Point-Set Topology

If $\gamma \in \mathrm{\Pi }(X,a,b)$, then the map $F:{[0,1]}^2\to X$ given by $F(s,t)=\gamma (s)$ for all $s,t\in [0,1]$ is a path homotopy from $\gamma$ to itself, so $\sim$ is reflective. If ${\gamma }^{\prime }\in \mathrm{\Pi }(X,a,b)$ with $\gamma \sim {\gamma }^{\prime }$ and $F:{[0,1]}^2\to X$ is a path homotopy from $\gamma$ to ${\gamma }^{\prime }$, then $G(s,t)=F(s,1-t)$ is a path homotopy from ${\gamma }^{\prime }$ to $\gamma$, so ${\gamma }^{\prime }\sim \gamma$. Thus, $\sim$ is symmetric. Finally, if ${\gamma }^{″}\in \mathrm{\Pi }(X,a,b)$ as well and we have both $\gamma \sim {\gamma }^{\prime }$ and ${\gamma }^{\prime }\sim {\gamma }^{″}$, with $F$ a path homotopy from $\gamma$ to ${\gamma }^{\prime }$ and $G$ a path homotopy from ${\gamma }^{\prime }$ to ${\gamma }^{″}$, then $H:{[0,1]}^2\to X$ defined by

is a path homotopy from $\gamma$ to ${\gamma }^{\prime }$, so $\sim$ is transitive. To see this, one should note that $F(s,1)={\gamma }^{\prime }(s)=G(s,0)$ for all $s\in [0,1]$. □

Let $F$ be a path homotopy from $\gamma$ to ${\gamma }^{\prime }$ and $G$ be a path homotopy from $\mu$ to ${\mu }^{\prime }$. Define $H:{[0,1]}^2\to X$ by

for $s,t\in [0,1]$. Then $H$ is a path homotopy from $\gamma \star \mu$ to ${\gamma }^{\prime }\star {\mu }^{\prime }$. For this, one should note that $F(1,t)=b=G(0,t)$ for all $t\in [0,1]$. □

a.

If $F$ is a homotopy from $\gamma$ to ${\gamma }^{\prime }$, then we define $G:{[0,1]}^2\to X$ by

$$G(s,t)=F(1-s,t),$$

and this is a homotopy from $\tilde{\gamma }$ to $\widetilde{{\gamma }^{\prime }}$.

b.

Define $F_a:{[0,1]}^2\to X$ by

$$F_a(s,t)=\{\begin{array}{cc}a\hfill & s\in [0,\frac{1-t}{2}],\hfill \\ \gamma (\frac{2s-1+t}{1+t})\hfill & s\in [\frac{1-t}{2},1].\hfill \end{array}$$

Note that for $s=\frac{1-t}{2}$, the second case yields $\gamma (0)=a$, so we have continuity. We then check that $F_a(s,0)=a=e_a(2s)$ for $s\in [0,\frac{1}{2}]$ while $F_a(s,0)=\gamma (2s-1)$ for $s\in [\frac{1}{2},1]$, so $F_a(s,0)=(e_a\star \gamma )(s)$ for all $s\in [0,1]$. We also have $F_a(s,1)=\gamma (s)$ for all $s\in [0,1]$ and $F_a(0,t)=a$ and $F_a(1,t)=\gamma (1)=b$ for all $t\in [0,1]$. Thus $F_a$ is a path homotopy from $e_a\star \gamma$ to $\gamma$.

If we replace $\gamma$ by $\tilde{\gamma }$, we get $e_b\star \tilde{\gamma }\sim \tilde{\gamma }$, and then

$$\gamma \star e_b=\widetilde{e_b\star \stackrel{~}{\gamma }}\sim \widetilde{\stackrel{~}{\gamma }}=\gamma .$$

c.

Define $H:{[0,1]}^2\to X$ by

$$H(s,t)=\{\begin{array}{cc}\gamma (2s(1-t))\hfill & s\in [0,\frac{1}{2}],\hfill \\ \gamma ((2-2s)(1-t))\hfill & s\in [\frac{1}{2},1].\hfill \end{array}$$

The $H$ is a path homotopy from $\gamma \star \tilde{\gamma }$ to $e_a$. We have $\tilde{\gamma }\star \gamma \sim e_b$ by replacing $\gamma$ by $\tilde{\gamma }$.

d.

Define $I:{[0,1]}^2\to X$ by

$$I(s,t)=\{\begin{array}{cc}\gamma (4{(1+t)}^{-1}s)\hfill & s\in [0,\frac{1}{4}(1+t)],\hfill \\ \mu (4s-1-t)\hfill & s\in [\frac{1}{4}(1+t),\frac{1}{4}(2+t)],\hfill \\ \nu ({(2-t)}^{-1}(4s-2-t))\hfill & s\in [\frac{1}{4}(2+t),1].\hfill \end{array}$$

We leave it to the reader to check that $I$ is a path homotopy from $(\gamma \star \mu )\star \nu$ to $\gamma \star (\mu \star \nu )$.

The operation in question is given on the classes of loops $\gamma ,\mu \in \mathrm{\Pi }(X,x_0)$ by $[\gamma ]\cdot [\mu ]=[\gamma \star \mu ]$. That this makes $G$ into a group follows from the various parts of Lemma 5.2.3: that is, the operation is associative by part d, the identity element is $[e_a]$ by part b, and the inverse of $[\gamma ]$ is $[\tilde{\gamma }]$ by part c. □

Since $f$ is a bijection, its inverse $f^{-1}$ is as well. We must show that $f^{-1}$ is a homomorphism. Let $a^{\prime },b^{\prime }\in G^{\prime }$, and note that there exist unique $a,b\in G$ with $f(a)=a^{\prime }$ and $f(b)=b^{\prime }$. We then have

□

Let $G$, $H$ and $K$ be groups. Then $\mathit{𝐺\cong 𝐺}$ via the identity map. If $\mathit{𝐺\cong 𝐻}$, then $\mathit{𝐻\cong 𝐺}$ by Lemma 5.2.13. If $\mathit{𝐺\cong 𝐻}$ and $\mathit{𝐻\cong 𝐾}$, then we have isomorphisms $f:G\to H$ and $f^{\prime }:H\to K$. Then $f^{\prime }\circ f$ is still a bijection, and $f^{\prime }(f(a\cdot b))=f^{\prime }(f(a)\cdot f(b))=f^{\prime }(f(a))\cdot f^{\prime }(f(b))$ for $a,b\in G$, so $f^{\prime }\circ f$ is a homomorphism as well, and therefore $\mathit{𝐺\cong 𝐾}$. □

Let $\lambda :[0,1]\to X$ be a path with $\lambda (0)=x_0$ and $\lambda (1)=x_1$. Define a map

for $\gamma \in \mathrm{\Pi }(X,x_0)$. If $\gamma \sim {\gamma }^{\prime }$, then $(\tilde{\lambda }\star \gamma )\star \lambda \sim (\tilde{\lambda }\star {\gamma }^{\prime })\star \lambda$, so this map induces the function ${\psi }_{\lambda }$. This is a bijection since it has an inverse induced by $\mu \mapsto (\lambda \star \mu )\star \tilde{\lambda }$. It is then an isomorphism, since

□

Let $\gamma ,{\gamma }^{\prime }\in \mathrm{\Pi }(X,a,b)$. Then ${\gamma }^{-1}\star {\gamma }^{\prime }\in \mathrm{\Pi }(X,a)$. Since $X$ is simply connected, ${\gamma }^{-1}\star {\gamma }^{\prime }\sim e_a$, so $\gamma \star ({\gamma }^{-1}\star {\gamma }^{\prime })\sim e_a\star \gamma$, from which it follows that ${\gamma }^{\prime }\sim \gamma$. □

If $\gamma \sim {\gamma }^{\prime }$ for some ${\gamma }^{\prime }\in \mathrm{\Pi }(X,x_0)$, and $F:{[0,1]}^2\to X$ is a path homotopy from $\gamma$ to ${\gamma }^{\prime }$, then $f\circ F$ is a homotopy from $f\circ \gamma$ to $f\circ {\gamma }^{\prime }$. Part b is immediate from the definition of composition of paths. □

Since $f^{-1}\circ f={\mathrm{id}}_X$, we have ${(f^{-1})}_{\ast }\circ f_{\ast }={\mathrm{id}}_{{\pi }_1(X,x_0)}$ by Lemma 5.2.25, and similarly for the other composition. □

Let $\gamma$ be a loop in $A$ based at $a_0$. Then $\gamma$ is also a loop in $X$ based at $a_0$, and $r\circ \gamma =\gamma$, so $r_{\ast }$ is surjective. Since $r\circ \iota ={\mathrm{id}}_A$, we have the last equality of the statement, which forces ${\iota }_{\ast }$ to be injective. □

By definition, the restriction $p{|}_B$ is a continuous surjective map. Let $x\in Y$, and let $U$ be an open neighborhood of $x$ in $Y$ (hence in $X$) contained in $p(B)$. By Remark 5.3.2, the open set $p^{-1}(U)$ is a disjoint union of open sets $V_c$ in $C$ for each $c\in p^{-1}(x)$ with $p{|}_{V_c}:V_c\to U$ a homeomorphism. For $b\in p^{-1}(x)\subseteq B$, since $p(V_b)=U\subseteq Y$, we have $V_b\subseteq B$. That is, $p{|}_B$ takes $V_b$ homeomorphically onto its image $U$. □

For $(x,x^{\prime })\in X\times X^{\prime }$, let $U$ and $U^{\prime }$ be open neighborhoods of $x$ and $x^{\prime }$ in $X$ and $X^{\prime }$ respectively such that $p^{-1}(U)$ and ${(p^{\prime })}^{-1}(U^{\prime })$ are disjoint unions of open neighborhoods of the points in the inverse images of $x$ and $x^{\prime }$, respectively, such that the images of these open neighborhoods map homeomorphically under $p$ and $p^{\prime }$ to $U$ and $U^{\prime }$, again respectively. Then $P^{-1}(U\times U^{\prime })$ is a disjoint union of all products of these sets, one for each point in $P^{-1}(x,x^{\prime })$, and again, they each map homeomorphically to $U\times U^{\prime }$ under $P$ by construction. □

Since $p$ is a covering map, there exists an open cover $𝒰$ of $X$ by sets that are evenly covered by $p$. Then $𝒱=\{{\gamma }^{-1}(U)\mid U\in 𝒰\}$ is an open cover of $[0,1]$. By Lemma 3.3.8, there exists $N\ge 1$ such that the intervals $A_i=[\frac{i}{N}:\frac{i+1}{N}]$ with $0\le i\le N-1$ are each contained in some element of $𝒱$, which has the form ${\gamma }^{-1}(U_i)$ for some $U_i\in 𝒰$.

We define $\tilde{\gamma }$ with $\tilde{\gamma }(0)=c_0$ on each $A_i$ recursively. Suppose we have defined $\tilde{\gamma }$ on $[0,\frac{i}{N}]$ for some $i\ge 0$. Let $c_i=\tilde{\gamma }(\frac{i}{N})$. Since $p$ evenly covers $U_i$, we have an open neighborhood $V_i$ of $c_i$ which maps homeomorphically to $U_i$ under $p$. Let $f_i={(p{|}_{V_i})}^{-1}:U_i\to V_i$ be the inverse homeomorphism, and define $\tilde{\gamma }(s)=f_i\circ \gamma (s)$ for $s\in A_i$. The map $\tilde{\gamma }$ on $[0,\frac{i+1}{N}]$ is then continuous by Lemma 2.1.12. Note that this is the only continuous extension of $\tilde{\gamma }(s)$ from $[0,\frac{i}{N}]$ to $[0,\frac{i+1}{N}]$. That is, $\tilde{\gamma }(\frac{i}{N})\in V_i$ and $f^{-1}(U_i)$ is the disjoint union of $V_i$ and its complement in the inverse image, so in that $\tilde{\gamma }{|}_{A_i}$ must have connected image, its entire image must lie in $V_i$. But then the map $p{|}_{V_i}:V_i\to U_i$ is bijective, so $f_i\circ \gamma {|}_{A_i}$ is the only continuous lift of $\gamma {|}_{A_i}$ to $C$ in $V_i$. Thus, we have defined our unique path $\tilde{\gamma }:[0,1]\to C$ lifting $\gamma$ with $\tilde{\gamma }(0)=c_0$. □

Let $F$ be a path homotopy between $\gamma$ and ${\gamma }^{\prime }$, and use Lemma 5.3.13 to uniquely lift it to $\tilde{F}:{[0,1]}^2\to C$ with $\tilde{F}(0,0)=c_0$. We repeatedly use the uniqueness of Proposition 5.3.12. Then $\tilde{F}(s,0)$ is a path lifting $\gamma$ with initial point $c_0$, so must be $\tilde{\gamma }$. Similarly, $\tilde{F}(0,t)$ is a path lifting the constant path $e_{x_0}$ with initial point $c_0$, so must be the constant path $e_{c_0}$. Then $\tilde{F}(0,1)=c_0$, and $\tilde{F}(s,1)$ is a path lifting ${\gamma }^{\prime }$ with initial point $c_0$, so must be ${\tilde{\gamma }}^{\prime }$. Finally, $\tilde{F}(1,t)$ is a path lifting the constant path $e_{\gamma (1)}$ with $\tilde{F}(1,t)=\tilde{\gamma }(1)$, so must also be constant. Thus, $\tilde{\gamma }$ and ${\tilde{\gamma }}^{\prime }$ have the same final point, and $\tilde{F}$ is a path homotopy from $\tilde{\gamma }$ to ${\tilde{\gamma }}^{\prime }$. □

That ${\varphi }_{c_0}$ is well-defined is an immediate consequence of Proposition 5.3.14. If $C$ is path connected and $c_1\in p^{-1}(x_0)$, then choose a path $\lambda \in \mathrm{\Pi }(C,c_0,c_1)$. Then $\lambda$ is a lift of $\gamma =p\circ \lambda \in \mathrm{\Pi }(X,x_0)$, and ${\varphi }_{c_0}([\gamma ])=c_1$.

Suppose that $C$ is simply connected, and let $\gamma ,\mu \in \mathrm{\Pi }(X,x_0)$ with ${\varphi }_{c_0}([\gamma ])={\varphi }_{c_0}([\mu ])$. Let $c_1$ denote the latter point. Let $\tilde{\gamma }$ and $\tilde{\mu }$ be the unique lifts to $C$ of $\gamma$ and $\mu$, respectively, with initial point $c_0$. Then have final point $c_1$ by assumption. Since $C$ is simply connected, there then exists a homotopy $\tilde{F}$ from $\tilde{\gamma }$ to $\tilde{\mu }$, and then $p\circ \tilde{F}$ is a homotopy from $\gamma$ to $\mu$. In other words, we have $[\gamma ]=[\mu ]$, so ${\varphi }_{c_0}$ is injective. □

The map $p:ℝ\to S^1$ given by $p(x)=e^{2\mathit{𝜋𝑖𝑥}}$ is a covering. Since $ℝ$ is simply connected and the inverse image of $1$ is $\mathbb{Z}\subset ℝ$, Theorem 5.3.15 tells us that the map ${\varphi }_0:{\pi }_1(S^1,1)\to \mathbb{Z}$ is a bijection.

We need only show that ${\varphi }_0$ is a homomorphism. So, let $\gamma ,\mu \in \mathrm{\Pi }(S^1,1)$, and set $n={\varphi }_0([\gamma ])$ and $m={\varphi }_0([\mu ])$. Let $\tilde{\gamma }$ be a lift of $\gamma$ with initial point $0$; its final point is then $n$. Let $\tilde{\mu }$ be the unique lift of $\mu$ with initial point $0$, and note that the function ${\tilde{\mu }}^{\prime }=n+\tilde{\mu }$ is a lift of $\mu$ with initial point $n$ and final point $n+\tilde{\mu }(1)=n+m$. Then $\tilde{\gamma }\star \tilde{\mu }$ lifts $\gamma \star \mu$ and has final point $n+m$, so ${\varphi }_0([\gamma ][\mu ])=n+m={\varphi }_0([\gamma ])+{\varphi }_0([\mu ])$. □

Suppose $f:D\to D$ has no fixed point. Define $g:D\to S^1$ by letting $g(x)$ for $x\in D$ be the unique point of $S^1$ on the ray from $f(x)$ to $x$ in ${ℝ}^2$. Then $g$ is a retract from $D$ onto $S^1$, as the reader will verify. But this contradicts Example 5.3.17. □