If we start with the **category of finite complexes** and continuous maps, and then identify two morphisms iff they are homotopic, we get the **homotopy category of finite complexes**, and it is trivial to observe that a continuous map $f:X\to Y$ is an isomorphism in this category iff it is a homotopy equivalence. Does there exist an equivalence relation on continuous maps (necessarily finer than that of being homotopic) compatible with composition such that the resulting quotient category has the property that a map $f:X\to Y$ is an isomorphism iff it is a *simple* homotopy equivalence?


I would also be interested in an answer to the following variant, which may be slightly easier. Fix a group $\pi$, and consider the category of bounded complexes of finite-dimensional free $\mathbb Z[\pi]$-modules. Again, we may identify morphisms iff they are chain homotopic, and the resulting homotopy category has the property that a chain map $f:A\_\bullet\to B\_\bullet$ is an isomorphism iff it is a chain homotopy equivalence. Is there a finer equivalence relation on morphisms such that in the resulting quotient category, a chain map is an isomorphism iff it is a chain homotopy equivalence and has vanishing Whitehead torsion?



Is there a category whose isomorphisms are precisely the simple homotopy equivalences?
Unfortunately not. Let's say $D$ is a category and $F$ is a functor from finite complexes to $D$ that takes simple homotopy equivalences to isomorphisms.


We note that for any finite complex $X$, the inclusion $i\_0: X \to [0,1] \times X$ is a simple homotopy equivalence. We can factor it as a composition of elementary expansions, one cell $e^n$ at a time, by inductively gluing in $[0,1] \times e^n \cong D^{n+1}$ along $[0,1] \times \partial e^n \cup \{0\} \times e^n \cong D^{n}\_{-}$.


Therefore, $F(i\_0)$ is an isomorphism. This fact, all by itself, forces $F$ to factor through the homotopy category. This is a fairly common manipulation; here is how it goes.


If we consider the other inclusion $i\_1: X \to [0,1] \times X$ and the projection $p: [0,1] \times X \to X$, we have
$$id = F(p) F(i\_0) = F(p) F(i\_1)$$
which implies first that $F(p) = F(i\_0)^{-1}$ and then that $F(i\_1) = F(p)^{-1}$, which forces $F(i\_0) = F(i\_1)$.


If $H: [0,1] \times X \to Y$ is a homotopy between two maps $f$ and $g$, then
$$
F(f) = F(H) F(i\_0) = F(H) F(i\_1) = F(g)
$$
so $F$ factors through the homotopy category, as desired.


