If two topological spaces are weak homotopy equivalent to each other, are their Cech cohomology groups the same?

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
2 Answers
$$T = \left\{ \left( x, \sin \frac{1}{x} \right ) : x \in (0,1] \right\} \cup \{(0,y)\mid y\in[1,1]\}$$
This has trivial homotopy groups in degrees $\ge1$ but according to Wikipedia nontrivial Čech cohomology in degree 1.
To give a more enlightening answer to the question:
Cech cohomology is not the same as singular cohomology. However it is on CWcomplexes. But there is CW approximation for topological spaces and singular cohomology is a weak homotopy invariant, so Cech cohomology can't be.