Homotopy between two paths implies triviality of the loop they form

Question

If $X$ admits a universal covering space and $\alpha$ and $\gamma$ are to homotopic paths between $x$ and $p(y)$, then $\alpha*\gamma^{-1}$ is nullhomotopic?

Answer 1

This is a basic fact about homotopy that doesn't require covering spaces.

Let $X$ be any space and $\alpha,\gamma:[0,1]\to X$ two equivalent paths with endpoints $x_0=\alpha(0)=\gamma(0)$ and $x_1=\alpha(1)=\gamma(1)$. Since $\alpha*\alpha^{-1}$ is equivalent to $\alpha*\gamma^{-1}$ (can you find a homotopy between them, given a homotopy $F:[0,1]^2\to X$ between $\alpha$ and $\gamma$?) to see that $\alpha*\gamma^{-1}$ is nullhomotopic it is enough to show that $\alpha*\alpha^{-1}$ is. The map $H:[0,1]^2\to X$

$$ H(s,t):=\begin{cases} x_0 & s\in\left[0,\frac{t}{2}\right]\\ \alpha(2s-t) & s\in\left[\frac t 2,\frac 1 2\right]\\ \alpha^{-1}(2s+t-1) & s\in\left[\frac 1 2,\frac{2-t}2\right]\\ x_0 & s\in\left[\frac{2-t}2,1\right] \end{cases} $$

is a homotopy between $\alpha*\alpha^{-1}$ and the constant path based at $x_0$.