Consider a homotopy $h: I \times X \rightarrow Y$ and denote
$$\begin{array}{rcl}
h_1: X & \rightarrow & Y\\
x & \mapsto & h(1,x).
\end{array}$$
Define the constant homotopy
$$\begin{array}{rcl}
k:I \times X & \rightarrow & Y\\
(t,x) & \mapsto & h_1(x)=h(1,x)
\end{array}$$
Then $h$ and $k$ are homotopic. Indeed the continuous map
$$\begin{array}{rcl}
I \times I \times X & \rightarrow & Y\\
(s,t,x) & \mapsto & h((1-s)t + s,x)
\end{array}$$
restricts to
$$\begin{array}{rcl}
\{0\} \times I \times X & \rightarrow & Y\\
(0,t,x) & \mapsto & h(t,x)
\end{array} \;\;\;\text{and}\;\;\;\begin{array}{rcl}
\{1\} \times I \times X & \rightarrow & Y\\
(1,t,x) & \mapsto & h(1,x).
\end{array}$$
Is this argument correct?
I feel unsure because of the following. I wanted to reconstruct the homotopy 2-category of topological spaces. Usually it is a lengthy but straightforward calculation (citing Borceux – Handbook of Categorical Algebra I – p. 286) to check that vertical and horizontal composition of homotopies is defined up to homotopy. But using my argument above it seems like there is nothing left to show, since as long as two homotopies restrict to the same target, they are already homotopic.
Another thing confusing me is that I naively expected the homotopy 2-category to have many different 2-cells (despite being equivalence classes of homotopies). But again, my argument above seems to imply that given two morphisms $f_0,f_1:X\rightarrow Y$ there is at most one 2-cell $f_0 \Rightarrow f_1$.
On the other hand it is a standard fact that any two paths with say common target are freely homotopic and it seems to me like this is what could be happening here as well. But e.g. Borceux doesn't mention the homotopies being relative something, so I don't see that being a contradiction.
As always: thank you for your time!
Best Answer
A 2-cell in the homotopy 2-category is a homotopy rel endpoints of homotopies. That is, the map $H:I\times I\times X\to Y$ is required to satisfy not just $H(0,t,x)=h(t,x)$ and $H(1,t,x)=k(t,x)$, but also $H(s,0,x)=h_0(x)$ and $H(s,1,x)=h_1(x)$, independent of $s$. The point here is that $h$ and $k$ are supposed to be homotopies between the same two maps $h_0=k_0$ and $h_1=k_1$, and $H$ is a homotopy of homotopies between those same two maps, so every stage $H(s,-,-)$ should be such a homotopy.