Proposition

A space $X$ is simply-connected iff there is a unique homotopy class of paths connecting any two points in $X$.

Solution

Suppose that $X$ is simply-connected.

Let $x, y \in X$ be given. Since $X$ is path-connected, $x, y$ are connected by a path. Let $f, g$ be two paths connecting $x$ and $y$. Then $f \cdot \overline{g}$ is a loop with a basepoint $x$. Since $X$ is simply-connected, $\pi(X, x)$ is the trivial group. Let $e$ denote the constant loop at $x$. Then $f \cdot \overline{g} \simeq e$.

\[\begin{align*} f \cdot \overline{g} \simeq e &\implies (f \cdot \overline{g}) \cdot g \simeq e \cdot g \\ &\implies f \cdot (\overline{g} \cdot g) \simeq g \\ &\implies f \cdot e \simeq g \\ &\implies f \simeq g. \end{align*}\]

Therefore, $f$ and $g$ are path homotopic, so there is a unique homotopy class of paths connecting any two points in $X$.

On the other hand, suppose that there is a unique homotopy class of paths connecting any two points in $X$.

Then $X$ is path-connected. Let $x \in X$. Let $[f] \in \pi(X, x)$. Then $f$ is a path connecting $x$ with $x$. Let $e$ be the constant loop at $x$. Then $f$ and $e$ are both paths connecting $x$ to $x$, so $f \simeq e$. Therefore, $\pi(X, x) = \{ [e] \}$s, so $X$ is simply-connected.