Is it true that if two spaces X,Y have the same CW chain complex i.e. ${C^{CW}(X)}_n={C^{CW}(Y)}_n$ and also the same differential maps $d_n$ at each stage, then $X \simeq Y$ ?
If the CW chain complexes match, it readily follows that the homology groups are the same. The CW chain complexes are related with the number of each cell we attach and the differentials are also obtained using the attaching maps. So is it true?
Let $X= S^2 \vee S^1 \vee S^1$. Then $X$ and $T^2$ have the same chain groups (if you give the torus the skeleton with 1 0-cell, 2 1-cells, and 1 2-cell). Clearly the boundary maps from the first chain group to the zeroth chain group are the same. The boundary map from the second chain group to the first is trivial on $X$, obviously, and trivial on $T_2$ because we attached the 2-cell with the commutator which ends up being trivial when we collapse one of the circles since one of the generators is disappearing.
This is a classic example where homology and cohomology are not enough to distinguish to spaces, but with the cup product you can. I should say, $X$ and $T_2$ are not homotopy equivalent because they have different $\pi_2$.