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Let $f: Y \rightarrow X$ be a map of topological spaces such that for any $x \in X, f^{-1}(x)$ has trivial cohomology for some cohomology theory (in my case, cohomology with rational coefficients is enough). Is it true that $f$ induces an isomorphism on cohomology for the same theory?

If not, does this become true if we assume $f$ is a map of CW-complexes, or in some similar way nice?

edited to add some context: I'm hoping for a similar result to Homotopy equvalence from contractibility of fiber.; however, I'm hoping to replace weak homotopy equivalence with some cohomology theory, both in the assumption and the conclusion.

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  • $\begingroup$ With no hypotheses on $X, Y$ there are silly counterexamples like $Y$ the circle with the usual topology and $X$ the circle with the indiscrete topology. Even aside from that sort of thing, if $f$ isn't a fibration there's no reason a priori to expect its genuine fibers to have much to do with its homotopy theory. In the positive direction, if $f$ is a fibration and $X$ is simply connected then I think the Serre spectral sequence does it. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 20, 2016 at 23:21
  • $\begingroup$ I don't have much background in algebraic topology, so I may be asking a stupid question here: it doesn't seem as if $f$ being a fibration is used in mathoverflow.net/questions/126425/…, so is it fully necessary? $\endgroup$
    – user44191
    Commented Jul 20, 2016 at 23:42
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    $\begingroup$ Yes, that theorem is quite surprising to me; I have no idea how it's proved. ("No reason a priori to expect X" doesn't mean "X is false"; it means what it says.) There's a similar result for cohomology here (but its hypotheses are pretty strong): en.wikipedia.org/wiki/Vietoris%E2%80%93Begle_mapping_theorem $\endgroup$
    – Qiaochu Yuan
    Commented Jul 20, 2016 at 23:46
  • $\begingroup$ That does seem like it answers the question - the case I'm looking at has X and Y as compact affine real varieties, and so compact metric spaces. $\endgroup$
    – user44191
    Commented Jul 21, 2016 at 0:04
  • $\begingroup$ @QiaochuYuan: If $f$ is a fibration, we don't have to go as far as the Serre spectral sequence, just to Leray-Hirsch. (The class $1$ already restricts to a basis of cohomology of each fiber.) $\endgroup$
    – Allen Knutson
    Commented Jul 26, 2016 at 13:01

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