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.