Munkres §52: The Fundamental Group

◄ Munkres §53: Covering Spaces | Munkres §51: Homotopy of Paths ►

Claim: If $A$ is star convex, then $A$ is simply connected.

Proof: $A$ is clearly path-connected. Let $a\in A$ be the star point, let $\alpha$ and $\beta$ be two loops in $A$, and define $F:I\times I\to A$ by

$$(x,t)\mapsto \{\begin{array}{ll}(1-2t)\alpha (x)+2ta& t\le 1/2\\ 2(1-t)a+(2t-1)\beta (x)& t>1/2\end{array}.$$

Then $F$ is a path homotopy between $\alpha$ and $\beta$, implying that ${\pi }_1(A,a)=0$.

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Claim: If $\gamma =\alpha \ast \beta$, then $\hat{\gamma }=\hat{\beta }\circ \hat{\alpha }$.

Proof:

$$\begin{array}{rl}\hat{\gamma }([f])& =[\overline{\alpha \ast \beta }]\ast [f]\ast [\alpha \ast \beta ]\\ & =[\bar{\beta }\ast \bar{\alpha }]\ast [f]\ast [\alpha \ast \beta ]\\ & =[\bar{\beta }]\ast [\bar{\alpha }]\ast [f]\ast [\alpha ]\ast [\beta ]\\ & =[\bar{\beta }]\ast \hat{\alpha }([f])\ast [\beta ]\\ & =\hat{\beta }(\hat{\alpha }([f]))\\ ◼& & =(\hat{\beta }\circ \hat{\alpha })([f]).\end{array}$$

Claim: ${\pi }_1(X,x_0)$ is abelian if and only if $\hat{\alpha }=\hat{\beta }$ for all paths $\alpha ,\beta$ from $x_0$ to $x_1$, where $X$ is path-connected.

Proof: Suppose that ${\pi }_1(X,x_0)$ is abelian and recall that ${\pi }_1(X,x_1)$ is isomorphic to it. Then

$$\begin{array}{rl}\hat{\alpha }([f])& =[\bar{\alpha }]\ast [f]\ast [\alpha ]\\ & =[\bar{\alpha }]\ast [f]\ast [\beta ]\ast [\bar{\beta }\ast \alpha ]\\ & =[\bar{\beta }\ast \alpha ]\ast [\bar{\alpha }]\ast [f]\ast [\beta ]\\ & =[\bar{\beta }]\ast [f]\ast [\beta ]\\ & =\hat{\beta }([f]).\end{array}$$

Conversely, suppose that $\hat{\alpha }=\hat{\beta }$ for all paths $\alpha ,\beta$ from $x_0$ to $x_1$, let $\alpha$ be a path from $x_0$ to $x_1$, let $f$ and $g$ be loops based at $x_0$, and note that $\gamma :=f\ast \alpha$ is a path from $x_0$ to $x_1$. Then

$$\begin{array}{rl}\hat{\gamma }([g])& =[\bar{\gamma }]\ast [g]\ast [\gamma ]\\ & =[\overline{f\ast \alpha }]\ast [g]\ast [f\ast \alpha ]\\ & =[\bar{\alpha }\ast \bar{f}]\ast [g]\ast [f\ast \alpha ]\\ & =[\bar{\alpha }]\ast [\bar{f}\ast g\ast f]\ast [\alpha ]\\ & =[\bar{\alpha }]\ast [g]\ast [\alpha ]\\ & =\hat{\alpha }([g])\end{array}$$

implies that $[\bar{f}\ast g\ast f]=[g]$, which in turn implies that $[g]\ast [f]=[f]\ast [g]$.

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Claim: If $a\in A\subset X$ and $r$ is a retraction of $X$ onto $A$, then

$$r_{\ast }:{\pi }_1(X,a)\to {\pi }_1(A,a)$$

is surjective.

Proof:

$$r_{\ast }\circ {\iota }_{\ast }=(r\circ \iota {)}_{\ast }={\mathrm{id}}_{{\pi }_1(A,a)},$$

where $\iota :A\hookrightarrow X$, implies that $r$ has a right inverse, which in turn implies that it is surjective.

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Claim: If $a\in A\subset {\mathbb{R}}^n$, $y\in Y$, $h:{\pi }_1(A,a)\to {\pi }_1(Y,y)$, and $h$ is extendable to a continuous $\tilde{h}:{\mathbb{R}}^n\to Y$, then $h_{\ast }$ is trivial.

Proof:

$$h=\tilde{h}\circ \iota ⟹h_{\ast }={\tilde{h}}_{\ast }\circ {\iota }_{\ast },$$

where $\iota :A\hookrightarrow {\mathbb{R}}^n$. However, the domain of ${\tilde{h}}_{\ast }$ is ${\pi }_1({\mathbb{R}}^n,a)=0$.

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Claim: If $X$ is path-connected, $h:X\to Y$ is continuous, $h(x_0)=y_0$, $h(x_1)=y_1$, $\alpha$ is a path from $x_0$ to $x_1$, and $\beta =h\circ \alpha$, then

$$\hat{\beta }\circ (h_{x_0}{)}_{\ast }=(h_{x_1}{)}_{\ast }\circ \hat{\alpha }.$$

This is equivalent to saying that $h_{\ast }$ is independent of base point up to isomorphism.

Proof:

$$\begin{array}{rl}\hat{\beta }\circ (h_{x_0}{)}_{\ast }([f])& =[\bar{\beta }]\ast (h_{x_0}{)}_{\ast }([f])\ast [\beta ]\\ & =[h\circ \bar{\alpha }]\ast [h\circ f]\ast [h\circ \alpha ]\\ & =(h_{x_1}{)}_{\ast }([\bar{\alpha }]\ast [f]\ast [\alpha ])\\ & =(h_{x_1}{)}_{\ast }\circ \hat{\alpha }.\end{array}$$

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Let $G$ be a topological group with operation $\cdot$ and identity $x_0$, let $\mathrm{\Omega }(G,x_0)$ be the set of loops in $G$ based at $x_0$, and let

$$(f\otimes g)(s):=f(s)\cdot g(s)$$

for all $f,g\in \mathrm{\Omega }(G,x_0)$.

Note that $\mathrm{\Omega }(G,x_0)$ equipped with $\otimes$ is a group with identity $x_0(s)=x_0$.

Claim: $\otimes$ induces a group operation $\otimes$ on ${\pi }_1(G,x_0)$.

Proof: Let $[f]\otimes [g]:=[f\otimes g]$ and note that it is well-defined since $(s,t)\mapsto F(s,t)\cdot G(s,t)$ is a homotopy between $s\mapsto F(s,0)\otimes G(s,0)$ and $s\mapsto F(s,1)\otimes G(s,1)$.

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Claim: $\ast$ and $\otimes$ on ${\pi }_1(G,x_0)$ are the same.

Proof: Note that

$$\begin{array}{rl}[f]\otimes [g]& =[f\ast e_{x_0}]\otimes [e_{x_0}\ast g]\\ & =[(f\ast e_{x_0})\otimes (e_{x_0}\ast g)]\\ & =[f\ast g]\\ & =[f]\ast [g]\end{array}$$

because

$$\begin{array}{rl}((f\ast e_{x_0})\otimes (e_{x_0}\ast g))(s)& =(f\ast e_{x_0})(s)\cdot (e_{x_0}\ast g)(s)\\ & =\{\begin{array}{ll}f(s)& s\le 1/2\\ g(s)& s>1/2\end{array}\\ & =(f\ast g)(s).\end{array}$$

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Claim: ${\pi }_1(G,x_0)$ is abelian.

Proof:

$$\begin{array}{rl}[f]\ast [g]& =[f]\otimes [g]\\ & =[e_{x_0}\ast f]\otimes [g\ast e_{x_0}]\\ & =[(e_{x_0}\ast f)\otimes (g\ast e_{x_0})]\\ & =[g\ast f]\\ & =[g]\ast [f].\end{array}$$

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