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I am hung up on part of the proof of the following theorem:

Theorem: Let $X$ be a path connected space and $x_0$, $x_1$ points of $X$. Then the fundamental groups $\pi_1(X,x_0)$ and $\pi_1(X,x_1)$ are isomorphic.

The first part of the proof goes like this:

Let $\gamma: I \to X$ be a path from $\gamma(0)=x_0$ to $\gamma(1)=x_1$. Then for $[\alpha]$ in $\pi_1(X,x_0)$, $(\bar{\gamma} \ast \alpha) \ast \gamma$ is a loop based at $x_1$. Thus we define a function $f:\pi_1(X,x_0) \to \pi_1(X,x_1)$ by $f([\alpha])=[(\bar{\gamma} \ast \alpha) \ast \gamma]$, $[\alpha] \in \pi_1(X,x_0)$.

For some reason I'm just not seeing why $(\bar{\gamma} \ast \alpha) \ast \gamma$ is a loop based at $x_1$. The way we take the product of paths, it seems that we would have $\bar{\gamma}(1)=\alpha(0)$ and $\alpha(1)=\gamma(0)$. I understand that $\bar{\gamma}(t)=\gamma(1-t)$ is a path from $x_1$ to $x_0$, then $\alpha$ is a loop at $x_0$, but I don't see how $\gamma$ transports the loop back to $x_1$.

How am I looking at this incorrectly? Thanks.

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It may depend on definitionh: from what book/paper/link are you taking the above? This is usually a rather easy proof... –  DonAntonio Feb 6 '13 at 5:12
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Possibly getting hung up on word choice, but maybe this is your confusion: it doesn't "transport" the loop back to $x_1$, it just concatenates itself the what you've already got. With $\bar\gamma$, you're going from $x_1$ to $x_0$, then $\alpha$ takes you around your space and back to $x_0$, and then $\gamma$ takes you back to $x_1$. –  gmoss Feb 6 '13 at 7:36
    
@DonAntonio: It's from Principles of Topology by Croom. But I'm fine with the proof if I assume the above to be true, it's just that detail that got me. My question has been answered though. –  Alex Petzke Feb 6 '13 at 17:53
    
@gmoss: Thanks! That is indeed what was confusing me. I see now that the product of those paths is in fact a loop, i.e. a function $\alpha: [0,1] \to X$ with $\alpha(0)=\alpha(1)=x_1$, it just doesn't look like a loop as we (or I) usually picture them. Since you hit on my source of confusion so well, I'll accept that as an answer if you'd like to make it one. –  Alex Petzke Feb 6 '13 at 17:56
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There is a nice picture/write up at mathprelims.wordpress.com/2009/03/18/… –  Juan S Feb 6 '13 at 21:47
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up vote 2 down vote accepted

Possibly getting hung up on word choice, but maybe this is your confusion: it doesn't "transport" the loop back to $x_1$, it just concatenates itself the what you've already got. With $\bar\gamma$, you're going from $x_1$ to $x_0$, then $\alpha$ takes you around your space and back to $x_0$, and then $\gamma$ takes you back to $x_1$.

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For $\alpha : [0, 1] \to X$ with $\alpha(0) = x_0$ and $\alpha(1) = x_1$, and $\beta : [0, 1] \to X$ with $\beta(0) = x_1$ and $\beta(1) = x_2$, we have

$$(\alpha\ast\beta)(t) = \begin{cases} \alpha(2t) & t \in \left[0, \frac{1}{2}\right]\\ \beta(2t-1) & t \in \left(\frac{1}{2}, 1\right] \end{cases}.$$

So in your situation we have

$$(\bar{\gamma}\ast\alpha)(t) = \begin{cases} \bar{\gamma}(2t) & t \in \left[0, \frac{1}{2}\right]\\ \alpha(2t-1) & t \in \left(\frac{1}{2}, 1\right] \end{cases}$$

so

\begin{align} ((\bar{\gamma}\ast\alpha)\ast\gamma)(t) &= \begin{cases} (\bar{\gamma}\ast\alpha)(2t) & t \in \left[0, \frac{1}{2}\right]\\ \gamma(2t-1) & t \in \left(\frac{1}{2}, 1\right] \end{cases}\\ &= \begin{cases} \bar{\gamma}(4t) & 2t \in \left[0, \frac{1}{2}\right]\\ \alpha(4t-1) & 2t \in \left(\frac{1}{2}, 1\right]\\ \\ \gamma(2t-1) & t \in \left(\frac{1}{2}, 1\right] \end{cases}\\ &= \begin{cases} \bar{\gamma}(4t) & t \in \left[0, \frac{1}{4}\right]\\ \alpha(4t-1) & t \in \left(\frac{1}{4}, \frac{1}{2}\right]\\ \\ \gamma(2t-1) & t \in \left(\frac{1}{2}, 1\right] \end{cases}. \end{align}

Now note that \begin{align} ((\bar{\gamma}\ast\alpha)\ast\gamma)(0) &= \bar{\gamma}(4\times 0)\ \text{as $t \in \left[0, \frac{1}{4}\right]$}\\ &=\bar{\gamma}(0)\\ &= \gamma(1-0)\\ &= \gamma(1)\\ &= x_1 \end{align} and \begin{align} ((\bar{\gamma}\ast\alpha)\ast\gamma)(1) &= \gamma(2\times 1 - 1)\ \text{as $t \in \left(\frac{1}{2}, 1\right]$}\\ &= \gamma(2-1)\\ &= \gamma(1)\\ &= x_1. \end{align}

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