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?
A question about homology group
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algebraic-topology
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1A more difficult version of this was asked on MO here http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent (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. – 2012-07-27
1 Answers
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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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0A 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. – 2012-07-27
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0Dont Lens spaces have finite cyclic fundamental groups? – 2012-07-27
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1@mland, I should have been more clear: they are example of what the OP wants, not of the approach I suggested. – 2012-07-27
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0So I figured. But thanks for the clarification. Lens spaces are great :) – 2012-07-27
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0I have not found any simple example like you suggested, but I've found an acyclic space which is not contractible. – 2012-07-29