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 $\bR\times \bZ\times \bZ\cup \bZ\times \bR\times \bZ\cup \bZ\times \bZ\times \bR$.