Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
(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 '12 at 2:15
    
Thank you, this comment is useful to me! –  Jan Weidner Mar 14 '12 at 8:42

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.