If two topological spaces are weak homotopy equivalent to each other, are their Cech cohomology groups the same?
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2$\begingroup$ The obvious and well known (I am sure :-) answer is NO (regardless of is or are :-). $\endgroup$– Włodzimierz HolsztyńskiFeb 21, 2015 at 12:02
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2 Answers
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$$T = \left\{ \left( x, \sin \frac{1}{x} \right ) : x \in (0,1] \right\} \cup \{(0,y)\mid y\in[-1,1]\}$$
This has trivial homotopy groups in degrees $\ge1$ but according to Wikipedia nontrivial Čech cohomology in degree 1.
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To give a more enlightening answer to the question:
Cech cohomology is not the same as singular cohomology. However it is on CW-complexes. But there is CW approximation for topological spaces and singular cohomology is a weak homotopy invariant, so Cech cohomology can't be.
