I am trying to find a function $f:X \to Y$, where $X$ and $Y$ are path connected, such that $f$ induces isomorphisms on all homology groups, but $X$ and $Y$ have non-isomorphic cohomology rings. It may be that this isn't possible - and that would be very helpful to know- but I think I have an example.

I've thought of using $X = \mathbb{R}P^2 \vee S^3$ and $Y = \mathbb{R}P^3$, since these are classic examples of spaces with isomorphic homology groups and non-isomorphic cohomology rings. Then $f: X \to Y$ would be the inclusion map on $\mathbb{R}P^2$ (thinking of it as a CW-subcomplex of $\mathbb{R}P^3$) and the covering map on $S^3$.

It seems clear to me that this map will induce isomorphisms on homology groups for $i=0, 1, 2$, since for $X$ those homology groups depend only on $\mathbb{R}P^2$ and the inclusion of a subcomplex plays nicely with homology. However, I'm stuck on showing that $f_*:H_3(X) \to H_3(Y)$ is an isomorphism. I think it should be one, but I'm not sure how exactly to justify it.


Here are two reasons that you can’t find an example of such a function (when the coefficient group is the integers):

By the five lemma and the universal coefficient theorem, a map inducing isomorphism on homology groups induces isomorphism on cohomology groups. This implies that the cup product is the same since induced maps preserve cup products.

If you knew your space was simply connected, you could also proceed by the homology version of the Whitehead theorem. This implies your map is a homotopy equivalence which implies the cup products are the same.

The difference in your question and the one you list is that you give an induced isomorphism while the other only presumes an abstract isomorphism.

The reason your example does not work is because the covering map does not actually induce an isomorphism between the top homology of the sphere and projective space, it is actually degree 2.

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  • $\begingroup$ great explanation, @Connor Malin. $\endgroup$ – Allan Ramos Aug 14 '19 at 19:45

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