theorem 1.  Any finite graph $G$ of average degree at least $4k$ has a  $(k+1)-$connected subgraph.

A result of the article is the following:

Theorem 2. Let $k\in\bN$ and let $G$ be a graph such that each vertex has degree at least $2k$ and each end has edge-degree at least $2k$. Then $G$ has a $(k + 1)-$edge connected region.

Theorem 1 doesn't guarantee by itself the existence of such a subgraph only given a minimum degree of $2k$ even in the finite case. The source of the theorem, which, unfortunately for me, is in German, may have such a result, but fortunately a simple proof of the following proposition follows from the steps used by Stein to prove Theorem 2.

Proposition 1. Let $k\in\bN$ and let $G$ be a finite graph with minimum degree $\delta(G)=2k$. Then $G$ has a $(k + 1)-$edge connected subgraph.

To understand a bit of Stein's proof, we have to know some concepts, like ray, end, region, boundary, and to adapt the proof to our finite case we must make sure none of those concepts are relevant for the finite case.

So, given a graph $G$ and a subgraph $H$, the boundary of $H$ is just the set $N(G-H)$ of all vertices of $G$ adjacent to some vertex in $G-H$. A region is a connected induced subgraph with finite boundary. An infinite path starting at some vertex is called a ray, and the ends are the equivalence classes of rays given a specific equivalence relation for them. The edge-degree of an end is the maximum cardinality of a set of edge-disjoint rays in it. So, indeed, since a finite graph has no rays, then it has no ends, and the corresponding hypothesis becomes a void one for the finite case. Also in a finite graph the boundary of any subgraph is obviously finite, so a region in a finite graph is just an induced subgraph.

With this we can say that the same proof should work in the finite case, but we also should be able to make it into a simpler proof. First, the proof is based on the following lemma:

Lemma. Let $D\neq \varnothing$ be a region of a graph $G$ so that $|E(D, G − D)| < m$ and so that $|E(D' , G − D')| ≥ m$ for every non-empty region $D' \subseteq D − \partial D$ of $G$. Then there is an inclusion-minimal region $H \subseteq  D$ with $|E(H, G − H)| < m$ and $H\neq \varnothing$.

The lemma is used to find a minimal region $C$ such that $|E(C, G − C)| < 2k$. But this lemma is just a big nuke for our finite case: If $G$ is not $(k+1)-$edge connected such a set $C$ exists: Take a minimum separating set $S$ of $G$ with $k$ edges, and a component $H$ of $G-S$. Such a component is a choice for the set $C$, and a good one since $|E(H,G-H)\leq k<2k$. The existence of a minimal one is also immediate: Any finite poset has a minimal element. Even more, we can take one of minimum size. So, for the following proof, which is in fact the last paragraph of the proof in the article (with a few more details), this set will be called $H$.

Proof of Proposition 1. Let $H$ be as in the comments above. We claim that $H$ is $(k+1)-$edge connected. Indeed, if it were not, then $|H|\geq 2$ (If $|H|=1$ then $V(H)=\{v\}$ for some $v\in V(G)$ and since $d(v)\geq 2k$, then $|E(H,G-H)|=d(v)\geq 2k$). Furthermore, $H$ has a minimum separating set $K'$ with $|K'|\leq k$. Then $H-K'$ has at least two components, say, $H',H''\subsetneq H$. If $$|E(H,G-H)\cap E(H',G'-H')|> \frac{E(H,G-H)}{2}$$
then $$|E(H'',G-H'')\cap E(H,G-H)|\leq \frac{E(H,G-H)}{2}< k.$$

So, given that, the following holds
$$E(H'',G-H'')\leq |E(H'',G-H'')\cap E(H,G-H)|+|K'|<2k$$
otherwise, if $|E(H,G-H)\cap E(H',G'-H')|\leq \frac{E(H,G-H)}{2}<2k$
then
$$E(H',G-H')\leq |E(H',G-H')\cap E(H,G-H)|+|K'|<2k.$$
Both cases contradict the minimality of $H$. Therefore, $H$ is $(k+1)-$edge connected.

Exercises on graph theory

Just some exercises of Bondy's and Murty's book. I also added some from Trudeau's book and Bollobás book.

Exercise 1. Let $G$ be a simple graph. Then $||G||\leq \binom{|G|}{2}$ where $|G|:=|V(G)|$ and $||G||:=|E(G)|$. Determine when equality holds.
Proof. First, the cardinality of the set of unordered pairs of a set $V$ is bounded by $\binom{|V|}{2}$, from which the first result follows. Now, a graph $G$ is complete iff it has exactly $\binom{|G|}{2}$ edges. Indeed, $G$ is complete if and only if every unordered pair ${u,v}$ of elements of $V(G)$ is in $E(G)$, if and only if there are $\binom{|G|}{2}$ of them.

Exercise 2. Every path is bipartite and a cycle is bipartite if and only if its length is even.
Proof. Let $G$ be a path and $n=|G|$. Partition $V=V(G)=\{v_1,\dots,v_n\}$ in the two sets $V_1=\{v_i\in V\mid i\text{ is odd.}\}$ and $V_0=\{v_i\in V\mid i\text{ is even.}\}$. Since every path in $G$ has the form $\{v_i,v_{i+1}\}$, and $i\not\equiv i+1\mod 2$ for every $i$, then the partition is adecuate for $G$ to be bipartite.

Now suppose $G$ is a cycle with even length. Then $n$ is also even. The partition given before still works so $G$ is bipartite. Now suppose that $G$ has odd length. If $V_1,V_2$ were a 'bipartition' of $V(G)$, then, renaming $V_1$ and $V_2$ appropiately we can suppose that $v_1\in V_1$. Then, since $V_1,V_2$ make a bipartition of $G$, the vertices adjacent to $v_1$, say, $v_{2},v_{n}$, must be in $V_2$. By applying the same reasoning repeatedly (starting with $v_2$ instead of $v_1$, in an ascending way) we conclude that $v_j\in V_1$ for every $j\equiv 1\mod 2$ and $v_j\in V_2$ for every $j\equiv 0\mod 2$. But applying the same reasoning (starting with $v_n$ in a descending way) we conclude that $v_j\in V_2$ for every $j\equiv n\equiv 1\mod 2$ and $v_j\in V_1$ for every $j\equiv 0\mod 2$. It follows that $V_1=V_2=V(G)$, a contradiction.

Exercise 3. Show that for any graph $G$, $\delta(G)\leq d(G)\leq \Delta(G)$, where $\delta(G),d(G),\Delta(G)$ denote the minimum, average, and maximum value between the degrees of the vertices of $G$ respectively.
Proof. For any set $A\subseteq \mathbb{Z}$, $\min A\leq \mu A\leq \max A$, where $\mu(A)$ denotes the average of the elements of $A$. So just take $A$ as the set of the degrees of the vertices of $G$.

Exercise 4.  Let $G$ be a graph. Then $G$ or $\overline{G}$ is connected.

Proof. Let's suppose $G$ is not connected and partition $G$ in it's more than one connected components, i.e. the partition associated to the equivalence relation $v\sim v'$ iff $v=v'$ or there is a path $v\leadsto v'$. Take two vertices $v,v'$. If they are in different connected components, then there is a path $v\leadsto v'$ in $\overline{G}$. If they are in the same one, take other vertex $w$ in another connected component. Then there are two paths $v\leadsto w,w\leadsto v'$. The composite path is a path $v\leadsto v'$. So, anyway, there is a path $v\leadsto v'$, thus $\overline{G}$ is connected.

Exercise 5. The graph $C_5$ is the only self-complementary cyclic graph.

Proof. If $C_5=[v_1,v_2,v_3,v_4,v_5,v_1]$ then $\overline{C}_5=[v_1,v_3,v_5,v_2,v_4,v_1]\cong C_5$, so $C_5$ is self-complementary. Furthermore, since the degree of all the vertices of $C_n$ is two, then if $n\neq 5$ the degree of the vertices of $\overline{C}_n$ is $n-3\neq 2$, so $\overline{C}_n$ can't be cyclic if $n\neq 5$.

Exercise 6. If $G$ is self-complementary then $|G|\equiv 0,1\mod 4$.

Proof. If $G$ is self-complementary then $||G||=||\overline{G}||$, so
$$\frac{|G|(|G|-1)}{2}=||K_{|G|}||=2||G||.$$
Then $|G|(|G|-1)=4||G||$. Suppose $|G|$ is odd. Then $|G|-1$ is even, and $\frac{4||G||}{|G|-1}=|G|$ is odd. This means that $4\mid |G|-1$. Otherwise, suppose $|G|$ is even. Then, by the same reasoning, $|G|-1=\frac{4||G||}{|G|}$ is odd, so $4\mid |G|$. The result follows.

An attempt to find the homology groups of $\bR P^n$ without using cellular homology

The homology groups of $P^n:=\bR P^n$ can be easily computed using cellular homology.
Here is an attempt to compute them by induction using the long exact sequence of homology. We want to prove that $H_m(P^n)=\mathbb{Z_2}$ for $m<n$ odd, $\mathbb{Z}$ for $m=0$, $0$ for $m$ even and $\mathbb{Z}$ if $m=n$ is odd. The case $n=2$ has already been proven here and $P^1\cong S^1$. My attempt goes like this:

 Let first $n>1$ be odd, and let's suppose for every $k<n$ the homology groups are as stated above. Then, using the fact that $P^{n-1}\subseteq P^n$ (As the projection of the points in $S^n$ which have a fixed coordinate null) and that $P^n/P^{n-1}\cong S^n$ we have a long exact sequence of (reduced) homology:
$$\cdots \to H_m(P^{n-1})\to H_m(P^n)\to H_m(S^n)\to H_{m-1}(P^{n-1})\to \cdots$$
and in particular, taking $m=n$ we have
$$\cdots \to H_n(P^{n-1})\to H_n(P^n)\to H_n(S^n)\to H_{n-1}(P^{n-1})\to \cdots$$
which gives us ($n-1$ is even), since $H_n(P^{n-1})=H_{n-1}(P^{n-1})=0$, the exact sequence
$$0\to H_n(P^n)\to \mathbb{Z}\to 0$$
and this gives an isomorphism between $\mathbb{Z}$ and $H_n(P^n)$.

Similarly, for $1\leq m<n$ with $m$ odd (If this case is possible), and since $H_m(P^{n-1})=\mathbb{Z}_2$ we have the exact sequence
$$0 \to \mathbb{Z}_2\to H_m(P^n)\to 0$$
and this gives an isomorphism between $\mathbb{Z}$ and $H_n(P^n)$.

Lastly, for $1<m<n$ even, and since $H_m(P^{n-1})=H_m(S^n)=0$ we have the exact sequence
$$0 \to H_m(P^n)\to 0$$
which means that $H_m(P^n)=0$.

So, for $n$ odd the job is done.

Similarly, take $n$ even and suppose that for every $k<n$ the homology groups are as stated above. Then take the long exact sequence:
$$\cdots \to H_m(P^{n-1})\to H_m(P^n)\to H_m(S^n)\to H_{m-1}(P^{n-1})\to \cdots$$
For $1\leq m< n-1$ odd we have $H_m(P^{n-1})=\mathbb{Z}_2$ and $H_m(S^n)=0$. Also $H_{m+1}(S^{n})=0$. So we have the exact sequence
$$0 \to H_m(P^{n-1})\to H_m(P^n)\to 0$$
which gives an isomorphism between $H_m(P^{n-1})$ and $H_m(P^n)$, so both of them are $\mathbb{Z}_2$.
For $1<m<n$ even we have, since $H_m(P^{n-1})=0$ and $H_m(S^n)=0$, the sequence
$$0\to H_m(P^n)\to 0$$
which results in $H_m(P^n)=0$.

The homology groups we haven't calculated yet are just the ones at the start of the long exact sequence, which is, since $H_n(P^{n-1})=0$,
$$0\to H_n(P^n)\to \mathbb{Z}\stackrel{\partial}{\to} \mathbb{Z}\to H_{n-1}(P^n)\to 0.$$

So what is left to prove is that the homomorphism $\partial$ is just multiplication by $2$. This is just a higher dimensional version of the problem here, and if can prove that, the result follows.

The singular homology groups of some surfaces

We want to compute the singular homology groups of  some surfaces, for example $T=S^1\times S^1$.
Remark. We'll give the same name to curves/loops, singular $1-$simplexes, and their image, for example $a$ can represent a curve $[0,1]\to X$, or a simplex $\Delta^1\to X$, or the image $a([0,1])$. The meaning will be taken from the context.

Now, take $a=S^1\times 1$ and $b=1\times S^1$. Then $A=a\cup b=a\vee b\cong S^1\vee S^1$. Furthermore, $T/A\cong S^2$. So, since $A$ is a deformation retract of the torus when removing a disk from it, we get the long exact sequence
$$\cdots \to \tilde{H}_n(A)\to \tilde{H}_n(T)\to \tilde{H}_n(S^2)\to \tilde{H}_{n-1}(A)\to\cdots$$
When $n>2$ we have $\tilde{H}_n(A)=\tilde{H}_n(S^2)=0$ so the exact sequence becomes
$$\cdots \to 0\to \tilde{H}_n(T)\to 0\to\cdots$$
so $H_n(T)=0$ for $n>2$. Since $T$ is path-connected then $H_0(T)=\bZ$.
Also, around $n=1$ we have
$$0\to H_2(T)\hookrightarrow \bZ\to \bZ\oplus \bZ\rightarrowtail H_1(T)\to 0$$ where $\hookrightarrow$ denotes a monomorphism and $\rightarrowtail$ denotes an epimorphism.
So, in the sequence $\ker(\bZ\to \bZ\oplus \bZ)=\im(H_2(T)\hookrightarrow \bZ)\cong H_2(T)$ and
$\coker(\bZ\to \bZ\oplus \bZ)\cong H_1(T)$. If we can prove $\bZ\to \bZ\oplus \bZ$ is $0$ we'll get $H_2(T)\cong\bZ$ and $H_1(T)\cong \bZ\oplus \bZ$. But this is the boundary morphism, so if $[\alpha]\in H_2(T,A)\cong H_2(T/A),$ then $\partial[\alpha]=[\partial \alpha]$ where the left class is in $H_2(T,A)$ and the right one in $H_1(A)$.

We'll look for a generator of $H_2(T,A)$ and then see it's mapped into $0$. To do this, we'll consider the following isomorphisms
$$H_2(T,A)\cong H_2(T,S)\cong H_2(N,S^1)\cong_{\partial}H_1(S^1)$$
where $S\cup N$ is $T$ and $S\cap N\simeq S^1$ (As in the picture)

which are really isomorphisms: The first one since $A$ is a deform-retraction of $S$, the second one by Excision Theorem: $H_2(T,S)\cong H_2(N,N\cap S)$. The third one by the exact sequence of the pair $(N,S^1)$. Now, take a generator $[g]\in H_1(S^1)$. There is a singular $2-$simplex $\Delta$ sent into $-[g]$ by $\partial$, which is described by the following picture:

This is, take the standard $2-$simplex, subdivide it as it shows, and then identify all the vertices, send $[1,3]$ into $g$, and collapse all which is not red in a point.

It's boundary is clearly $-g$. So $[\Delta]$ is a generator of $H_2(N,S^1)$, which, by the inclusion $(N,S^1)\hookrightarrow (T,S)$ is a generator of $H_2(T,S)$, which deforms to the generator of $H_2(T,A)$:

whose boundary in $A$ is just $-([a]-[b]-[a]+[b])=0$ (It's a commutator in $\pi_1(S^1\vee S^1)$). So, indeed, $\bZ\to \bZ\oplus\bZ$ is just the zero morphism. The process and result is the same when replacing $\bZ$ for any coefficient group $R$. For the surface $M_g$ of genus $g$ it's very similar. With the same notations, $A\cong \bigvee_{i=1}^g S^1$ and $M_g/A\cong S^2$. $H_n(M_g)=0$ for $n>2$ and around $n=1$ we have the exact sequence:

$$0\to H_2(M_g)\hookrightarrow \bZ\to \bigoplus_{i=1}^g \bZ \rightarrowtail H_1(T)\to 0,$$

in which the morphism $\bZ\to \bigoplus_{i=1}^g \bZ$ is still zero: the generator of $H_2(M_g,A)$ is sent into the image of a product of commutators in $\pi_1(A)$, which is zero in $H_1(A)$. So $H_2(M_g)=\bZ$ and $H_1(M_g)=\bZ^g$.

The case of the projective plane $P=\bR P^2$ it's similar, but now the space is the following one:

Thus $A$ is just $S^1$ instead of $S^1\vee S^1$. This changes the exact sequence when $n\leq 2$. It still holds that $P/A\cong S^2$. So, the sequence

$$\cdots \to \tilde{H}_n(A)\to \tilde{H}_n(P)\to \tilde{H}_n(S^2)\to \tilde{H}_{n-1}(A)\to\cdots$$

becomes, around $n=1$,

$$0\to \tilde{H}_2(P)\to \tilde{H}_2(S^2)\to \tilde{H}_{1}(S^1)\to \tilde{H}_1(P)\to 0$$

which is

$$0\to H_2(P)\to \bZ\to \bZ\to H_1(P)\to 0.$$

As before, the homomorphism in the fragment $\bZ\to\bZ$ will determine the homology groups.

The same way as before, the generator of $H_2(P,A)$ is just the singular simplex:

and it's boundary is $2[a]$, so $\bZ\to \bZ$ is just multiplication by $2$. From $\ker (\bZ\to \bZ)=0$ we can extract an exact triple $0\to H_2(P)\to 0$ so $H_2(P)=0$. Also $H_1(P)\cong \coker (\bZ\to \bZ)=\bZ/2\bZ$.

Of course, using cellular homology all these calculations become too easy and the homology groups can be easily computed even for more complex spaces, like all the projective spaces.

Exercises about Spheres

We start from the fact that any maps $f,g:S^n\to S^n$ such that $f(x)\neq g(x)$ for any $x$, satisfy $f\simeq a\circ g$ with $a$ the antipodal $a(s)=-s$, that homotopical maps have the same degree, and that $\deg (g\circ f)=\deg g\deg f$ for any $f,g:S^{n}\to S^n$. We also know that $\deg a=(-1)^{n+1}$. From this we can prove the following (From Greenberg and Harper's book on algebraic topology).

Exercise 16.8.  Let $f:S^n\to S^n$ be homotopic to a constant. Then $f$ has a fixed point and also a point $x$ at which $f(x)=-x$.
Proof. Since $f$ is homotopic to a constant, then $\deg f=0$, so $f$ must have a fixed point. Now take $g=a\circ f\simeq a\circ \text{cst}_{s_0}=\text{cst}_{-s_0}$ for some $s_0\in S^n$. By the same argument, $g$ has a fixed point, i.e. some $x$ such that $g(x)=-f(x)=x$, but this means $f(x)=-x$.

Exercise 16.9. Any map $f:S^{2n}\to S^{2n}$ has a fixed point or sends some point into its antipode.
Proof. Suppose $f$ doesn't have either a fixed point and a point sent into its antipode. Then $f\simeq a$, and since $\deg a=-1$, $\deg f=-1$. Also, since there are not points $x$ such that $f(x)=-x=a(x)$, $f\simeq a\circ a=1_{S^{2n}}$ so $\deg f=1$. This is a contradiction.

Exercise 16.12. Every map $\bR P^{2n}\to \bR P ^{2n}$ has a fixed point.
Proof. Take $f:\bR P^{2n}\to \bR P^{2n}$. Then $f$ lifts to $\tilde{f}:\bR P^{2n}\to S^{2n}$. Take $\overline{f}=\tilde f\circ \pi:S^{2n}\to S^{2n}$ where $\pi:S^{2n}\to \bR P^{2n}$ is the canonical projection. Then, by 16.9, $\overline{f}$ has a fixed point or a point $x$ such that $f(x)=-x$. In the first case, $\tilde{f}([x])=x$ and in the second one $\tilde{f}([x])=-x$ where $[x]={x,-x}\in \bR P^n$. But this means that $f([x])=\pi\circ \tilde{f}([x])=[x]$, so $[x]$ is a fixed point.

Exercise 16.13. Prove that the projection $\pi:S^n\to \bR P^n$ is not nullhomotopic.
Proof. Let's suppose it's nullhomotopic and $\sigma:S^n\to S^n$ a lift of $\pi$, which is, by . Then $\sigma$ is nullhomotopic by the lifting criterion (If $H$ is a homotopy between $\pi$ and the constant map and since $\pi$ lifts to $\sigma$, then $H$ lifts to a homotopy between $\sigma$ and a lift to the constant map, which is itself constant). Then, by Lemma 1234567, and the fact that $S^{2n}$ is simply connected, $\sigma$ is a deck transformation, which, being surjective, cannot be nullhomotopic, a contradiction.

Homology Groups of $((S^1\vee S^1)\vee \bR P^2)\vee \bR P^3$

The following is a problem from an exam. We'll solve it supposing we can calculate the homology groups of $\bR P^n$ at least for $n\leq 3$ (Using cellular homology they aren't hard to calculate, and the homology of $\bR P^2$ can be easily calculated using simplicial homology). So, the reduced homology groups are just $H_n(\bR P^2)=\bZ_2$ if $n=1$ and $0$ otherwise. And $H_n(\bR P^3)=\bZ_2$ if $n=1$, $\bZ$ if $n=3$ and $0$ otherwise.

Exercise. Find the fundamental group, the homology groups, and the cohomology ring mod 2 of the space $X$ given by pasting a circle, a real projective plane, and a three-dimensional real projective space to a circle along segments as follows (Ignore the blue $1-$cells):

Solution. The fundamental group is easy. $X$ is just the wedge sum $((S^1\vee S^1)\vee \bR P^2)\vee \bR P^3$ so, inductively, $\pi_1(X)\cong \bZ*\bZ*\bZ_2*\bZ_2$. Also, since the first homology group is just the abelianization of the fundamental group, we get $H_1(X)=\bZ\oplus\bZ\oplus\bZ_2\oplus\bZ_2$.

The homology group will also be computed inductively (We could use that the homology of a wedge sum is the direct sum of the homologies but we won't). To do this, take first just the wedge $S^1\vee S^1$. By the long homology sequence we have, for $n\geq 2$, an exact sequence of reduced homology groups:

$$H_n(S^1)\to H_n(S^1\vee S^1)\to H_n(S^1)$$

which is 

$$0\to  H_n(S^1\vee S^1) \to 0 $$

so $H_n(S^1\vee S^1)=\bZ\oplus \bZ$ if $n=1$ and $0$ otherwise.

In the same way, for $n\geq 2$ we get exact sequences:

$$H_n(\bR P^2)\to H_n ((S^1\vee S^1)\vee\bR P^2)\to H_n(S^1\vee S^1)$$

which are

$$0\to H_n ((S^1\vee S^1)\vee\bR P^2)\to 0$$

so for $n\geq 2$, $H_n ((S^1\vee S^1)\vee\bR P^2)=0$, and for $n=1$, $H_n((S^1\vee S^1)\vee\bR P^2)=\bZ\oplus \bZ\oplus \bZ_2$.

In a similar way, but with a slightly different result, for $n=2$ we get the exact sequence

$$H_2(\bR P^3)\to H_2(X)\to H_2(X/\bR P^3)$$

which makes $H_2(X)=0$. We also get, for $n=3$,

$$0\to H_3(\bR P^3)\to H_3(X)\to H_3(X/\bR P^3)$$

which is

$$0\to \bZ\to H_3(X)\to 0$$

which means that $H_3(X)\cong\bZ$.

For $n\geq 3$ we again get sequences of the form $0\to H_n(X)\to 0$.

Therefore, $H_n(X)=\bZ\oplus\bZ\oplus\bZ_2\oplus\bZ_2$ for $n=1$, $\bZ$ for $n=0,3$ and $0$ otherwise.

Exercises about Residues

These are just exercises from this list (The file is likely to disappear or change someday so if that happens I'll remove the link, too).

Exercise 1. The analytic function
$$f(z)=\frac{z e^{\sin(z+1)}}{(z-(4+2i)(z^2-19i))}$$
has a power series around $z=0$. Find the radius of convergence ot this series.

Solution. The radius of convergence of this series is just $\min |z_0|$ where $z_0$ ranges over the nonremovable singularities of $f$. These singularities are just where $z=4+2i$ and $z^2=19i$ (since $ze^{\sin(z+1)}\neq 0$ for those z\neq 0). In the first case, $|z|=2\sqrt{5}=\sqrt{20}$. In the second case, since $|z^2|=|z|^2=19$ then $|z|=\sqrt{19}$. Since $\sqrt{19}<\sqrt{20}$ then the radius of convergence of this series is just $\sqrt{19}$.

Exercise 2. Let $f$ be a meromorphic, periodic function: $f(z+c)=f(z)$. Prove that for all $z_0$, $\Res_{z_0}(f)=\Res_{z_0+c}(f)$.

Proof. Take a circle $C$ small enough around $z_0$ such that $f$ is holomorphic in $\text{Int}C-\{z_0\}$. Then the function $f(z+c)$ is holomorphic in $\text{Int}(C+c)-\{z_0+c\}$ and
$$\Res_{z_0}(f)=\frac{1}{2\pi i}\int_C f(z)dz=\frac{1}{2\pi i}\int_C f(z+c)dz=\frac{1}{2\pi i}\int_{C+c} f(z)dz=\Res_{z_0+c}(f).$$

Exercise 3. Let $g(z)=f(az)$. Compute $\Res_{z_0}(g)$ in terms of $f$.

Solution. $z_0$ is an isolated singularity of $g$, iff $az_0$ is an isolated singularity of $f$. So if
$$f(z)=\sum_{n\in \bZ} a_n (z-az_0)^n$$
in a neighborhood of $az_0$, then
$$g(z)=f(az)=\sum_{n\in\bZ} a_n(az-az_0)^n=\sum_{n\in\bZ} a^n a_n(z-z_0)^n$$
in a neighborhood of $z_0$. For $n=-1$, this gives the coefficient $\Res_{z_0} g=\frac{a_{-1}}{a}=\frac{\Res_{az_0} f}{a}$.

Exercise 4. Let $f$ be a holomorphic function with a triple zero at $z_0$. Compute $\Res_{z_0} \frac{1}{f}$ in terms of $f$.

Solution. Since $z_0$ is a triple zero, $f'(z_0)=f''(z_0)=0$, which means that a series expansion of $f$ around $z_0$ will be of the form
$$\sum_{n=3}^\infty a_n (z-z_0)^n.$$
The function $\frac{1}{f}$ will have a triple pole at $z_0$, so a series expansion around $z_0$ will have the form
$$\sum_{n=-3}^\infty b_n (z-z_0)^n.$$

Since $\frac{1}{f}f=1$, and multiplying the series we get $a_3 b_{-3}=1$ from which $b_{-3}=\frac{1}{a_3}$.
We also get $a_4 b_{-3}+a_3 b_{-2}=\frac{a_4}{a_3}+a_3 b_{-2}=0$, and $\frac{a_5}{a_{3}}+a_4 b_{-2}+a_3 b_{-1}=0$.

So
$$b_{-2}=-\frac{a_4}{a_3^2}$$
and
$$b_{-1}=-\frac{\frac{a_5}{a_3}+a_4 b_{-2}}{a_3}=-\frac{\frac{a_5}{a_3}- \frac{a_4^2}{a_3^2}}{a_3}.$$

But $b_{-1}$ is the residue and $a_n=\frac{f^{(n)}}{n!}$, so we're done.

Exercise 5 is about calculating some integrals.
The first one is
$$I_1=\int_{\partial B_1(0)} \frac{e^{iz}}{z^2}dz.$$
Since $e^{iz}=1+iz+\cdots$, then $\frac{e^{iz}}{z}=\frac{1}{z^2}+\frac{i}{z}+\cdots$, so by the residue theorem $I_1=-2\pi$.

The next one is
$$\int_{\partial B_r(a)}\frac{dz}{z-a}$$
which by a Riemann sum argument is known to be $2\pi i$.

The next one is $I_3=\int_{\partial B_1(0) }\frac{\sin z}{z^3}dz$. Since all the even coefficients in the Maclaurin series of $\sin z$ are $0$, then all the odd coefficients of the Laurent series of $ \frac{\sin z}{z^3}$ are $0$, in particular the coefficient of $z^{-1}$. Thus, $I_3=0$.

When is $G\to G/H$ a covering map of topological groups?

The following, or something similar will appear in Monday's exam so I better work on it. Also, here, every neighborhood is declared to be open.

Exercise. Let $G$ be a topological group and $H$ a discrete, closed subgroup of $G$. Then $G\to G/H$ is a covering space.

The proof can be done manually, but we'll use the following proposition.

Proposition (1.40 from Hatcher's book). If an action of $H$ on a space $Y$ satisfies that each $y\in Y$ has a neighborhood $U$ such that $hU\cap U\neq \varnothing$ implies $h=1$ (A strong hypothesis of a properly discontinuous action), then

• The quotient map $Y\to Y/H$ is a normal covering space.
• $H$ is the group of deck transformations of $Y\to Y/H$ if $Y$ is path-connected.
• $H$ is isomorphic to $\pi_1(Y/H)/\pi_1(Y)$ (Where $\pi_1(Y)$ is its identification as a subgroup of $\pi_1(Y/H)$) if $Y$ is path-connected and locally path-connected.

Easy Lemma. Let $G$ be a topological group. Then there for each neighborhood $U$ of $1$ there exists a neighborhood $V\subseteq U$ of $1$ such that $V=V^{-1}$ and $V^2=\{vv'\mid v,v'\in V\}\subseteq U$.

Proof of Easy Lemma. By continuity of the product $\pi$ and the product topology of $G\times G$, there exist open sets $V_1,V_2\subseteq G$ such that $1\in \pi(V_1\times V_2)\subseteq U$. Now take $V=(V_1\cap V_2)\cap (V_1\cap V_2)^{-1}$. Then $1\in V$, $V=V^{-1}$ and $V\subseteq V_1 \subseteq V_1 V_2\subseteq U$. Furthermore $V^2=\pi(V\times V)\subseteq \pi(V_1\times V_2)\subseteq U$.

Proof of Exercise. We only need to prove that the action of $H$ on $G$ by left multiplication is a covering space action. This can be done as follows.

Let $U$ be a neighborhood of $1$ such that $U\cap H=\{1\}$. Take a second neighborhood $V\subseteq U$ of $1$ such that $V=V^{-1}$ and $V^2=\{vv'\mid v,v'\in V\}\subseteq U$. It also holds that $V\cap H=\{1\}$.

Now, for each $g\in G$, take $V_g=Vg$, and let's suppose that $z\in hVg\cap Vg\neq \varnothing$ for some $h\in H$. Then $z=hvg=v'g$ for some $v,v'\in V$. This means that $hv=v'$ or $h=v'v^{-1}\in V^2\cap H$ thus $h=1$.

So, the action of $H$ on $G$ is a covering space action and $G$ is a covering space of $G/H$ via the canonical projection.

There are more consequences: $H$ is the group of deck transformations of this cover, and isomorphic to $\pi_1(G/H)/\pi_1(G)$, given that $G$ is path-connected and locally path-connected.

Easy Example. Take $G=S^1$ and $H=U_n$, the group of the $n$th roots of unity. $G/H$ is just $S^1$ with the projection defined as $z\mapsto z^n$, and $U_n\cong \bZ_n$ is the group of deck transformations.

About the Homology Groups of the Parachute Triangle

Just a handwritten exercise in Spanish from Hatcher's book.

A note: When you take $n(\gamma-\alpha+\beta)=0$ you can inmediately say $n=0$ without the intermediate step, since these abelian groups are free.

About the Retractions of a surface of genus $g$ onto circles.

Exercise. In the surface $M_g$ of genus $g$, let $C$ be a circle that separates $M_g$ into two compact subsurfaces $M_h', M_k'$ obtained from the surfaces $M_h,M_k$ by deleting an open disk from each. Show that $M_{h'}$ doesn't retract onto its boundary circle $C$, and hence $M_g$ doesn't retract onto $C$.

Solution.  When we deal with $M_g$, are dealing with cell complexes like the following:

built by pasting $2g$ $1-$cells $\alpha_1,\beta_1,\alpha_2,\beta_2,\dots, \alpha_g,\beta_g$ to a $0-$cell, and then pasting a $2-$cell by the map $S^1\to X^1$ given by $[\alpha_1,\beta_1][\alpha_2,\beta_2]\cdots [\alpha_g, \beta_g]$.

When a disk is removed, the resulting surface $M_g'$

deform-retracts to the $1-$skeleton

whose fundamental group is $\Large{*}_{\normalsize i=1}^{\normalsize2g}\normalsize\bZ$. Now, let's suppose that $M_h'$ retracts onto $C$. Then, the homomorphism $i_*:\bZ\to \Large{*}_{\normalsize i=1}^{\normalsize2h}\normalsize\bZ$ is injective. We can then, quotient everything over its commutator (Since $i_*$ is injective) to get a homomorphism $i^*:\bZ\to \bigoplus_{i=1}^{2h} \bZ$ which, since $\bigoplus_{i=1}^{2h} \bZ$ has no torsion, is also injective. This is a contradiction, since $i_*(1)=[e_1,e_2][e_3,e_4]\cdots [e_{2h-1},e_{2h}]\mapsto 0$ when passing to the quotient.

Universal Abelian Covering Space

This exercise from Hatcher's book is just an application of group theory:

Exercise (Part 1). For a path-connected, locally path-connected, and semilocally simply-connected space $X$, call a path-connected covering space $E\to X$ abelian if it is normal and has abelian deck transformation group. Show that $X$ has an abelian covering space that a covering space of every other abelian covering space of $X$, and that such a 'universal' abelian covering space is unique up to isomorphism.

Proof. To prove existence, take $N=\pi_1(X)'=[\pi_1(X),\pi_1(X)]$. Then there exists a covering space $p':X'\to X$ uniquely determined by $N$. It's normal since $X$ is path-connected and $N\trianglelefteq \pi_1(X)$, and it's abelian since $G/G'$ is abelian for any group $G$. Furthermore every subgroup $H\trianglelefteq G$ such that $G/H$ is abelian contains the commutator $G'$, so if $p:E\to X$ is an abelian covering space, then
$$H=\im p'_{*}\leq \im p_{*}=\pi_1(E)$$
so $p'$ lifts to a map $\tilde{p}':X'\to E$ which, by the lemma 1234567 here is a covering map. We're now done with the existence. To prove uniqueness, let's suppose $Y$ is such a covering space. Then since $X'$ is abelian, $Y$ is a covering space of $X$, so that $\pi_1(Y)\leq N$. Furthermore $Y$ is abelian, so $N\leq \pi_1(Y)$. Thus $N=\pi_1(Y)$ and by uniqueness of $X'$ as the covering space associated by the commutator, $X'\cong Y$.

Exercise (Part 2). Describe this covering space explicitly for $X=S^1\vee S^1$ and $X=S^1\vee S^1\vee S^1$.

Solution. When $X=S^1\vee S^1$, take $E=\bR\times \bZ\cup \bZ\times \bR$ the integer grid of $\bR^2$ and let $\bZ\times \bZ$ act on $E$ by $(m,n)(x,y)=(x+m,y+n)$. Furthermore, each $(x,y)\in E$ has a neighborhood $U$ such that $(m,n)U\cap U\neq \varnothing$ only when $(m,n)=(0,0)$, so this action is a covering space action. So the quotient map $p:E\to E/\bZ\times \bZ$ is a normal covering map and $\bZ\times \bZ$ is the group of deck transformations of $p$. But $E/\bZ\times \bZ\cong X$. This proves $E$ is an abelian cover of $X$. The generating loops $a,b$ of $X$ are lifted to $E$ at each point $(i,j)$ of the fiber by $\tilde{a}:(i,j)\leadsto (i+1,j)$ and $\tilde{b}: (i,j)\leadsto (i,j+1)$. By a quick use of Van-Kampen's Theorem each loop in $E$ can be seen as a product of loops in Tic-Tac-Toe-shaped open sets, which are homotopic to loops in unit squares. But each of these is just of the form $(\tilde{\alpha}\tilde{\beta}\tilde{\overline{\alpha}}\tilde{\overline{\beta}})^n$ where $\tilde{-}$ means a lift of $-$ at some point, and from this follows that the elements of $p_*\pi_1(E)$ are products of elements of the form $\alpha\beta\overline{\alpha}\overline{\beta}=[\alpha,\beta]$. Therefore $p_*\pi_1(E)=\pi_1(X)'$. Analogously, it can be proved that the corresponding covering space of $S^1\vee S^1\vee S^1$ is $\bR\times \bZ\times \bZ\cup \bZ\times \bR\times \bZ\cup \bZ\times \bZ\times \bR$.

Just Calculate the Residues

In an exam for a course of Complex Analysis last year they asked to compute the following integral:
$$\int_{\partial B_1 (0)} \frac{dz}{(e^{2\pi z}+1)^2}.$$ To compute it, we only have to calculate the residues at $\pm \frac{i}{2}$, the zeros of $(e^{2\pi z}+1)^2$ inside $B_1(0)$.

To do this we'll calculate the relevant coefficient $d_{-1}$ of the Laurent series of $\frac{1}{(e^{2\pi z}+1)^2}$ at $\pm \frac{i}{2}$. It's easy enough. Let $f(x)=e^{2\pi z}+1$.  Let's suppose we know (And it's true that we know) how to calculate its Maclaurin series:
$$f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n$$
where the $z_0$ in the right side can be replaced by $\pm \frac{i}{2}$. Let $g(z)=\frac{1}{f(z)}$ have a Laurent series expansion
$$g(z)=\sum_{n=-1}^{\infty} b_n (z-z_0)^n$$
in a neighborhood $U$ of $z_0$ (We can do this since $z_0$ is a simple zero of $f$).

Then $f(z)g(z)=1$ in $U$. With this fact we'll compute the terms $b_{-1}=\Res_{z_0} g$ and $b_0$. In the product of the series, the coefficient of $z_0$ is $1=c_{0}=a_0b_0+a_1b_{-1}$. Since $a_0=0$, then $1=a_1 b_{-1}$ so that $b_{-1}=\frac{1}{a_1}=\frac{1}{f'(z_0)}$(This fact is well known). In the same way we can see that
$$a_0 b_1+a_1 b_0+a_2 b_{-1}=0$$
which results in
$$a_1 b_0+a_2 b_{-1}=0$$
or
$$b_0=-\frac{a_2}{a_1^2}=-\frac{1}{2}\frac{f''(z_0)}{f'(z_0)^2}.$$
We have now calculated enough terms. Now, knowing that
$$g(z)=\sum_{n=-1}^\infty b_n z^n$$
we must only calculate the residues of $g(z)^2$ at $z_0$.
But, as before, the coefficient $d_{-1}$ of $z^{-1}$ will be
$$d_{-1}=b_0 b_{-1}+b_{-1}b_0=2b_0 b_{-1}=-\frac{2}{f'(z_0)}\frac{1}{2}\frac{f''(z_0)}{f'(z_0)^2}=-\frac{f''(z_0)}{f'(z_0)^3}.$$

Now we're ready. See that $f'(z)=2\pi e^{2\pi z},f''(z)=(2\pi)^2 e^{2\pi z}$, and since $e^{2\pi z_0}=-1$, $f'(z_0)=-2\pi,f''(z_0)=-4\pi^2$.

So $\Res_{z_0} \frac{1}{e^{2\pi z}+1}=d_{-1}=-\frac{-4\pi ^2}{-2\pi (4\pi ^2)}=-\frac{1}{2\pi}$.

And last, by the residue theorem,
$$\int_{\partial B_1(0)} \frac{dz}{(e^{2\pi z}+1)^2}=2\pi i \frac{-1}{2\pi}+2\pi i \frac{-1}{2\pi}=-2i.$$

About the Meromorphic functions on the Riemann Sphere

Proposition. Let $f:\hat{\bC}\to \hat{\bC}$ be a meromorphic function. Then $f$ is a rational function.

To prove this, we'll use some lemmas.

Lemma 1. Let $f$ be such a meromorphic function. Then it has only a finite number of poles.
Proof.  A meromorphic function is just a holomorphic function in $\hat{\bC}$ (minus a discrete set of poles). If $a=\infty$ then, since the set $S$ of singularities is discrete and closed, then it's discrete and compact, thus finite.

Proof of Proposition. Let $a_1,\cdots, a_n$ be the poles of $f$ and take
$$g(z)=\prod_{i=1}^n (z-a_i)^{o(a_i)} f(z).$$
Then $f(z)$ is holomorphic in $\bC$, and still meromorphic in $\hat{\bC}$. So,
$$\lim_{z\to\infty} g(z)=\in \hat{\bC}.$$
If $\lim_{z\to\infty} g(z)\in \bC$, then by Liouville theorem $g$ is constant and $f(z)=\frac{g(12)}{\prod_{i=1}^n(z-a_i)^{o(a_i)}}$.

If $\lim_{z\to\infty} g(z)=\infty$ then $\infty$ is a pole of $g$, and by the lemma shown here, $g$ is a polynomial. In both cases we're done.

About Entire Functions Which are Polynomials

Let $f$ be an entire function and suppose that $\infty$ is a pole of order $n$, so that $g(z)=f(\frac{1}{z})$ has a pole of order $n$ at $0$. Then, in a punctured neighborhood of $0$, $g$ has a Laurent series expansion
$$g(z)=\sum_{k=-n}^\infty a_n z^n.$$
Furthermore
$$\lim_{z\to\infty}g(z)=\lim_{z\to\infty} f(\frac{1}{z})=\lim_{z\to 0} f(z)$$
so
$$g(z)=\sum_{k=-n}^0 a_n z^n.$$
Therefore, in a punctured neighborhood of $\infty$, given by $|z|>R$ for some $R>0$ , $f(z)=g(\frac{1}{z})=\sum_{k=0}^n a_{-n} z^{n}$. Thus, $f$ agrees with a polynomial in a big enough set, and by the identity theorem $f$ is a polynomial. The next result follows:

Lemma. If $f$ is an entire function with $\infty$ as a pole, then $f$ is a nonconstant polynomial.

The reciprocal is true: Every nonconstant polynomial has $\infty$ as a pole.

Proposition.  Let $f,g$ be entire functions such that $f\circ g$ is a nonconstant polynomial. Then both $f$ and $g$ are polynomials.

Proof. If $f\circ g$ is a nonconstant polynomial then $f\circ g(z)\to \infty$ when $z\to \infty$. Let's suppose that $g$ is not a polynomial. Then $\infty$ is a removable singularity or an essential singularity of $g$. In the former case $g$ is entire in the extended plane, and by Liouville theorem, $g$ is constant, thus $f\circ g$ is also constant, a contradiction. So $\infty$ is an essential singularity of $g$. In this case, by Casorati-Weierstrass theorem, there exists a sequence $s_n\to\infty$ such that $g(s_n)\to a$ where $a$ is any complex number. Then $f\circ g(s_n)\to f(a)\neq \infty$, a contradiction. So, $g$ must be a nonconstant polynomial. Similarly, if $f$ is not a polynomial, and $\infty$ is an essential singularity of $f$ (If it's removable we can argue as above to show $f\circ g$ is constant) then by Casorati-Weierstrass theorem there is some sequence $s_n\to\infty$ such that $f(s_n)\to a'$. Also, by the Fundamental Theorem of Algebra, the equations $g(z)=s_n$ always have solutions, so we get some choice of sequence $r_n\to\infty$ such that $g(r_n)=s_n$, so that $f\circ g(r_n)\to a'$, another contradiction. So $f$ is a polynomial.

Note: The nonconstant condition cannot be removed: You can take any entire function as $g$ and any constant polynomial as $f$ and $f\circ g$ will be a constant polynomial.

About the singularities of $\sin \frac{1}{\cos\frac{1}{z}}$

Exercise. Find and classify the singularities of $f(z)=\sin\frac{1}{\cos\frac{1}{z}}$.

Solution. Whenever $\cos \frac{1}{z}$ is holomorphic and doesn't vanish $f(z)$ is holomorphic. Since all the zeros of $\cos z$ are its real zeros, i.e. $z=(2k+1)\frac{\pi}{2}$, then the only singularities of $f(z)$ are when  $z=0$, and when $z=\frac{1}{(2k+1)\frac{\pi}{2}}$. It may happen that one or more of them are removable, but this is not the case.

Let's first consider $z_0=0$. In this case, the singularity is not isolated.

Now consider $z_0=(2k+1)\frac{\pi}{2}$ for some $k$. Then if $z\to z_0$ along the real axis, $\cos \frac{1}{z}\to 0$ along the real axis, and if $z\to z_0$ in such a way that $\frac{1}{z}\to \frac{1}{z_0}$ along the line $\frac{1}{z_0}+ti$, then $\cos\frac{1}{z}\to 0$ along the imaginary axis. This, because $$\cos \frac{1}{z}=\cos\left(\frac{1}{z_0}+ti\right)=\cos\left(\frac{1}{z_0}\right)\cosh(t)-i\sin\left(\frac{1}{z_0}\right)\sinh(t)=\pm i\sinh(t)$$ where the sign depends on $k$.

So, in these cases, $\frac{1}{\cos\frac{1}{z}}\to \infty$ along the real axis and the imaginary axis respectively. Along the real axis, $f(z)$ oscilates, and in particular, is bounded, but along the imaginary axis, $f(z)\to\infty$. So these singularities are essential.

An Application of Schwarz Lemma

Schwarz Lemma (Theorem 4.3.4.13 of Ahlfors Complex Analysis).  Let $f$ be holomorphic on $B_1(0)$ such that $|f(z)|\leq 1$ and $f(0)=0$. Then, for all $z$, $|f(z)|\leq |z|$ and  $|f'(0)|\leq 1$. If $|f(z)|=|z|$ for some $z\neq 0$, or if $|f'(0)|=1$, then $f(z)=cz$ with a constant $c$ of absolute value $1$.
Lemma. Let $\mu(z)=\frac{z-a}{\overline{a}z-1}$ for a fixed $a\in B_1(0)$ and each $z\in \bC$. Then $\mu(B_1(0))= B_1(0)$.
Proof. First note that  $|\mu(0)|=|a|<1$.  Furthermore,
$$|\mu(1)|=\left|\frac{1-a}{\overline{1-a}}\right|=1,|\mu(-1)|=\left|\frac{1+a}{\overline{a+1}}\right|=1$$
and
$$|\mu(i)|=\left|\frac{i-a}{\overline{a}i-1}\right|=\left|\frac{i-a}{-\overline{a}-i}\right|=\left|\frac{i-a}{\overline{i-a}}\right|=1.$$
So $\mu$ is a Möbius transformation sending $\partial B_1(0)$ to $\partial B_1(0)$. Since $0$ is sent into $B_1(0)$, then $\mu(B_1(0))=B_1(0)$.
Exercise. Let $g:B_1(0)\to B_1(0)$ be a holomorphic function such that $g(\frac{1}{2})=0$. Then $|g(0)|\leq \frac{1}{2}$.
Proof. We'll use part of the proof of the Schwarz-Pick theorem, along with the facts that the Möbius transformations used are involutions.
Let $$\mu(z)=\frac{\frac{1}{2}-z}{1-\frac{z}{2}}.$$
Take $f=g\circ\mu$. Then $f(B_1(0))\subseteq B_1(0)$
and $f(0)=g\left(\frac{1}{2}\right)=0.$
Applying Schwarz Lemma, $|f(z)|\leq |z|$ for all $z$. But then,
$$\left|g\left(\frac{\frac{1}{2}-z}{1-\frac{z}{2}}\right)\right|\leq |z|,$$
which for $z=\frac{1}{2}$ gives
$$\left|g(0)\right|\leq \frac{1}{2},$$
the desired result.


Note: In the proof of the Schwarz-Pick Theorem the inverse $\mu^{-1}$ is used instead of $\mu$, and there is a second Möbius transformation which in this case is $\lambda(z)=-z$. So in this case $\mu=\mu^{-1}$ and $\lambda$ can be removed.

The Fundamental Group of a Sphere with Two Points Identified.

This is exercise 1.2.7. from Hatcher's book.
Lemma (Proposition 1.26.(a), Page 50 of Hatcher's Algebraic Topology).
If $Y$ is a cell complex obtained from a cell complex $X$ by attaching $2-$cells $e_{\alpha}^2$, $\alpha\in \Lambda$ by attaching maps $\phi_\alpha: S^1\to X$, then the inclusion $X\hookrightarrow Y$ induces a surjection $\pi_1(X,x_0)\to \pi_1(Y,x_0)$ whose kernel is $N=\langle[\gamma_\alpha \phi_\alpha\overline{\gamma}_\alpha]\mid \alpha\in \Lambda, \gamma_\alpha:x_0\to \phi_\alpha(1)\rangle$  where the $\phi_{\alpha}$ are seen as loops.
Exercise. Let $X$ be the quotient space of $S^2$ obtained by identifying the north and south poles to a single point. Put a cell complex structure on $X$ and use this to compute $\pi_1(X)$.

Solution. We can give $S^2$ the usual cell structure with one $0-$cell $a$ and one $2-$cell attached by the only map $S^2\to \{a\}$. Instead, we can give $S^2$ another cell structure from $S^1$ pasting two $2-$cells to the $1-$cell structure of $S^1$. We can also define a $2-$cell structure on $S^2$ by starting with two points $n,s$, two $1-$cells $\gamma,\delta$ between $n$ and $s$, and two $2-$cells pasted into $\alpha\beta$. This last approach works, since then we are giving $X$ the structure of the quotient complex $S^2/\{n,s\}$. It also gives a cell complex structure to $S^1$ starting with two points and attaching two $1-$cells.

We can instead think of it geometrically. When identifying the poles of the sphere with a point the resulting figure seems like a torus with a generating circle identified to a point. This determines the cell structure shown by the next drawing:

Here we have $X$ built from $S_1=\text{im}\alpha$ by attaching a $2$-cell by the map $\phi=\alpha\overline{\alpha}$ which is nullhomotopic. So, by the lemma, the homomorphism

$$\pi_1(S^1,v_1)\to \pi_1(X,v_1)$$

induced by the inclusion is an isomorphism, and $\pi_1(X,v_1)\cong \bZ$.

Monodromy

This is almost totally taken from Hatcher's book.

Let $p:\cX\to X$ is a covering space of a path-connected, locally path-connected space which admits a universal cover $q:U\to X$, and let $\gamma:x_0\leadsto x_1$ be a curve on $X$. Then there is a map $L_\gamma: p^{-1}(x_1)\to p^{-1}(x_0)$ given by $L_\gamma(\cx_1)=\cx_0$ where $\cx_0=\tilde{\gamma}(0)$ and $\tilde{\gamma}$ is a lift of $\gamma$ ending at $\cx_1$, which is a bijection, since it has an inverse $L_\overline{\gamma}$. If $c_{x_0}$ is the constant loop at $x_0$, then $L_{c_{x_0}}=1_{p^{-1}(x_0)}$, and $L_{\gamma*\gamma'}=L_\gamma L_{\gamma'}$. Furthermore, if $\gamma$ is a loop with endpoint $x_0$, then $L_\gamma\in S_{p^{-1}(x_0)}$. The map $[\gamma]\mapsto L_\gamma$ is then (Since it doesn't depend on the representative of the homotopy class of $\gamma$) an homomorphism; i.e. a permutation representation of an action $\pi_1(X,x_0)$ on $p^{-1}(x_0)$, given naturally by
$$[\gamma]\cdot \cx=L_\gamma (\cx).$$

It can be proved (See Hatcher's Algebraic Topology, pages 68-70) that a covering space is uniquely determined by this action (up to isomorphism). We'll prove a part of it, as an exercise.

Proposition 1. Let now $p:\cX\to X, r:\cY\to X$ be isomorphic covering spaces. Then the group actions induced by them are isomorphic.

Proof. Let $\phi:\cX\to \cY$ be an isomorphism, i.e. a homeomorphism such that $p=r\circ \phi$. Let $\cF=p^{-1}(x_0),\cG=r^{-1}(x_0)$. Then $\phi|_{\cF}$ is a bijection $\cF\to\cG$. Indeed, if $f\in \cF$, then $r(\phi(f))=p(f)=x_0$, so $\phi(f)\in \cG$. Analogously, $\im\phi^{-1}|_{\cG}\subseteq \cF$, so $\phi^{-1}_{\cG}$ is its inverse, and $\phi|_{\cF}$ is a bijection. Let $[\gamma]\in \pi_1(X,x_0)$. We'll prove that $\phi([\gamma]f)=[\gamma]\phi(f)$. Indeed, $[\gamma]f=L^\cX_\gamma(f)=\tilde{\gamma}(0)$ where $\tilde{\gamma}$ is a lift of $\gamma$ in $\cX$, ending at $f$. But $\phi\circ\tilde\gamma$ is a lift of $\gamma$ in $\cY$ ending at $\phi(f)$, thus $\phi(\tilde\gamma(0))=L^{\cY}_\gamma(\phi(f))=[\gamma]\phi(f)$. So $\phi$ is an isomorphism of $\pi_1(X,x_0)-$sets, the required result.

It can be seen, in Hatcher's book, that given a group action of $\pi_1(X,x_0)$ on a set $F$ and a fixed point $u_0$ such that $q(u_0)=x_0$, we can build the covering space $U\times F/\sim$ where $(u,f)\sim (u',f')$ if $q(u)=q(u')$ and $[(q\circ\alpha')(\overline{q\circ\alpha})]f=f'$ and $\alpha:u_0\leadsto u,\alpha':u_0\leadsto u'$. Also, the covering map $p:U\times F/\sim \to X$ is given by $p([u,f])=q(u)$, and the fiber $p^{-1}(x_0)$ is in a 1-1 correspondence with $F$ such that the actions are essentially equivalent.

This said, we can prove the converse of Proposition 1.

Proposition 2. Let $F,F'$ be isomorphic $\pi_1(X,x_0)-$sets. Then the associated covering spaces are isomorphic.

Proof. Take $\phi:F\to F'$ a $\pi_1(X,x_0)-$isomorphism (Which is obviously a homeomorphism, taking discrete topologies), and define the obviously homeomorphism $\varphi:U\times F\to U\times F'$ given by $(u,f)\mapsto (u,\phi(f))$.

See that $(u,f)\sim_F (u',f')$ iff $q(u)=q(u')$ and $[(q\circ \alpha')(\overline{q\circ\alpha})]f=f'$ for $\alpha:u_0\leadsto u, \alpha':u_0\leadsto u'$. But since $\phi$ is a $\pi_1(X,x_0)-$isomorphism, this occurs iff $q(u)=q(u')$ and $[(q\circ \alpha')(\overline{q\circ\alpha})](\phi (f))=\phi (f')$, which occurs iff $(u,\phi(f))\sim_{F'}(u',\phi (f'))$. Thus $\varphi$ becomes a homeomorphism of the quotients:
\begin{CD} U\times F @>{\varphi}>> U\times F'\\ @VVV @VVV\\ U\times F/\sim_F@>{\varphi}>> U\times F'/\sim_{F'}\end{CD}

Furthermore, $p_{F'}\circ\varphi([u,f])=p_F([u,\phi (f)])=q(u)=p_F([u,f])$, so $\varphi$ is really a covering space isomorphism, which is the desired result.

Using Rouché's Theorem for Counting Roots in some Annulus.

Let $f(z)=z^4-6z+1$. We want to find the number of roots of $f$ in the annulus given by $1<|z|<2$.
Remember that Rouché's theorem says, in particular, that if $f,g$ are holomorphic in a domain $D$, and $\gamma$ is a Jordan curve such that $\gamma$ and its interior are contained in $D$, such that for any $z\in \gamma$,
$$|f(z)-g(z)|<|g(z)|.$$
Then $n_{f,\gamma}=n_{g,\gamma}$ where $n_{-,\gamma}$ is the number of zeros of $-$ in the interior of $\gamma$.

So let's first consider $D=\mathbb{C}$ and $\gamma$ the circle of radius $2$. Take $g(z)=z^4$.
Then, over $\gamma$ we have
$$|f(z)-g(z)|=|-6z+1|\leq |6z|+|1|=13<16=|z|^4.$$
Since $n_{g,\gamma}=4$, then $n_{f,\gamma}=4$.

Now take $\delta$ as the circle of radius $1$, and $h(z)=-6z+1$. Then, over $\delta$,
$$|h(z)|=|1-6z|\geq ||1|-|6z||=|1-6|=5>1=|z^4|=|f(z)-h(z)|.$$

Then, since $h$ has exactly one zero in the interior of $\delta$, say $z=\frac{1}{6}$, $f$ must also have exactly one. So, in the annulus, $1<|z|<2$, $f$ has exactly $3$ zeros.

Using the Argument Principle for Counting Roots

Let $f(z)=z^{11}-2z^6+z^4+10i$. We want to find the number of roots of $f$ which are in the upper semiplane. By the Argument Principle, if $\gamma$ is a curve which has no roots of $f$ and $n_f$ is the number of roots of $f$ in the upper plane counting multiplicities, then
$$n_f=\frac{1}{2\pi i} \int_\gamma \frac{f''(z)}{f(z)}dz=\int_{f(\gamma)}\frac{dw}{w}=n(f\circ \gamma,0)$$
where $n(-,0)$ is the winding number of $-$ around $0$.

So, now take $R$ so big, that $f$ doesn't vanish when $|z|\geq R$, and such that $f(z)\sim z^{11}$.
Take $\gamma$ as the upper half circle of radius $R$ followed by the segment between $-R$ and $R$:

Also note that there are no real roots of $f$, since for $z$ real, $f(z)=0$ means $10i=0$. So, now we can use the Argument Principle, since there are no roots of $f$ on $\gamma$. Furthermore, $R$ is big enough to say that every root of $f$ is inside $\gamma$. Next we must see how $f\circ \gamma$ behaves.

When $\gamma$ does a half-twist around $0$, $f\circ \gamma$ does five twists and almost an extra half, ending at some point $x+10i$ where $x\in\mathbb{R}$. Then $f\circ\gamma$ goes somehow, around the line $y=10i$, from $x+10i$ to the starting point. Since that's done over the real axis, it doesn't make the extra twist around $0$. So $n(f\circ\gamma,0)=5$: That's the number of roots.

In the picture it can be seen the number of twists, and how the last one doesn't complete itself.

About $\pi_1(\mathbb{R P}^n)$.

The proof of this is based on the idea given on page 21 here. It's about finding the fundamental groups of the sphere and the Projective Plane without using Van Kampen's theorem. It still seems like a weak form of it, though.
Let $\nu=(0,\dots,1),\xi=-\nu\in S^n$. Also let $I=[0,1]$.
Lemma 1. Let $\gamma$ be a loop in $S^n$ starting at $\xi$ such that $\gamma$ doesn't pass through $\nu$. Then $\gamma$ is nullhomotopic.
Proof. $\im\gamma\subseteq S^2-\nu\cong \bR^n$. Since $\bR^n$ is simply connected, $\gamma$ is nullhomotopic.

Lemma 2. Let $\gamma$ a loop in $S^n$ starting at $\xi$ which passes through $\nu$. Then $\gamma$ is homotopic to a loop in $S^n-\nu$.
Proof. Let $U$ be a small open ball around $\nu$ and $U'$ a bigger ball, such that $\overline{U}\subseteq U'$. Since $\gamma^{-1}(v)$ is closed in $I$ then it's compact. Now, $\gamma^{-1}(U)$ is a union of open, disjoint intervals which cover $\gamma^{-1}(\nu)$; take a finite number of them which still cover $\gamma^{-1}(\nu)$, say $\{I_k\}_{k=1}^n$ such that $\overline{I_k}=[a_k,b_k]$ for each $k$, and note that $\gamma(I_k)\subseteq \gamma(\gamma^{-1}(U))\subseteq U$. Also $\gamma(\overline{I_k})\subseteq U'$. Since $U'$ is simply connected, then, for each $k$, the curve $\gamma|_{\overline{I_k}}$ is homotopic to any curve $\gamma_k$ with the same endpoints not passing through $\xi$.
Then
$$[\gamma]=[\gamma|_{[0,a_1]}*\gamma_1*\gamma|_{[b_1,a_2]}*\cdots*\gamma_n*\gamma|_{[b_n,1]}]$$
where
$$\gamma|_{[0,a_1]}*\gamma_1*\gamma|_{[b_1,a_2]}*\cdots*\gamma_n*\gamma|_{[b_n,1]}$$
is a loop starting at $\xi$, and not passing through $\nu$.

Corollary. $S^n$ is simply connected.
Proof. Every loop on $S^n$ starting at $\xi$ is homotopic to a loop not passing through $\nu$, which is nullhomotopic.

Proposition. $\pi_1(\bR\bP^n,\overline{x})\cong \bZ/2\bZ$ for any $\overline{x}\in\bR\bP^n$.
Proof. Let $\overline{x}=\{x,-x\}$ with $x\in S^n$. Let $\gamma$ be a loop in $\bR\bP^n$ starting at $\overline{x}$. Then it lifts to a curve $\tilde{\gamma}$ in $S^n$ starting at $x$. If $\tilde\gamma$ is itself a loop, then $\tilde\gamma$ is nullhomotopic and then $\gamma$ is also nullhomotopic. Then, let's suppose $\tilde\gamma$ is not a loop; so that $\tilde\gamma(1)=-x$ and $\gamma$ is not nullhomotopic. Let $\delta$ be another loop at $\overline{x}$ and $\tilde\delta$ its lift at $-x$. Then $\tilde\gamma*\tilde\delta$ is a loop, and also lifts $\gamma*\delta$ at $x$. So $\tilde\gamma*\tilde\delta$, and $\gamma*\delta$ are both nullhomotopic. So every element of $\pi_1(\bR\bP^n,\overline{x})$ is an inverse of $[\gamma]$, thus $\pi_1(\bR\bP^n,\overline{x})$ has at most two elements. Since $\gamma$ is not nullhomotopic, $\pi_1(\bR\bP^n,\overline{x})$ has exactly two elements. So $\pi_1(\bR\bP^n,\overline{x})\cong \bZ/2\bZ$.

Note. We can also use an argument about the number of sheets of the covering space $S^2$: The cardinal of the fiber of $\overline{x}$ is in a $1-1$ correspondence to the cardinal of $\pi_1(\bR\bP^n,\overline{x})$. Since the former has two elements, the later too. I like the proof above more, though.

About the Fundamental Group of a Topological Group

Proposition. Let $(G,\cdot)$ be a topological group. Let $C_e$ be the path-connected component of the identity $e$. Then

1. $C_e$ is a topological group.
2. $\pi_1(C_e,e)$ is abelian.
3. For any $g\in G$, $\pi_1(G,g)$ is abelian.

Lemma. Let $(G,\cdot,*)$ be a set with two binary operations on it. Also suppose that both operations have a common identity $e$ and for any $g,h,j,k\in G$,

$$(g\cdot h)*(j\cdot k)=(g*j)\cdot(h*k).$$

Then $*=\cdot$ and $*$ is commutative.

Proof of Lemma. If $g,h\in G$ then $$g * h=(e\cdot g)* (h\cdot e)=(e*h)\cdot(g*e)=h\cdot g=(h*e)\cdot (e*g)=(h\cdot e)*(e\cdot g)=h*g.$$

The result follows.

Proof of Proposition. To prove 1. remember that for $u\in G$, $\phi_u:G\to G$ given by $x\mapsto xu$ is a homeomorphism. Also denote $*$ as the product of paths or homotopy classes of paths. Now take paths $\alpha:e\leadsto g$ and $\beta:e\leadsto h$. Then since $\phi_h$ is continuous, $\phi_h\circ \alpha:h\leadsto gh$, so $gh\in C_e$. Also $\phi_{g^{-1}}\circ \alpha: g^{-1}\leadsto e$, so $g^{-1}\in G$. So $C_e\leq G$ and since the restrictions of $*$ and $\square ^{-1}$ are well defined in $C_e$, then $C_e$ is a topological group.

To prove 2. define $[\alpha]\cdot[\beta]=[\alpha\cdot \beta]$ where $(\alpha\cdot\beta)(s)=\alpha(s)\beta(s)$. This operation is well defined on homotopy classes, since $\alpha\simeq_{\text{Rel}0,1}^H\alpha'$ and $\beta\simeq_{\text{Rel}0,1}^J\beta'$ then the function $H\cdot J$ is a homotopy, relative to $0,1$ between $\alpha\cdot\beta$ and $\alpha'\cdot\beta'$.

So now there are two binary operations on $\pi_1(C_e,e)$, given by $*$ and $\cdot$. The homotopy class $[c_e]$ of the constant loop $c_e$ is the identity of $(\pi_1(C_e,e),*)$. But it's also an identity in $(pi_1(C_e,e),\cdot)$. Indeed, if $\alpha$ is a loop starting at $e$, then $$(c_e\cdot \alpha)(s)=c_e(s)\alpha(s)=e\alpha(s)=\alpha(s)=\alpha(s)e=\alpha(s)c_e(s)=(\alpha\cdot c_e)(s).$$

Also, if $\alpha,\beta,\gamma,\delta$ are loops in $e$, then for $0\leq s\leq 1$
$$\begin{align*}(\alpha\cdot \beta)*(\gamma\cdot \delta)(s) &=\left\{\begin{array}{cc}\alpha(2s)\beta(2s) & 0\leq s\leq \frac{1}{2}\\\gamma(2s-1)\delta(2s-1) & \frac{1}{2}\leq s\leq 1\end{array}\right.\\ &= (\alpha *\gamma)\cdot(\beta *\delta)(s).\end{align*}(s)$$

so taking homotopy classes

$$([\alpha]\cdot[\beta])*([\gamma]\cdot [\delta])=([\alpha]*[\gamma])\cdot([\beta] *[\delta]).$$

As a result of the lemma $*=\cdot$ and $*$ is commutative, so $\pi_1(C_e,e)$ is abelian.

To prove 3. note that $\pi_1(G,e)=\pi_1(C_e,e)$ and for $g\in G$, $\phi_{g*}:\pi_1(G,e)\to \pi_1(G,g)$ is an isomorphism. The result follows. 

Universal Covers of Homotopy-Equivalent Spaces

Proposition 123. Let $p:\cX\to X,q:\cY\to Y$ be universal covers where $X,Y$ are path-connected and locally path-connected. If $X\simeq Y$ then $\cX\simeq \cY$.

Lemma 1234. Let $f\in C(X,Y)$. Then $f$ is a homotopy equivalence if there exist maps $g,h\in C(Y,X)$ such that $fg\simeq 1_Y$ and $hg\simeq 1_X$.

Proof of Lemma 1234. 
Given that $fg\simeq 1_Y$ then $h(fg)\simeq h1_Y=h$. But $h(fg)=(hf)g\simeq 1_X g=g$. So $h\simeq g$, thus $1_X\simeq hf\simeq gf$, so $f$ is a homotopy equivalence.

Lemma 1234567.  Let $\fX\stackrel{p}{\to}\cX\stackrel{q}{\to} X$ a sequence such that $q$ and $qp$ are covering maps. If $X$ is locally path-connected then $p$ is a covering map.

Proof of Lemma 1234567. Let $e\in E$. Take a neighborhood $U$ of $q(e)$ which is path-connected and evenly covered with respect to both $q$ and $qp$. Then
$$q^{-1}(U)=\bigsqcup_{i\in I} V_i, (qp)^{-1}(U)=\bigsqcup_{j\in J}W_j$$
such that $q|_{V_i},(q p)|_{W_j}$ are homeomorphisms for each $i,j$. Fix $i$ such that $e\in V:=V_i$. Since $W$ is path-connected, we have $p(W_j)\subseteq V_{i_j}$ for each $j$ and some $i_j\in I$. But since $q|_{V_{i_j}}$ and $qp|_{W_j}$ are homeomorphisms then $p|_{W_j}:W_j\to V_{i_j}$ is a homeomorphism. If $p^{-1}(V)=\varnothing$ we're done. If not, then, since $p^{-1}(V)\subseteq (qp)^{-1}(U)$, if, for some $j$, $p^{-1}(V)\cap W_j\neq \varnothing$, since $p|_{W_j}$ is now a homeomorphism onto $V$ we get $p^{-1}(V)\cap W_j=W_j$. But then $p^{-1}(V)$ is just the union of such $W_j$'s, so $p$ is a covering map.


Proof of Proposition 123. If $X\simeq Y$ then there exist $f:X\to Y$ and $g:Y\to X$ such that $fg\simeq 1_Y$ and $gf\simeq 1_X$.
We have the following diagrams:

and applying the functor $\pi_1$, since $\cX$ and $\cY$ are simply connected we get the commutative diagrams

where $e$ is the trivial group. From this, $(f\circ p)_*=(g\circ q)_*=e=\pi_1(\cX)=\pi_1(\cY)$, so there exists lifts $F,G$ of $fp,gq$, i.e. maps $F:\cX\to\cY, G:\cY\to \cX$ such that the following diagram commutes:

Note that $GF$ is a lift of $gfp$. We also have a homotopy $h:I\times \cX\to X$ between $gfp$ and $p$ which lifts to an homotopy $H$ starting at $GF$.

Note also that the final branch $\sigma=H(1,-)$ has the property that $p\circ \sigma=h(1,-)=p$, i.e. the following triangle commutes

so $\sigma$ is a morphism of $\cX$.
So we have maps
$$\cX\stackrel{\sigma}{\to}\cX\stackrel{p}{\to} X$$
such that $p$ and $p\sigma$ are covering maps. Then, since $X$ is locally path-connected, $\sigma$ is a covering map. But since $\cX$ is simply connected, $\sigma$ is really an isomorphism.


So, since $GF\simeq \sigma$, then $(\sigma^{-1}G)F\simeq 1_\cX$. Analogously there is some isomorphism $\tau:\cY\to\cY$ such that $F(G\tau^{-1})\simeq 1_\cY$. Applying the lemma, we get the result.

About Compositions of Covering Maps

Proposition. Let $\fE\stackrel{q}{\to}\cE\stackrel{p}{\to}X$ be covering spaces. Suppose that for every $x\in X$, $|p^{-1}(x)|<\infty.$ Then $p\circ q$ is a covering map.

Proof.
Let $U$ be an open neighborhood of a point $x$ such that
$$p^{-1}(U)=\bigsqcup_{i=1}^n \cV_i$$
where $\{\cV_i\}_{i=1}^n$ is a family of open sets in $\cE$, each of them mapped homeomorphically by $p$ onto $U$.
For each $i\in\{1,\dots,n\}$ take the unique element $e_i\in \cV_i$  such that $p(e_i)=x$.
For each $i$ take an evenly covered neighborhood $\cU_i$ of $e_i$ such that $\cU_i\subseteq \cV_i$ (If not, take the intersection). Take
$$U'=\bigcap_{i=1}^n p(\cU_i).$$
Then $U'$ is an open set containing $x$ (This, since each $\cU_i$ has a preimage of $x$).
Also, since $p_i=p|_{\cV_i}$ is a homeomorphism, then for each $i$, $\cU_i'=p_i^{-1}(U')$ is an open subset of $\cU_i$ mapped homeomorphically by $p$ onto $U'$. Also note that fixing $i$,
$$q^{-1}(\cU_i')=\bigsqcup_{j\in J}\fV_j$$
such that for each $j$, $q|_{\fV_j}$ is a homeomorphism onto $\cU_i'$. But then, since $p:\cU_i'\to U'$ is a homeomorphism, then $p\circ q:\fV_j\to U'$ is also a homeomorphism. And last,
$$p\circ q^{-1}(U')=\bigsqcup_{i=1}^n U_i'=\bigsqcup_{j_i\in J_i}\bigsqcup_{i=1}^n \fV_{j_i},$$
from which the result follows.

About a Path-connected Covering Space of a Subspace

Definition (Covering Space).
Let $X$ be a topological space. Then an arrangement $p:\cE\to X$ is a covering space of $X$ iff each point $x\in X$ has an evenly covered neighborhood $U$, i.e. an open neighborhood of $x$ such that
$$p^{-1}(U)=\bigsqcup_{i\in I} \cV_i$$
where $\{\cV_i\}_{i\in I}$ is a family of open sets in $\cE$, each of them mapped homeomorphically by $p$ into $U.$

A simply connected covering space is called an universal cover.
Proposition Nyaa.
Let \(p:\cE\to X\) be the universal cover of a (path-connected, locally path-connected and semilocally simply connected) space \(X\), and \(A\subseteq X\) path-connected and locally path-connected. Let \(\cA\subset\cE\) be a path-connected component of  \(p^{-1}(A)\). Then \(p:\cA\to A\) is the covering space corresponding to the subgroup \(\ker i^*\), where \(i^*:\pi_1(A,a)\to \pi_1(X,a)\) is the morphism induced by the inclusion $i:A\to X$.

Remark. Note that \(\cA\) is the covering space corresponding to a subgroup \(H\leq G\) when it can be said that $[\alpha]\in H$ iff \(\alpha\) lifts to a loop in \(\cA\). Note also that in the definition of a covering space (Taken from Hatcher's book, for example) it isn't said that a covering space is surjective.

Lemma 1. Let $p:\cE\to X$ be a covering space. Then $p$ is an open map.

Proof of Lemma 1.
Let $\cU$ be an open set in $\cE$. Let $x\in p(\cU)$. Let $U$ be an evenly covered neighborhood of $x$, and $\cV$ an open set of $\cE$ mapped homeomorphically by $p$ into $U$, such that $\cU\cap \cV\neq \varnothing$. Then since $p|_{\cV}$ is an homeomorphism, and $\cU\cap \cV$ is open in $\cV$, we get $p(\cU\cap\cV)$ is open in $U$, and also contained in $p(\cU)$. But since $U$ is open in $X$, $p(\cU\cap\cV)$ is also open in $X$: It's a neighborhood of $x$ contained in $p(\cU)$. Therefore, $p(\cU)$ is open, and we're done.

Lemma 2.
Let $p:\cE\to X$ a covering space and $A\subseteq X$ path-connected and locally path-connected. Let $\cU$ be a path-connected component of $p^{-1}(A)$. Then the restriction $\overline{p}:\cU\to A$ is a covering space, which also is surjective.

Proof of  Lemma 2.
It's easy to see that $\overline{p}:\cU\to A$ is a covering space. Indeed, take an evenly covered open neighborhood $W$ of a point $x\in A$ and intersect it with $A$. Then $\p^{-1}(W)=p^{-1}(W)\cap \cU$ is a (possibly empty) disjoint union of open sets in $\cU$ mapped homeomorphically by $p$ to $W$. Note that the same reasoning shows that $p:p^{-1}(A)\to A$ is a covering space.

Now let's see that $\p$ is surjective. Since a covering map is open, $\p(\cU)$ is an open set. Let's now suppose that $x\in A-\p(\cU)$. Let $U$ be a path-connected evenly covered neighborhood of $x$. Then $p^{-1}(U)$ is a disjount union of a family $\{\cV_i\}_{i\in I}$ of path-connected open sets of $\cE$, each of them mapped homeomorphically into $U$, each of them contained in a path-connected component of $p^{-1}(A)$, and each of them contained in $p^{-1}(A)-\cU$. Then $\p(\cU)\cap A=\varnothing$. So $A-\p(\cU)$ is open. Since $A$ is path-connected, in particular it's connected and $\p(\cU)=A$.

Proof of Proposition Nyaa.
We have that $[\alpha]\in\ker i^*$ iff $i^*([\alpha])=[\alpha]_X=0$ iff $\alpha$ is nullhomotopic in $X$ iff the lifting $\tilde{\alpha}$, starting at any point of $p^{-1}(a)$ in $\cA$, is nullhomotopic. But to be nullhomotopic, it must be a loop (More formally, the homotopy between $\alpha$ and a constant loop lifts to a homotopy between $\tilde{\alpha}$ and the constant loop starting at the same point, which makes $\tilde{\alpha}$ a loop), and since $\cE$ is simply connected, $\tilde{\alpha}$ is nullhomotopic iff it's a loop, which happens iff $[\alpha]\in \pi_1(E,\alpha(0))$. The result follows.