Proposition

Let $X$ be a path-connected topological space.

Let $f, g: I \rightarrow X$ be two paths from $p$ to $q$. Show that $f \sim g$ if and only if $f \circ \overline{g} \sim c_p$.
Show that $X$ is simply connected if and only if any two paths in $X$ with the same initial and terminal points are path-homotopic.

Solution

1

$\begin{align*} f \sim g &\iff [f] = [g] \\ &\iff [f] \cdot [\overline{g}] = [g] \cdot [\overline{g}] \\ &\iff [f \cdot \overline{g}] = [g \cdot \overline{g}] \\ &\iff [f \cdot \overline{g}] = c_p \\ &\iff f \cdot \overline{g} \sim c_p. \end{align*}$

2

Suppose that $X$ is simply connected. Let $f, g$ be two paths in $X$ with the same initial and terminal points $p, q$. Then $f \cdot \overline{g}$ is a loop based at $p$. Since $X$ is simply connected, $f \cdot \overline{g} \sim c_p$. From 1, we know that this implies $f \sim g$.

On the other hand, suppose that any two paths in $X$ with the same initial and terminal points are path-homotopic. Let $[f] \in \pi_1(X, p)$. Then $f$ is path-homotopic to $c_p$. Thus $[f] = [c_p]$, so $\pi_1(X, p)$ is the trivial group.