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I realise that one could define two spaces to be "homology equivalent" if they have the same homology, as people say, i.e., if they have all homology groups isomorphic. But I find this unsatisfactory because this is not how it goes for homotopy. Homotopy equivalence is not defined as the equality of all the homotopy groups, but as the existence of continuous maps to and fro that compose to maps homotopic to identity maps. And we know that this is strictly stronger than the equality of all the homotopy groups together.

Is there a similar, natural notion of spaces being homologous, which then turns out to be at least as strong as all the homology groups being isomorphic?

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    $\begingroup$ Look up the notion of quasi-isomorphism. $\endgroup$
    – Diego Artacho
    2 days ago
  • $\begingroup$ Requiring the existence of a continuous map that induces an isomorphism on all homology groups?` $\endgroup$
    – Thorgott
    2 days ago

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I realise that one could define two spaces to be "homology equivalent" if they have the same homology, as people say, i.e., if they have all homology groups isomorphic.

Yeah, such notion would not be useful, except for the special case when all homology groups are trivial.

Homotopy equivalence is not defined as the equality of all the homotopy groups, but as the existence of continuous maps to and fro that compose to maps homotopic to identity maps.

That's true, but there is a notion of weak homotopy equivalence, which is a continuous map $f:X\to Y$ such that $\pi_n(f):\pi_n(X)\to\pi_n(Y)$ is an isomorphism for any $n$. This is strictly stronger than requiring $\pi_n(X)$ to be isomorphic to $\pi_n(Y)$, we want the isomorphism to arise from a continuous map.

This idea already is quite useful, and in fact Whitehead showed that every weak homotopy equivalence between (homotopy) CW complexes is actually a strong homotopy equivalence. For more information on weak homotopy equivalences see the wiki entry.

So, by analogy we could define a "homology equivalence" as a continuous map $f:X\to Y$ such that $H_n(f):H_n(X)\to H_n(Y)$ is an isomorphism for any $n$. Although I don't think such maps are widely studied, they don't behave as well as weak homotopy equivalences. One thing to note is that by combining Whitehead's theorem with Hurewicz theorem we get that a homology equivalence between simply connected CW complexes must be a strong homotopy equivalence. Although without "simply connected" assumption this is not true for homology, see here.

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    $\begingroup$ Ok, so this'd be the analogue of weak homotopy equivalence for homology. Will it now be totally arbitrary to expect some other, natural notion of homology equivalence which is at least formally stronger than this one? $\endgroup$
    – Vijay Das
    yesterday
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    $\begingroup$ @VijayDas this is maths, so you limited only by logic and imagination. You can define things however you want. The only question is: how would such thing be useful? I don't think anything more can be done with the idea, although I might be wrong. $\endgroup$
    – freakish
    yesterday
  • $\begingroup$ Homology equivalences like this are prevalent throughout geometric and algebraic topology. One keyword to look up would be "group completion" (in the context of homotopy theory). One could of course define a ''weak homology equivalence'' versus a ''homology equivalence'' where the latter is required to have an inverse. Unlike for homotopy, these are genuinely different in the case of CW complexes. $\endgroup$
    – Connor Malin
    10 hours ago

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