3. Covering and Fibration

Covering and Lifting

Definition 3.1. Let $p:E\to B$ be in $\mathsf{Top}$. A trivialization of $p$ over an open set $U\subset B$ is a homeomorphism $\phi :p^{-1}(U)\to U\times F$ over $U$, i.e. , the following diagram commutes$p$ is called locally trivial if there exists an open cover $\mathcal{U}$ of $B$ such that $p$ has a trivialization over each open $U\in \mathcal{U}$. Such $p$ is called a fiber bundle, $F$ is called the fiber and $B$ is called the base. We denote it by$$F\to E\to B,$$where there is no ambiguity from the context. If we can find a trivialization of $p$ over the whole $B$, then $E$ is homeomorphic to $F\times B$ and we say $p$ is a trivial fiber bundle.

Example 3.2. The projection map$$\begin{array}{c}\begin{array}{rcl}{\mathbb{R}}^{m+n}& \to & {\mathbb{R}}^n,\\ (x_1,\cdots ,x_n,\cdots ,x_{n+m})& \mapsto & (x_1,\cdots ,x_n)\end{array}\end{array}$$is a trivial fiber bundle with fiber ${\mathbb{R}}^m$.

Example 3.3. A real vector bundle of rank $n$ over a manifold is a fiber bundle with fiber ${\mathbb{R}}^n$.

Example 3.4. We identify $S^{2n+1}$ as the unit sphere in ${\mathbb{C}}^{n+1}$ parametrized by$$S^{2n+1}=\{z_0,z_1,\cdots ,z_n\in {\mathbb{C}}^{n+1}||z_0{|}^2+|z_1{|}^2+\cdots +|z_n{|}^2=1\}.$$There is a natural $S^1$-action on $S^{2n+1}$ given by$$e^{i\theta }:(z_0,\cdots ,z_n)\mapsto (e^{i\theta }{z}_0,\cdots ,e^{i\theta }{z}_n),e^{i\theta }\in S^1.$$This action is free, and the orbit space can be identified with the $n$-dim complex projective space $\mathbb{C}{\mathbb{P}}^n$$$S^{2n+1}/S^1\cong \mathbb{C}{\mathbb{P}}^n=({\mathbb{C}}^{n+1}-\{0\})/{\mathbb{C}}^{\ast }.$$Then the projection map $S^{2n+1}\to \mathbb{C}{\mathbb{P}}^n$ is a fiber bundle with fiber $S^1$. It is a nontrivial fact that they are not trivial fiber bundles. The case when $n=1$ gives the Hopf fibratioin$$S^1\to S^3\stackrel{p}{\to }{S}^2=\mathbb{C}{\mathbb{P}}^1$$which is particularly interesting.

In this case, the projection sends $(z_0,z_1)\in S^3\subset {\mathbb{C}}^2$ to $z_0/z_1\in S^2=\mathbb{C}\cup \{\infty \}$. In polar coordinates, we have$$r_0^2+r_1^2=1\text{and}p(z_0,z_1)=(r_0/r_1)e^{\mathbf{i}({\theta }_0-{\theta }_1)}$$for $z_j=r_j{e}^{\mathbf{i}{\theta }_j}$. For a fixed $\rho =r_0/r_1$, we obtain a torus $T_{\rho }$ in $S^3$.

When identifying $S^3$ with the compatification of ${\mathbb{R}}^3$ (or considering the stereographic projection $S^3\to {\mathbb{R}}^3$), we can visualize (Figure 1) the foliation of ${\mathbb{R}}^3$ by these tori $T_{\rho }$, where $T_0$ degenerates to the unit circle on the $xy$-plane of ${\mathbb{R}}^3$ and $T_{\infty }$ degenerates to the $z$-axis. Each $S^1$-fiber is a slope 1 simple closed curve on one of the tori $T_{\rho }$, and the image of the projection is exactly the compatification $S^2$ of any half plane$$\{(r\cos (\phi ),r\sin (\phi ),z)\mid r>0,z\in \mathbb{R}\}$$of ${\mathbb{R}}^3$ (for a fixed $\phi \in \mathbb{R}$).

Figure 1. A visualization of Hopf fibration

Moreover, the disjoint union of any two fibers over two distinct points of $S^1$ is in fact the so called Hopf link, shown in Figure 2. Such the link is nontrivial, the Hopf fibration is not a trivial fiber bundle.

Figure 2. Hopf link

Definition 3.5. A covering (space) is a locally trivial map $p:E\to B$ with discrete fiber $F$. A covering map which is a trivial fiber bundle is also called a trivial covering. If we would like to specify the fiber, we call it a $F$-covering. If the fiber $F$ has $n$ points, we also call it a $\mathbf{n}$-fold covering.

Figure 3. Trivialization (left) and covering (right)

Example 3.6. The map $\exp :{\mathbb{R}}^1\to S^1,t\to e^{2\pi \mathbf{i}t}$ is a $\mathbb{Z}$-covering.

Figure 4. The $\mathbb{Z}$-covering of $S^1$

If $U=S^1-\{-1\}$, then$${\exp }^{-1}(U)=\underset{n\in \mathbb{Z}}{⨆}(n-\frac{1}{2},n+\frac{1}{2}).$$

Example 3.7. The map $S^1\to S^1,e^{2\pi \mathbf{i}\theta }\mapsto e^{2\pi \mathbf{i}n\theta }$ is an $|n|$-fold covering, for any $n\in \mathbb{Z}-\{0\}$.

Example 3.8. The map $\mathbb{C}\to \mathbb{C}$, $z\mapsto z^n$, is not a covering (why?). But

•	the map ${\mathbb{C}}^{\ast }\to {\mathbb{C}}^{\ast }$, $z\mapsto z^n$, is an $|n|$-fold covering, where ${\mathbb{C}}^{\ast }=\mathbb{C}-\{0\}$and $n\in \mathbb{Z}-\{0\}$.
•	the map $\exp :\mathbb{C}\to {\mathbb{C}}^{\ast }$, $z\mapsto e^{2\pi iz}$ is a $\mathbb{Z}$-covering.

Example 3.9 (From Hatcher [??]). The figure-8has two coverings as follows (the left is a $2$-fold (or double) covering and the right is a $3$-fold covering).

The 4-regular tree is its universal cover (a covering which is simply connected), see Figure 5.

Figure 5. 4-regular tree

Example 3.10. Recall that the number of holes (genus) and the number of boundary components determine the homeomorphism type of a compact oriented topological surface. Let $S_{g,b}$ be a surface of genus $g$ and with $b$ boundary components.

•	$S_{4,0}$ admits a $7$-fold covering from $S_{22,0}$, cf. Figure 6.
•	In general, $S_{g,b}$ admits a $m$-fold covering from $S_{mg-m+1,mb}$. Figure 6. A 7-covering

Example 3.11. Denote by $\mathbb{R}{\mathbb{P}}^n$ the real projective space of dimension $n$$$\mathbb{R}{\mathbb{P}}^n={\mathbb{R}}^{n+1}-\{0\}/(\underline{x}\sim t\underline{x}),\forall t\in \mathbb{R}-\{0\},\underline{x}\in {\mathbb{R}}^{n+1}-\{0\}.$$Let $S^n$ be the $n$-sphere. Then there is a natural double cover $S^n\to \mathbb{R}{\mathbb{P}}^n$.

Example 3.12 (Branched double cover). Figure 7 shows a branched double cover of a disk$$\iota :{\Sigma }_n\to D^2,\text{branching at n points.}$$

Figure 7. The Birman-Hilden double cover via the twisted surface

Namely, when deleting those $n$ (red) points (denoted by $\Delta$) from both ${\Sigma }_n$ and $D^2$, we obtain a $2$-fold covering:$$\iota :{\Sigma }_n\backslash \Delta \stackrel{2:1}{\to }{D}^2\backslash \Delta$$with a bijection $\iota :\Delta \to \Delta$ on $\Delta$. This can be used to show the homeomorphism in Figure ?? (it is a special case of Figure 7 for $n=3$. In general, it has genus $g=\lfloor \frac{n}{2}\rfloor -1$ with $b=n-2g$ boundary components.)

Figure 8. The normal view of the branched double cover of the punctured disk

Definition 3.13. Let $p:E\to B,f:X\to B$. A lifting of $f$ along $p$ is a map $F:X\to E$ such that $p\circ F=f$

Lemma 3.14. Let $p:E\to B$ be a covering. Let$$\begin{array}{rl}D& =\{(x,x)\in E\times E|x\in E\}\\ Z& =\{(x,y)\in E\times E|p(x)=p(y)\}.\end{array}$$Then $D\subset Z$ is both open and closed.

Proof. Exercise.

$\square$

Theorem 3.15 (Uniqueness of lifting). Let $p:E\to B$ be a covering. Let $F_0,F_1:X\to E$ be two liftings of $f:X\to B$. Suppose $X$ is connected and $F_0,F_1$ agree somewhere. Then $F_0=F_1$.

Proof. Let $D,Z$ be defined in Lemma 3.14. Consider the map $\tilde{F}=(F_0,F_1):X\to Z\subset E\times E$. By assumption, we have $\tilde{F}(X)\cap D\ne \varnothing$. Moreover, Lemma 3.14 implies that ${\tilde{F}}^{-1}(D)$ is both open and closed. Since $X$ is connected, we find ${\tilde{F}}^{-1}(D)=X$ which is equivalent to $F_0=F_1$.

$\square$

Fibration

Definition 3.16. A map $p:E\to B$ is said to have the homotopy lifting property (HLP) with respect to $X$ if for any maps $\tilde{f}:X\to E$ and $F:X\times I\to B$ such that $p\circ \tilde{f}=F{|}_{X\times \{0\}}$, there exists a lifting $\tilde{F}$ of $F$ along $p$ such that $\tilde{F}{|}_{X\times \{0\}}=\tilde{f}$, i.e., the following diagram is commutative

Definition 3.17. A map $p:E\to B$ is called a fibration (or Hurewicz fibration) if $p$ has HLP for any space.

Theorem 3.18. A covering is a fibration.

Proof. Let $p:E\to B,f:X\to B,\tilde{f}:X\to E,F:X\times I\to B$ be the data as in Definition 3.16. We only need to show the existence of ${\tilde{F}}_x$ for some neighbourhood $N_x$ of any given point $x\in X$.In fact, for any two such neighbourhoods $N_x$ and $N_y$ with $N_x\cap N_y=N_0\ne \varnothing$, we have ${\tilde{F}}_x{\mid }_{N_0}$ and ${\tilde{F}}_y{\mid }_{N_0}$ agree at some point on $\tilde{f}{\mid }_{N_0}$ and hence agree everywhere in $N_0$ by the uniqueness of lifting (Theorem 3.15). Thus $\{{\tilde{F}}_x\mid x\in X\}$ glue to give the required lifting $\tilde{F}$.

Since $I$ is compact, given $x\in X$ we can find a neighbourhood $N_x$ and a partition$$0=t_0<t_1<\cdots <t_m=1$$such that $p$ has a trivialization over open sets $U_i\supset F(N_x\times [t_i,t_{i+1}])$. Now we construct the lifting ${\tilde{F}}_x$ on $N_x\times [t_0,t_k]$, for $1\le k\le m$, by induction on $k$.

•	For $k=1$, the lifting ${\tilde{F}}_x$ on $N_x\times [t_0,t_1]$ to one of the sheets of $p^{-1}(U_1)$ is determined by $\tilde{f}{\mid }_{N_x\times \{0\}}$:
•	Assume that we have constructed ${\tilde{F}}_x$ on $N_x\times [t_0,t_k]$ for some $k$. Now, the lifting ${\tilde{F}}_x$ on $N_x\times [t_k,t_{k+1}]$ to one of the sheets of $p^{-1}(U_k)$ is determined by $\tilde{f}{\mid }_{N_x\times \{t_k\}}$, which can be glued to the lifting on $N_x\times [t_0,t_k]$ by the uniqueness of lifting again. This finishes the inductive step.

We obtain a lifting ${\tilde{F}}_x$ of $F$ on $N_x\times I$ as required.

$\square$

Corollary 3.19. Let $p:E\to B$ be a covering. Then for any path $\gamma :I\to B$ and $e\in E$ such that $p(e)=\gamma (0)$, there exists a unique path $\tilde{\gamma }:I\to E$ which lifts $\gamma$ and $\tilde{\gamma }(0)=e$.

Proof. Apply HLP to $X=\mathrm{pt}$.

$\square$

Corollary 3.20. Let $p:E\to B$ be a covering. Then ${\Pi }_1(E)\to {\Pi }_1(B)$ is a faithful functor. In particular, the induced map ${\pi }_1(E,e)\to {\pi }_1(B,p(e))$ is injective.

Proof. Let ${\tilde{\gamma }}_i:I\to E$ be two paths and $[{\tilde{\gamma }}_i]\in {\mathrm{Hom}}_{{\Pi }_1(E)}(e_1,e_2)$. Let ${\gamma }_i=p\circ {\tilde{\gamma }}_i$. Suppose $[{\gamma }_1]=[{\gamma }_2]$ and we need to show that $[{\tilde{\gamma }}_1]=[{\tilde{\gamma }}_2]$.

Let $F:{\gamma }_1\simeq {\gamma }_2$ be a homotopy. Consider the following commutative diagram with the lifting $\tilde{F}$ by HLPThen the uniqueness of lifting implies $\tilde{F}{\mid }_{I\times \{1\}}={\tilde{\gamma }}_2$. Thus, $\tilde{F}:{\tilde{\gamma }}_1\simeq {\tilde{\gamma }}_2$.

$\square$

Transport functor

Let $p:E\to B$ be a covering. Let $\gamma :I\to B$ be a path in $B$ from $b_0$ to $b_1$. It defines a map$$\begin{array}{rl}{T}_{\gamma }:p^{-1}(b_0)& \to p^{-1}(b_1)\\ e_0& \mapsto e_1=\tilde{\gamma }(1)\end{array}$$where $\tilde{\gamma }$ is a lift of $\gamma$ with initial condition $\tilde{\gamma }(0)=e_0$.

Figure 9. The transportation

Assume $[{\gamma }_1]=[{\gamma }_2]$ in $B$. HLP implies that $T_{{\gamma }_1}=T_{{\gamma }_2}$. We find a well-defined map:$$\begin{array}{rl}T:{\mathrm{Hom}}_{{\Pi }_1(B)}(b_1,b_2)& \to {\mathrm{Hom}}_{\mathsf{Set}}(p^{-1}(b_1),p^{-1}(b_2))\\ [\gamma ]& \mapsto T_{[\gamma ]}.\end{array}$$

This leads to the following definition (check the functor property!).

Definition 3.21. The following data$$\begin{array}{rl}T:{\Pi }_1(B)& \to \mathsf{Set}\\ b& \to p^{-1}(b)\\ [\gamma ]& \mapsto T_{[\gamma ]}\end{array}$$define a functor, called the transport functor. In particular, we have a well-defined map$${\pi }_1(B,b)={\mathrm{Aut}}_{{\Pi }_q(B)}(b)\to {\mathrm{Aut}}_{\mathsf{Set}}(p^{-1}(b)).$$Here ${\mathrm{Aut}}_{\mathsf{Set}}(S)$ consists of all set isomorphisms in ${\mathrm{Hom}}_{\mathsf{Set}}(S,S)$ for $S\in \mathrm{Obj}(\mathsf{Set})$.

Example 3.22. Consider the covering map$$\mathbb{Z}\to {\mathbb{R}}^1\stackrel{\exp }{\to }{S}^1.$$Consider the following path representing an element of ${\pi }_1(S^1)$$${\gamma }_n:I\to S^1,t\to \exp (nt)=e^{2\pi int},n\in \mathbb{Z}.$$

Start with any point $m\in \mathbb{Z}$ in the fiber, ${\gamma }_n$ lifts to a map to ${\mathbb{R}}^1$$${\tilde{\gamma }}_n:I\to R^1,t\mapsto m+nt.$$We find $T_{[{\gamma }_n]}(m)=\tilde{\gamma }(1)=m+n$. Therefore $T_{[{\gamma }_n]}\in {\mathrm{Aut}}_{\mathsf{Set}}(\mathbb{Z})$ is$$T_{[{\gamma }_n]}:\mathbb{Z}\to \mathbb{Z},m\mapsto m+n.$$

Proposition 3.23. Let $p:E\to B$ be a covering, $E$ be path connected. Let $e\in E,b=p(e)\in B$. Then the action of ${\pi }_1(B,b)$ on $p^{-1}(b)$ is transitive, whose stabilizer ${\mathrm{Stab}}_e({\pi }_1(B,b))$ at $e$ is ${\pi }_1(E,e)$. In other words,$$p^{-1}(b)\cong {\pi }_1(B,b)/{\pi }_1(E,e)$$as a coset space, i.e. we have the following short exact sequence$$\begin{array}{rl}1\to {\pi }_1(E,e)\to {\pi }_1(B,b)& \stackrel{{\partial }_e}{\to }{p}^{-1}(b)\to 1.\\ [\gamma ]& \mapsto T_{\gamma }(e)\end{array}$$

Proof. For any point $e^{\prime }\in p^{-1}(b)$, let $\tilde{\gamma }:e\to e^{\prime }$ be a path in $E$ and $\gamma =p\circ \tilde{\gamma }$. Then $e^{\prime }={\partial }_e([\gamma ])$. This shows the surjectivity of ${\partial }_e$.

HLP implies that $p_{\ast }:{\pi }_1(E,e)\to {\pi }_1(B,b)$ is injective and we can view ${\pi }_1(E,e)$ as a subgroup of ${\pi }_1(B,b)$. By definition, for $\tilde{\gamma }\in {\pi }_1(E,e)$, we have ${\partial }_e([p\circ \tilde{\gamma }])=\tilde{\gamma }(1)=e$, i.e. ${\pi }_1(E,e)\subset {\mathrm{Stab}}_e({\pi }_1(B,b))$. On the other hand, if $T_{\gamma }(e)=e$, then the lift $\tilde{\gamma }$ of $\gamma$ is a loop, i.e. $\tilde{\gamma }\in {\pi }_1(E,e)$. Therefore, ${\pi }_1(E,e)\supset {\mathrm{Stab}}_e({\pi }_1(B,b))$. This implies ${\pi }_1(E,e)={\mathrm{Stab}}_e({\pi }_1(B,b))$, which finishes the proof.

$\square$

Lifting Criterion

Theorem 3.24 (Lifting Criterion). Let $p:E\to B$ be a covering. Consider a continuous map $f:X\to B$, where $X$ is path connected and locally path connected. Let $e_0\in E,x_0\in X$ such that $f(x_0)=p(e_0)$. Then there exists a lift $F$ of $f$ with $F(x_0)=e_0$ if and only if$$f_{\ast }({\pi }_1(X,x_0))\subset p_{\ast }({\pi }_1(E,e_0)).$$

Proof. If such $F$ exists, then$$f_{\ast }({\pi }_1(X,x_0))=p_{\ast }(F_{\ast }({\pi }_1(X,x_0)))\subset p_{\ast }({\pi }_1(E,e_0)).$$Conversely, let$$\tilde{E}=\{(x,e)\in X\times E|f(x)=p(e)\}\subset X\times E$$and consider the following commutative diagramThe projection $\tilde{p}$ is also a covering. We have an induced commutative diagram of functorswhich induces natural group homomorphisms$${\pi }_1(X,x_0)\stackrel{f_{\ast }}{\to }{\pi }_1(B,b_0)\to \mathrm{Aut}(p^{-1}(b_0))=\mathrm{Aut}({\tilde{p}}^{-1}(x_0)),b_0=f(x_0)=p(e).$$Let ${\tilde{e}}_0=(x_0,e_0)\in \tilde{E}$. The condition $f_{\ast }({\pi }_1(X,x_0))\subset p_{\ast }({\pi }_1(E,e_0))$ says that ${\pi }_1(X,x_0)$ stabilizes ${\tilde{e}}_0$. By Proposition 3.23, this implies we have a group isomorphism$${\tilde{p}}_{\ast }:{\pi }_1(\tilde{E},{\tilde{e}}_0)\cong {\pi }_1(X,x_0).$$Since $X$ is locally path connected, $\tilde{E}$ is also locally path connected. Then path connected components and connected components of $\tilde{E}$ coincide. Let $\tilde{X}$ be the (path) connected component of $\tilde{E}$ containing ${\tilde{e}}_0$, then ${\pi }_1(\tilde{E},\tilde{e})\cong {\pi }_1(X,x_0)$ implies that $\tilde{p}:\tilde{X}\to X$ is a covering with fiber a single point, hence a homeomorphism. Its inverse defines a continuous map $X\to \tilde{E}$ whose composition with $\tilde{E}\to E$ gives $F$.

$\square$

$G$-principal covering

Definition 3.25. Let $G$ be a discrete group. A continuous action $G\times X\to X$ is called properly discontinuous if for any $x\in X$, there exists an open neighborhood $U$ of $x$ such that$$\begin{array}{c}g(U)\cap U=\varnothing ,\forall g\ne 1\in G.\end{array}$$We define the orbit space $X/G=X/\sim$ where $x\sim g(x)$ for any $x\in X,g\in G$.

Proposition 3.26. Assume $G$ acts properly discontinuously on $X$, then the quotient map $X\to X/G$ is a covering.

Proof. For any $x\in X$, let $U$ be the neighbourhood satisfying $g(U)\cap U=\varnothing ,\forall g\ne 1\in G$. Then$$p^{-1}\left(p(U)\right)=\underset{g\in G}{⨆}gU$$is a disjoint union of open sets. Thus, $p$ is locally trivial with discrete fiber $G$, hence a covering.

$\square$

Definition 3.27. A left (right) $\mathbf{G}$-principal covering is a covering $p:E\to B$ with a left (right) properly discontinuous $G$-action on $E$ over $B$such that the induced map $E/G\to B$ is a homeomorphism.

Example 3.28. $\exp :{\mathbb{R}}^1\to S^1$ is a $\mathbb{Z}$-principal covering for the action $n:t\to t+n,\forall n\in \mathbb{Z}$.

Example 3.29. $S^n\to \mathbb{R}{P}^n\cong S^n/{\mathbb{Z}}_2$ is a ${\mathbb{Z}}_2$-principal covering.

Proposition 3.30. Let $p:E\to B$ be a $G$-principal covering. Then the transport is $G$-equivariant, i.e.,$$T_{[\gamma ]}\circ g=g\circ T_{[\gamma ]},\forall g\in G,\gamma \text{a path in}B.$$

Proof. Let $\gamma :b_0\to b_1$ and $e_0\in p^{-1}(b_0)$. Then $\tilde{\gamma }:e_0\to e_1=T_{[\gamma ]}(e_0)$ for some $e_1\in p^{-1}(b_1)$. If we apply the transformation $g$ to the path $\tilde{\gamma }$, we find another lift of $\gamma$ but with endpoints $g(e_0)$ and $g(e_1)$. Therefore$$T_{[\gamma ]}(g(e_0))=g(e_1).$$It follows that $T_{[\gamma ]}(g(e_0))=g(e_1)=g(T_{[\gamma ]}(e_0))$. This proves the proposition.

Figure 10. Transport commutes with $G$-action

$\square$

Theorem 3.31. Let $p:E\to B$ be a $G$-principal covering, $E$ path connected, $e\in E,b=p(e)$. Then we have an exact sequence of groups$$1\to {\pi }_1(E,e)\to {\pi }_1(B,b)\to G\to 1.$$In other words, ${\pi }_1(E,e)$ is a normal subgroup of ${\pi }_1(B,b)$ and $G={\pi }_1(B,b)/{\pi }_1(E,e)$.

Proof. Let $F=p^{-1}(b)$. The previous proposition implies that ${\pi }_1(B,b)$-action and $G$-action on $F$ commute. It induces a ${\pi }_1(B,b)\times G$-action on $F$. Consider its stabilizer at $e$ and two projections${\mathrm{pr}}_1$ is an isomorphism and ${\mathrm{pr}}_2$ is an epimorphism with $\ker ({\mathrm{pr}}_2)={\text{Stab}}_e({\pi }_1(B,b))={\pi }_1(E,e)$.

$\square$

Apply this theorem to the covering $\exp :{\mathbb{R}}^1\to S^1$, we find a group isomorphism$$\mathrm{deg}:{\pi }_1(S^1)\stackrel{\cong }{\to }\mathbb{Z}$$which is called the degree map. An example of degree $n$ map is$$S^1\to S^1,e^{i\theta }\mapsto e^{in\theta }.$$

Applications

Definition 3.32. Let $i:A\subset X$ be an inclusion. A continuous map $r:X\to A$ is called a retraction if $r\circ i=1_A$. It is called a deformation retraction if furthermore we have a homotopy $i\circ r\simeq 1_X\mathrm{rel}A$. We say $A$ is a (deformation) retract of $X$ if such a (deformation) retraction exists.

Proposition 3.33. If $i:A\subset X$ is a retract, then $i_{\ast }:{\pi }_1(A)\to {\pi }_1(X)$ is injective.

Proof. Let $r:X\to A$ such that $r\circ i=1_A$. Then the composition$${\pi }_1(A)\stackrel{i_{\ast }}{\to }{\pi }_1(X)\stackrel{r_{\ast }}{\to }{\pi }_1(A)$$is the identity. It follows that $i_{\ast }:{\pi }_1(A)\to {\pi }_1(X)$ is injective.

$\square$

Corollary 3.34. Let $D^2$ be the unit disk in ${\mathbb{R}}^2$. Then its boundary $S^1$ is not a retract of $D^2$.

Proof. Since $D^2$ is contractible, we have ${\pi }_1(D^2)=1$. But ${\pi }_1(S^1)=\mathbb{Z}$. Then the corollary follows from the proposition above.

$\square$

Theorem 3.35 (Brouwer fixed point Theorem). Let $f:D^2\to D^2$. Then there exists $x\in D^2$ such that $f(x)=x$.

Proof. Assume $f$ has no fixed point. Let $l_x$ be the ray starting from $f(x)$ pointing toward $x$. Then$$D^2\to S^1,x\mapsto l_x\cap \partial D^2$$is a retraction of $\partial D^2=S^1\subset D^2$. Contradiction.

$\square$

Theorem 3.36 (Fundamental Theorem of Algebra). Let $f(x)=x^n+c_1{x}^{n-1}+\cdots +c_n$ be a polynomial with $c_i\in \mathbb{C},n>0$. Then there exists $a\in \mathbb{C}$ such that $f(a)=0$.

Proof. Assume $f$ has no root in $\mathbb{C}$. Define a homotopy of maps from $S^1$ to $S^1$$$F:S^1\times I\to S^1,F(e^{i\theta },t)=\frac{f(\tan (\frac{\pi t}{2})e^{i\theta })}{\left|f(\tan (\frac{\pi t}{2})e^{i\theta })\right|}.$$By construction, $\mathrm{deg}(F{|}_{S^1\times 0})=0$ and $\mathrm{deg}(F{|}_{S^1\times 1})=n$. But they are homotopic hence representing the same element in ${\pi }_1(S^1)$. Contradiction.

$\square$

Proposition 3.37 (Antipode). Let $f:S^1\to S^1$ be an antipode-preserving map, i.e. $f(-u)=-f(u)$. Then $\mathrm{deg}(f)$ is odd. In particular, $f$ is NOT null homotopic.

Proof. Let $F:{\mathbb{R}}^1\to {\mathbb{R}}^1$ be a lifting of $f\circ \exp :{\mathbb{R}}^1\to S^1$ for the covering map $\exp :{\mathbb{R}}^1\to S^1$. Then$$\begin{array}{r}F(x+1)=F(x)+\mathrm{deg}(f).\end{array}$$Since $f$ is antipode-preserving, $F(x+1/2)=F(x)+m$ for some $m\in \mathbb{Z}+1/2$. So $F(x+1)=F(x)+2m$ which implies $\mathrm{deg}(f)=2m$ is odd.

$\square$

Example 3.38. Let $\sigma :S^1\to S^1$ be the antipode map defined by $\sigma (u)=-u$. Then $\mathrm{deg}(\sigma )=1$.

Theorem 3.39 (Borsuk-Ulam). Let $f:S^2\to {\mathbb{R}}^2$. Then there exists $x\in S^2$ such that $f(x)=f(-x)$.

Proof. Assume $f(x)\ne f(-x),\forall x\in S^2$. Define$$\rho :S^2\to S^1,\rho (x)=\frac{f(x)-f(-x)}{\left|f(x)-f(-x)\right|}.$$Let $D^2$ be the upper hemi-sphere of $S^2$. It defines a homotopy between constant map and $\rho {|}_{\partial D^2}:S^1\to S^1$, hence $\mathrm{deg}(\rho {|}_{\partial D^2})=0$. On the other hand, $\rho {|}_{\partial D^2}$ is antipode-preserving: $\rho {|}_{\partial D^2}(-x)=-\rho {|}_{\partial D^2}(x)$, hence $\mathrm{deg}(\rho {|}_{\partial D^2})$ is odd. Contradiction.

$\square$

Theorem 3.40 (Ham Sandwich Theorem). Let $A_1,A_2$ be two bounded regions of positive areas in ${\mathbb{R}}^2$. Then there exists a line which cuts each $A_i$ into half of equal areas.

Proof. Let $A_1,A_2\subset {\mathbb{R}}^2\times \{1\}\subset {\mathbb{R}}^3$.

Given $u\in S^2$, let $P_u$ be the plane passing the origin and perpendicular to the unit vector $u$. Let $A_i(u)=\{p\in A_i|p\cdot u\le 0\}.$ Define the continuous map$$f=(f_1,f_2):S^2\to {\mathbb{R}}^2,f_i(u)=\text{Area}(A_i(u)).$$By Borsuk-Ulam, $\exists u$ such that $f(u)=f(-u)$. The intersection ${\mathbb{R}}^2\times \{1\}\cap P_u$ gives the required line since$$f(u)=f(-u)⟺f_i(u)=\frac{1}{2}\text{Area}(A_i).$$

$\square$