I am studying the concept of homotopy equivalent maps of complexes, and I was asking myself whether what I consider very natural properties were satisfied. Notably: let $F, G, H$ be complexes.
Suppose that $\alpha, \beta : F \longrightarrow G$ and $\phi, \psi : G \longrightarrow H$ are maps of complexes such that $\alpha$ is homotopy equivalent to $\beta$ and $\phi$ is homotopy equivalent to $\psi$. Is it true that $\phi \circ \alpha$ is homotopy equivalent to $\psi \circ \beta$?
By definition, we say that $F$ and $G$ are homotopy equivalent if there exist two maps of complexes $\alpha : F \longrightarrow G$ and $\phi : G \longrightarrow F$ such that $\phi \circ \alpha$ is homotopy equivalent to $id_F$ and $\alpha \circ \phi$ is homotopy equivalent to $id_G$. Let's define $\alpha$ to be a pseudo-inverse of $\phi$ if this property is satisfied (so I don't have to write everything again). Suppose that $\phi$ has two pseudo-inverses $\alpha$ and $\beta$. Is it true that they are homotopy equivalent?
In 1, yes, indeed $\phi\alpha$ is homotopy equivalent to $\phi\beta$ which is homotopy equivalent to $\psi\beta$. If $\theta$ is a homotopy between $\alpha $ and $\beta$, then $\phi\theta$ is a homotopy between $\phi\alpha $ and $\phi\beta$, etc.
In 2, consider $\alpha\phi\beta$. This is homotopy equivalent both to $\text{id}\,\beta$ and $\alpha\,\text{id}$, by 1.