I am working on a problem on path homotopy out of Mukres' topology (2ed., chapter 9 on Fundamental Group, section 51) that goes like this:
Given that $h, h': X \to Y$ and $k, k': Y \to Z$ are homotopic, prove that $k \circ h$ and $k' \circ h'$ are homotopic.
Here are what I knew and what I would like to know:
(1) If $h \simeq h'$, my folksy intuition says that $h$ can be "morphed" to $h'$ starting from $t =0$ and ending at $t = 1$. Thus a genteel will say that there exists mapping $H$,
$$\begin{align} H : X \times [0. 1] &\to Y \text{, such that }\\ H(x, 0) &= h(x)\\ H(x, 1) &= h'(x). \end{align}$$
Similarly the same will say that for $k \simeq k'$, there exists mapping $K$,
$$\begin{align} K : Y \times [0. 1] &\to Z \text{, such that }\\ K(y, 0) &= k(y)\\ K(y, 1) &= k'(y). \end{align}$$
(2) As for path composition $h \circ k$, again my homespun intuition says that one-half of time I "cruise" in $h$ and one-half time in $k$, or in a grandee's lingo:
$$h \circ k (t) = \begin{cases} h(2t), &t \in [0, \frac{1}{2}]\\ k(2t-1), &t \in [\frac{1}{2}, 1]. \end{cases}$$
(3) But for the other way around $k \circ h$ and $k' \circ h'$, are these correct?
$$ k \circ h (t) = \begin{cases} k^{-1}(2t), &t \in [0, \frac{1}{2}]\\ h^{-1}(2t-1), &t \in [\frac{1}{2}, 1], \end{cases} \qquad \text{ (let } f(t) := k \circ h(t))$$
$$ k' \circ h' (t) = \begin{cases} k'(2t), &t \in [0, \frac{1}{2}]\\ h'(2t-1), &t \in [\frac{1}{2}, 1]. \end{cases} \qquad \text{ (let } f'(t) := k' \circ h'(t))$$
(4) In order to prove $f \simeq f'$, am I to show that there exists a map $F$, such that
$$F : Z \times [0, 1] \to X\\ \begin{align} F(z, 0) &= f(z)\\ F(z, 1) &= f'(z)? \end{align}$$
If this step is correct, please let me know how I should go from there. Thank you for your time and help.
An extended discussion of where you're going wrong with a hint at the end:
Composition of paths is not the same as composition of functions.
You have a function $h: X \rightarrow Y$. That is, it takes inputs from $X$ and spits out values in $Y$. You have $k : Y \rightarrow Z$ similarly. So you get $k \circ h: X \rightarrow Z$. There's no "time" parameter, it doesn't "spend half its time" going from the input in $X$ to the intermediate output in $Y$ and then the rest on to the final output in $Z$.
The homotopies do have a time parameter. We have $H: X \times I \rightarrow Y$ a homotopy with $H(x,0) = h(x)$ and $H(x,1) = h'(x)$. A nice way to think about this is having $h_t(x) = H(x,t)$, so you see the map $h$ vary with time from $h = h_0$ to $h' = h_1$. Similarly for $K: Y \times I \rightarrow Z$ and $k_t$.
So your intuition is getting mixed up between the maps and the homotopies, but the intuition you have can work if you direct it toward the homotopies instead of the maps.
So you want to transform $k_0 \circ h_0$ to $k_1 \circ h_1$. That is, you want a map $L : X \times I \rightarrow Z$ such that $L(-,0) = k_0 \circ h_0$ and $L(-,1) = k_1 \circ h_1$. Can you come up with such a map? Can you prove it's continuous?