It is clear to me that if all paths (with the same endpoints) in a region are homotopic then that region is simply connected, however I am having difficultly proving the converse, that is, all paths with the same endpoints are homotopic in a simply connected region. Here is what I have so far
Given two paths $\alpha$ and $\beta$ which begin and end at the same point, we can define a loop at either of those points by
\begin{align}
\gamma(t) =
\begin{cases}
\alpha(2t) &\quad 0 \leq t \leq 1/2\\
\beta(1-2t) &\quad 1/2 \leq t \leq 1\\
\end{cases}
\end{align}
By assumption, the space is simply connected, and so all loops can be deformed into all other loops. The loop $\gamma$ can then be deformed to
\begin{align}
\gamma'(t) =
\begin{cases}
\alpha(2t) &\quad 0 \leq t \leq 1/2\\
\alpha(1-2t) &\quad 1/2 \leq t \leq 1\\
\end{cases}
\end{align}
Intuitively, it seems like there should then be a homotopy between $\alpha$ and $\beta$, however I cannot think of a rigorous way to show that one exists.
Any help would be appreciated, thanks.
Best Answer
There is also a purely algebraic solution. If the paths $\alpha$ and $\beta$ both start at $x$ and end at $y$, then
$[\alpha] = [\alpha * e_y] = [\alpha * (\bar{\beta} * \beta)]=[(\alpha * \bar{\beta}) * \beta] = [e_{x} * \beta] = [\beta]$,
where $e_x$ and $e_y$ denote the respective constant paths.