If $\alpha,\alpha'$ path homotopic and $\beta,\beta'$ path homotopic then $\alpha*\beta$ and $\alpha'*\beta'$ are path homotopic I want to prove the statement 

If $\alpha,\alpha'$ path homotopic and $\beta,\beta'$ path homotopic
  then $\alpha*\beta$ and $\alpha'*\beta'$ are path homotopic

$\alpha,\alpha'\beta,\beta':[0,1]\rightarrow X$ are loops based at $x_0\in X$. While I can setup separate homotopies I just don't know how to come up with a homotopy between $\alpha*\beta$ and $\alpha'*\beta'$. Any help is appreciated.
Here $\alpha*\beta(t)=\alpha(2t)$ when $0\leq t \leq 1/2$ and $\alpha*\beta(t)=\beta (2t-1)$ when $1/2\leq t \leq 1$
 A: Let $F \colon [0, 1] \times [0, 1] \to X, G \colon [0, 1] \times [0, 1] \to X$ be the homotopies between $\alpha, \alpha'$ and $\beta, \beta'$ respectively. To form the composite loops $\alpha \ast \beta$ and $\alpha' \ast \beta'$, you cut the unit interval into two pieces, and 'speed up' each path by a factor of two, joining them at their common basepoint $x_{0}$ at $s = 1/2$. We want to do a similar procedure with the homotopies: define $H(s, t) \colon [0, 1] \times [0, 1] \to X$ by 
$$H(s, t) = \begin{cases} F(2s, t) & \text{ if } s \in [0, 1/2] \\ G(2s-1, t) & \text{ if } s \in [1/2, 1] \end{cases}$$
It is a good exercise to check the details, so I leave it to you to verify that:
$(1)$ $H(s, t)$ is continuous (note that special care is needed at $s = 1/2$, but otherwise this should be straightforward). 
$(2)$ $H(0, t) = H(1, t) = x_{0}$
$(3)$ $H(s, 0) = (\alpha \ast \beta)(s), H(s, 1) = (\alpha' \ast \beta')(s)$ for all $s \in [0, 1]$. 
A: Why would a function like this not work as well?
$F(s, t)=
\begin{cases}
f_0\left(\frac{2s}{1-t}\right), & 0\leq s\leq \frac{1-t}{2}\\
g_0\left(\frac{2s-(1-t)}{1-t}\right), & \frac{1-t}{2}\leq s\leq 1-t\\
f_1\left(\frac{2s + 2t-2}{t}\right), & 1-t\leq s\leq 1-\frac{t}{2}\\
g_1\left(\frac{2s+t-2}{t}\right), & 1-\frac{t}{2}\leq s\leq 1
\end{cases}$
where $F(s,0) = [f_0\cdot g_0](s)$ and $F(s,1) = [f_1\cdot g_1](s)$, besides $F(0,t) = F(1,t) = x_0$.
