Suppose $X$ is a topological space. I would like to show that given a path $\gamma : [0,1] \to X$ the concatenation $\gamma \bullet -\gamma$ is homotopic to the constant path $e_{\gamma(0)}(s) = \gamma (0)$ where $-\gamma$ is the reverse path: $-\gamma(s) = \gamma(1-s)$. The solution in my notes has a slightly messy expression for the homotopy which I can't build much intuition upon. I think there is a slightly simpler one, the intuition is: traverse less and less of the path by moving at slower speed and then pick up from the point where it stops moving forward with the inverse path. This is the explicit formula for the homotopy:
$$
h(s,t)=
\begin{cases}
\gamma\big(2st\big) & 0\leq s \leq 1/2\\
-\gamma\big(1-2t(1-s)\big) & 1/2 \leq s \leq 1.
\end{cases}
$$
Is this correct? It seems to me like there is nothing wrong with it, since it gives the constant path at $t=0$ and the concatenation at $t=1$. But it seems strange that the notes used a more complicated solution.
Best Answer
This community wiki solution is intended to clear the question from the unanswered queue.
Your approach is correct. However, there are many other homotopies which may be more "complicated".