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For a pair of spaces $X,Y$ we have $H_*(X)=H_*(Y)$. Can we necessarily find a continuous function $f$ from $X$ to $Y$ or from $Y$ to $X$, such that $f_*$ induces the isomorphism of homology group?

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A more difficult version of this was asked on MO here mathoverflow.net/questions/53399/… (more difficult because there Dylan wanted spaces with the same homotopy and homology groups) The lens spaces I mentioned below also work for this variant. –  Mariano Suárez-Alvarez Jul 27 '12 at 5:17

1 Answer 1

No, we cannot. ${}{}{}{}{}{}{}{}$

Suppose, for example, that $X$ and $Y$ are spaces which have the homotopy type of CW-complexes which have the same homology groups and which are simply connected. If there is a map $f:X\to Y$ which induces an isomorphism in homology, then by the homology version of Whitehead's theorem, $f$ is in fact an homotopy equivalence.

So it is sufficient to exhibit an example of two simply connected finite CW-complexes with the same homology groups which are not homotopy equivalent. Can you do this?

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A beautiful example is the pair of lens spaces L(5,1) and L(5,2). It is somewhat non-trivial to prove this, though. I is done in Greenberg and Harper's book on algebraic topology, where I learned it from. –  Mariano Suárez-Alvarez Jul 27 '12 at 5:14
Dont Lens spaces have finite cyclic fundamental groups? –  mland Jul 27 '12 at 7:59
@mland, I should have been more clear: they are example of what the OP wants, not of the approach I suggested. –  Mariano Suárez-Alvarez Jul 27 '12 at 17:36
So I figured. But thanks for the clarification. Lens spaces are great :) –  mland Jul 27 '12 at 22:08
I have not found any simple example like you suggested, but I've found an acyclic space which is not contractible. –  Hezudao Jul 29 '12 at 22:50

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