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Problem of the Week #54 - June 10th, 2013

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Chris L T521

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Jan 26, 2012
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Here's this week's problem.

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Problem: Recall that the fundamental group $\pi_1(X,x_0)$ is the set of homotopy classes of base point preserving maps $(S^1,s_0)\rightarrow (X,x_0)$. Let $[S^1,X]$ be the set of free homotopy classes of maps without conditions on base points. There there exists a map $\Phi:\pi_1(X,x_0)\rightarrow [S^1,X]$ obtained by ignoring base points.

Show that:

(a) $\Phi$ is onto if $X$ is path connected.
(b) $\Phi([f])=\Phi([g])$ if and only if $[f]$ and $[g]$ are conjugate in $\pi_1(X,x_0)$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
No one answered this week's question. You can find my solution below.


Proof: Let $f,g\in\pi_1(X,x_0)$. Ignoring the base point, we will show that $fgf^{-1}\simeq g$ (here, $\simeq$ means "homotopic to"). WLOG, assume that $fgf^{-1}$ transverse $f$, $g$, and $f^{-1}$ on three intervals: $[0,1/3]$, $[1/3,2/3]$, and $[2/3,1]$, respectively. Thinking of $S^1$ as $\mathbb{R}/\mathbb{Z}$, we can instead start at $1/3$ and end at $4/3$ (this, in return, corresponds to a free homotopy that moves the base point). Hence, $fgf^{-1}$ is free homotopic to $gff^{-1}$, which is homotopic to $g$. Thus, the conjugacy classes map into homotopy classes of maps $S^1\rightarrow X$. Now, any homotopy class of maps $S^1\rightarrow X$ can be represented by some loop in $X$. If $X$ is path connected, this can be extended to a loop based at $x_0$, and such a loop will be mapped by $\Phi$ to this homotopy class. Therefore, $\Phi$ is surjective (onto).

To show that $\Phi$ is injective, we again let $f,g\in \pi_1(X,x_0)$ be elements that are homotopic if we ignore base points (i.e. $\Phi([f])=\Phi([g])$). Then there is a continuous map $H:[0,1]^2\rightarrow X$ such that $H(0,t)=H(1,t)$ for all $t$, and $H(t,0)=f(t)$ and $H(t,1)=g(t)$. Let $h:[0,1]\rightarrow X$ be defined by $h(t)=H(0,t)$ so that $h$ keeps track of the basepoint change along $H$. Then $h(0)=H(0,0)=f(0)$ and $h(1)=H(0,1)=g(0)$, so $h\in\pi_1(X,x_0)$.


Claim: $hgh^{-1}\simeq f$.


Proof of Claim: Write
\[f\simeq \begin{cases}h(3t) & \text{if $0\leq r\leq \frac{1}{3}$}\\ H(t,0) & \text{if $\frac{1}{3}\leq t\leq\frac{2}{3}$}\\ h^{-1}(3t-2) & \text{if $\frac{2}{3}\leq t\leq 1$}\end{cases}\]
and
\[hgh^{-1}\simeq \begin{cases}h(3t) & \text{if $0\leq t\leq \frac{1}{3}$}\\ H(3(t-\frac{1}{3}),1) & \text{if $\frac{1}{3}\leq t\leq \frac{2}{3}$}\\ h^{-1}(3t-2) & \text{if $\frac{2}{3}\leq t\leq 1$}.\end{cases}\]
This observation suggests using the following homotopy $\tilde{H})(t,s):[0,1]^2\rightarrow X$ from $f$ to $hgh^{-1}$:
\[\tilde{H}(t,s) = \begin{cases}h(3t) & \text{if $0\leq t\leq \frac{s}{3}$}\\ H((2s+1)(t-\frac{s}{3}),s) & \text{if $\frac{s}{3}\leq t\leq 1-\frac{s}{3}$}\\ h^{-1}(3t-2) & \text{if $1-\frac{s}{3}\leq t\leq 1$}\end{cases}\]
Then $\tilde{H}(t,0)=f(t)$ and $\tilde{H}(t,1)=hgh^{-1}$, and $\tilde{H}(0,s) =\tilde{H}(1,s) =h(0) =x_0$; thus $f$ and $g$ come from the same conjugacy class of $\pi_1 (X,x_0)$, and hence $\Phi$ is injective.$\hspace{.25in}\blacksquare$
 
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