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Is sheaf cohomology an invariant of the weak homotopy type? More precisely let $R$ be a commutative ring and $f:X\rightarrow Y$ a weak homotopy equivalence. Does it follow, that the induced maps $H^n(Y,\underline R) \rightarrow H^n(X,\underline R)$ are isomorphisms?

Edit: Since any space is weakly homotopy equivalent to a CW-complex, CW-complexes are locally contractible and for locally contractible spaces sheaf and singular cohomology coincide, a positive answer to this question would imply that sheaf cohomology and singular cohomology coincide for any space. This seems unlikely, but I don't know a counter example.

edited Mar 13 2012 at 21:15

asked Mar 13 2012 at 20:52

Jan Weidner
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(I'm not really adding anything to the answer already given) $H^0$ for cech cohomology detects connectedness, while $H^0$ for singular cohomology detects path-connectedness. – Yosemite Sam Mar 14 2012 at 2:15

Thank you, this comment is useful to me! – Jan Weidner Mar 14 2012 at 8:42

Karol Szumiło · Accepted Answer · 2012-03-13 21:24:50Z UTC

No. For paracompact spaces sheaf cohomology coincides with Čech cohomology. In particular it applies to the closed topologist's sine curve $C$. There is a map $C \to S^1$ inducing an isomorphism on Čech cohomology, but $C$ is weakly contractible.

Sheaf cohomology invariant of weak homotopy type?

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