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