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The question is pretty much contained in the title:

What are examples of equivalence relations of topological spaces which are neither stronger nor weaker than homotopy equivalence?

Something that comes to mind is cobordism, for instance, but I was wondering about the existence of equivalence relations which are fairly different than homotopy (and that work for all topological spaces), or perhaps somewhat related to homotopy equivalence but very different in motivation.

I guess that this can be reduced to asking for equivalence relations on maps since homotopy equivalence of spaces is defined as equivalence of compositions of maps to identity maps.

I have no particular area of mathematics in mind for the origin or use of such equivalences, only that they work for all topological spaces.

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    $\begingroup$ Having the same cardinality. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 5, 2015 at 21:15
  • $\begingroup$ @QiaochuYuan This is interesting! Can it be reduced to an equivalence of maps as in the question (we would ask for maps such that the two obvious compositions are related ---not necessarily equal--- to the respective identities)? Obviously we cannot ask for the compositions to be equal to the identities, because we then would have a homeomorphism. ("map" in the question means "continuous function"). $\endgroup$
    – Aldo Guzmán Sáenz
    Commented Nov 5, 2015 at 21:27
  • $\begingroup$ Same covering dimension $\endgroup$
    – David Lampert
    Commented Nov 5, 2015 at 23:43
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    $\begingroup$ Expanding on Qiaochu's response: en.wikipedia.org/wiki/… $\endgroup$
    – Todd Trimble
    Commented Nov 6, 2015 at 0:03

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Here is a simple-minded example that is probably not the sort of thing you are looking for since it is not defined in terms of maps. For a topological space $X$ define its "local homology spectrum" to be $$ \{ n\ |\ H_n(X,X-\{x\};{\mathbb Z})\neq 0 \hbox{ for some $x\in X$}\} $$ Then set $X\sim Y$ if $X$ and $Y$ have the same local homology spectrum. This is an equivalence relation under which all manifolds of the same dimension are equivalent, though they need not be homotopy equivalent. In the other direction, euclidean spaces of different dimensions are homotopy equivalent but are not equivalent in this new sense.

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  • $\begingroup$ This is a very interesting construcion! A further question: as you mention, it is not defined in terms of maps but, is it in fact impossible to define it in terms of maps? Nevertheless, this invariant seems quite useful, I'll think about it some more. Thanks! $\endgroup$
    – Aldo Guzmán Sáenz
    Commented Nov 5, 2015 at 22:02

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