Topology - Dunce Cap Homotopy Equivalent to $S^2$ So I'm trying to find two spaces with isomorphic homology groups but where the spaces aren't homotopy equivalent. From my work so far, taking the Dunce Cap as a triangle with the edges identified as $aa(a^{-1})$ if that makes sense, the homology group would be $\Bbb Z$ for $n=0,2$ and $0$ elsewhere. However how do I prove that it isn't homotopy equivalent to $S^2$? I'm relatively confident it isn't as it's homotopy equivalent to $D^2$?

A contractible space has no higher homology groups, so your example doesn't work. A working example would be $X=\mathbb CP^2$ and $Y=S^2\vee S^4$. They have the same homology, but are not homotopy equivalent. One way of seeing this, is looking at the cohomology.