Question on part of the proof that $\pi_1(X,x_0)$ is isomorphic to $\pi_1(X,x_1)$

Question

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.

Answer 1

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 $x_1$.

Answer 2

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}