Prove that $h$ and $p*q$ are homotopic relative to {$0,1$}

Question

Let $0<s<1$. Given paths $p$ and $q$ with $p(1)=q(0)$, define $h$ by the formula

$$h(t) = \begin{cases} p(t/s),& \text{if} \quad 0 \leq t \leq s \\ q((t-s)/(1-s)), &\text{if} \quad s\le t\le 1 \end{cases}$$

Prove that $h$ and $p*q$ are homotopic relative to $\{0,1\}$. ($*$ denotes path multiplication)

I'm really having a hard time with this one. My professor said that you can get the formula for the homotopy between $p$ and $p*h$ pretty much straight from the formula for $h$, and it's just a matter of re-parameterizing, but I honestly don't see it. Can someone please explain? Thanks

Answer 1

Let's equip the notation of $h$ with a subscript $s$, so $$h_s(t) = \begin{cases} p(t/s),& \text{if } 0 \leq t \leq s \\ q((t-s)/(1-s)), &\text{if } s\le t\le 1 \end{cases}$$ Then $h_\frac12$ is just $p*q$, and you want a homotopy $h_s\simeq h_\frac12$. The natural choice would be to let $H:I\times I\to X$ consist of $H(-,u)=h_{\frac u2+(1-u)s}(-)$. Can you show that $H$ is continuous?