Are all paths with the same endpoints homotopic in a simply connected region? 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.
 A: Hint: Let $S^1$ be the unit circle and $D$ be the disk $|z|\leq 1$. Then show  that if a map $f:S^1\to X$ is homotopic to a constant map, then there is a map $D\to X$ that agrees with $S^1\subset D$.
Once you have this $D\to X$, you can easily see that any two paths with the same end-points in $D$ are homotopic, because $D$ is convex, so $\gamma_1(x)(1-t)+\gamma_2(x)t$ is a homotopy between $\gamma_1$ and $\gamma_2$. If $\gamma_1$ and $\gamma_2$ have the same endpoints, this map has the same endpoints for all $t$. 
In particular, in your example, the map from $\alpha$ and $\beta$ put together become a loop, and then the two paths are represented by the two ways around the circle to the midpoint.
A: 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. 
