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What goes wrong in the axiomatic definition of a generalized (co)homology theory if one drops the axiom of homotopy invariance i.e. that homotopic maps should induce the same map in (co)homology?

Or do we have examples? Are there "interesting" or "useful" functors $\mathfrak{h}^{\cdot}:\mathrm{Spaces}\to \mathrm{Ab}$ that respect all Eilenberg-Steenrod axioms except homotopy invariance? Could such an $\mathfrak{h}$ be used to distinguish between two homotopy equivalent non-homeomorphic spaces?

(Take your favourite definition of admissible spaces)

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    $\begingroup$ I don't think I've ever checked how it relates to the Eilenberg-Steenrod axioms, but one natural cohomology theory (in the slightly more casual sense) that is not homotopy invariant is cohomology with compact support. This lack of invariance has nice applications to proofs of natural facts, for example, that the tangent bundle of the sphere is not homeomorphic to the trivial bundle. $\endgroup$
    – Minhyong Kim
    Commented Dec 29, 2011 at 21:57
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    $\begingroup$ Cohomology with compact support is not functorial either... unless you consider proper maps, but then it's proper homotopy invariant! $\endgroup$
    – Fernando Muro
    Commented Dec 29, 2011 at 23:01
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    $\begingroup$ Well, I guess it's covariant functorial for manifolds. Compact support cohomology should be thought of as closer to homology than cohomology, I think. This reminds me: intersection (co)homology is not a homotopy invariant either. $\endgroup$
    – Minhyong Kim
    Commented Dec 30, 2011 at 1:02
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    $\begingroup$ @Minhyong For oriented manifolds the covariant version is just homology by Poicaré duality, which is functorial and also homotopy invariant, so it does not produce a counterexample. And the contravariant version is not functorial, just think of $\mathbb{R}\rightarrow S^1$. $\endgroup$
    – Fernando Muro
    Commented Dec 30, 2011 at 14:51
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    $\begingroup$ Fernando: I'm sure you know all this, but perhaps I should be a bit more careful so as not to mislead anyone. The covariant functoriality I was referring to does indeed use duality, and hence, changes degree. Thus, I suppose any given $H^n_c$ is not even covariant functorial. As you mention, the preservation of degrees only happens when one identifies with homology. On the other hand, for these reasons, any fixed $H^n_c$ is not a homotopy invariant, in spite of duality. Anyways, all these subtleties was the reason for my original qualification that $H^n_c$ is only a 'casual' cohomology. $\endgroup$
    – Minhyong Kim
    Commented Dec 31, 2011 at 1:14

3 Answers 3

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ordinary differential cohomology roughly speaking satisfies the demands of the question

the homotopy axiom fails and is replaced by a variation statement

it is a functor defined for smooth manifolds and smooth maps

the suspension isomorphism is only true in a weaker version

and there are [single space] axioms

Simons& Sullivan first issue of topology

finally it is quite geometric and useful in differential geometry & quantum field theory

it extends to the generalized cohomology context essentially by dropping the homotopy axioms thus it pervades a natural class of examples to the spirit of the question this question is not obviously a good one, but it is and I salute the questioner.

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For a manifold $X$, define $H_0(X)$ to be the direct sum of all the tangent spaces to $X$. This extends in the obvious way to a functor on the category of manifolds and smooth maps. For a pair $(X,A)$, define $H_0(X,A)=H_0(X)/H_0(A)$. For $i>0$, set $H_i(X,A)=0$.

This would seem to satisfy all of the Eilenberg Steenrod axioms except homotopy.

(For cohomology, use cotangent spaces.)

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    $\begingroup$ Ok, there exists one, so this seems to answer some of the questions. It looks like an artificial example though; have some (perhaps more complicated) functors like this ever been studied or used? $\endgroup$
    – Qfwfq
    Commented Dec 29, 2011 at 20:58
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    $\begingroup$ But I think this shows that you should not expect useful examples because in the absence of the homotopy axiom, there's no reason to expect a long exact sequence associated to any map that's not an inclusion. $\endgroup$
    – Steven Landsburg
    Commented Dec 29, 2011 at 21:06
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    $\begingroup$ All examples will be artificial $\endgroup$
    – Fernando Muro
    Commented Dec 29, 2011 at 21:26
  • $\begingroup$ @SL: oh i see, so there's no long esct. seq. without homotopy invariance. @FM: the comment by MK to the question seems to indicate there might be interesting examples.. $\endgroup$
    – Qfwfq
    Commented Dec 29, 2011 at 22:41
  • $\begingroup$ @Qfwfq That depends on what you find interesting. I still think that all examples will be sort of artificial. $\endgroup$
    – Fernando Muro
    Commented Dec 29, 2011 at 23:02
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Fix an abelian coefficient group $B$. Given an inclusion of spaces $A \subset X$, you can let $H^n(X,A) = 0$ for $n \neq 0$, and let $H^0(X,A)$ be the set of all (possibly discontinuous) functions from the underlying set $X^\delta$ of $X$ to $B$ which restrict to zero on $A$. In particular, $H^0(X)$ is the group of all functions $X^\delta \to B$.

(If you like, the map $X \mapsto X^\delta$ is a functor from spaces to spaces which preserves inclusions, "excisive contexts", and takes a point to a (weakly) contractible space, but it does not preserve homotopies. If you have another such functor, you could compose it with cohomology with coefficients in $B$ and get another example.)

This is a little less silly than it sounds. The dual homology functor takes $X$ to the set of formal sums $\sum b_x [x]$ of finite sums of elements of $X$ with coefficients in $B$, and similarly for the relative version. This, as stated, just produces an abelian group. However, there is a natural topology that can be imposed, and (for CW-complexes) the resulting topological abelian group has homotopy groups naturally isomorphic to the singular homology groups of $X$ with coefficients in $B$.

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