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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?

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    $\begingroup$ It may be worth knowing that if you take the nerve of the simple homotopy equivalences (ie, abandoning interest non-equivalences, retreating from categories to groupoids), you get an interesting space, which is different from the nerve of the homotopy equivalences, and not just in its set of components. $\endgroup$
    – Ben Wieland
    Commented Nov 16, 2014 at 17:43
  • $\begingroup$ Relevant: math.harvard.edu/~lurie/281.html. Based on skimming through those I think there is indeed a category whose isomorphisms are precisely the simple homotopy equivalences, but you need to start with something more restrictive than continuous maps at the outset or something... $\endgroup$
    – Qiaochu Yuan
    Commented Nov 27, 2014 at 4:24
  • $\begingroup$ @Qiaochu Yuan: Could you be more specific? (I did look through some of the notes just now). Tyler Lawson's answer seems pretty robust to me. What sort of restrictions on maps did you have in mind? (I guess we do get a positive answer if we only allow maps which are simple homotopy equivalences, but that's rather useless). $\endgroup$
    – John Pardon
    Commented Nov 27, 2014 at 18:46
  • $\begingroup$ @John: in Lecture 7 Lurie constructs what appears to be an $\infty$-category, or maybe $\infty$-groupoid, of simple homotopy types. I think he avoids the argument in Tyler's answer by only restricting attention to invertible morphisms, or something like that. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 27, 2014 at 21:27

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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.

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    $\begingroup$ This seems to imply that the localization of finite complexes at the simple equivalences is the homotopy category (hence making my answer rubbish). $\endgroup$
    – Denis Nardin
    Commented Nov 16, 2014 at 4:35
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    $\begingroup$ @DenisNardin I haven't yet been able figure out where the incompatibility between our statements is. $\endgroup$
    – Tyler Lawson
    Commented Nov 16, 2014 at 4:36
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    $\begingroup$ @Denis: in your answer you claimed that simple homotopy equivalences are compositions of simple expansions and simple contractions, but the definition I'm familiar with is that a simple homotopy equivalence is a map which is homotopic to such a composition. Your argument doesn't work with this second definition. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 16, 2014 at 4:41
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    $\begingroup$ Yeah, I've checked it and you're right. I still think that part of the proof can be salvaged, but it's sort of a moot point since all I would prove is that the homotopy category is given by quotienting by the homotopy equivalence relation... $\endgroup$
    – Denis Nardin
    Commented Nov 16, 2014 at 4:44
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    $\begingroup$ Well, but I think that what Denis had in mind was something more like localizing the category of finite complexes with respect to the elementary expansions and collapses, and so I'd still like to see more explicitly how more maps become isomorphisms than composites of these. $\endgroup$
    – Tyler Lawson
    Commented Nov 16, 2014 at 4:46

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