Let $h,h':X\to Y$ be homotopic and $k,k':Y\to Z$ be homotopic. The idea is to show that $k\circ h$ and $k'\circ h'$ are homotopic.
Let $F$ be a homotopy between $h(x)$ and $h'(x)$ and $G$ a homotopy between $k(x)$ and $k'(x)$. Define $H:X\times[0,1]\to Z$ by $G(F(x,t),t)$. Then $H$ is continuous as a composition of continuous maps and for $x\in X$ we have
\begin{align}
H(x,0)&=G(F(x,0),0) = G(h(x),0) = k\circ h(x)\\ H(x,1) &= G(F(x,1),1) = G(h(x),1) = k'\circ h'(x),
\end{align}
so $H$ is a homotopy between $k\circ h$ and $k'\circ h'$.
My question is this: We cannot exactly compose $F$ and $G$, since both are functions of two inputs, and one output, but instead we compose $G\circ (F, t\mapsto t)$. It turns out that this is correct, but how did we know a priori to choose $t\mapsto t$? Why not, e.g. $t\mapsto t^2$? This turns out to be a homotopy as well, since
\begin{align}
G(F(x,0),0^2) = G(F(x,0),0) = k\circ h(x)\\
G(F(x,1),1^2) = G(F(x,1),1) = k'\circ h'(x).
\end{align}
But clearly $t\mapsto 1-t$ would not work. Is it the case that we can choose any continuous $\alpha : [0,1]\to[0,1]$ with $\alpha(0)=0$ and $\alpha(1)=1$ and $G\circ(F,\alpha)$ would be a homotopy?
Best Answer
Given a homotopy $H : A \times I \to B$ and $t \in I = [0,1]$ define $H_t : A \to B, H_t(a) = H(a,t)$.
Now let us denote your homotopy by $G * F$. This is certainly the most obvious solution because $(G * F)_t = G_t \circ F_t$, i.e. the homotopy $G * F$ is obtained by composing $G$ and $F$ levelwise.
Now let $\alpha, \beta : I \to I$ be continuous maps such that $\alpha(0) = \beta(0) = 0$ and $\alpha(1) = \beta(1) = 1$. Then $F^\alpha = F \circ (id_X \times \alpha)$ and $G^\beta = G \circ (id_Y \times \beta)$ are also homotopies between $h,h'$ and $k,k'$, respectively. Therefore also $G^\beta * F^\alpha$ is a homotopy as desired. But it definitely has no benefit to invoke $\alpha, \beta$ as additional ingredients.