Does homology detect chain homotopy equivalence? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:25:53Z http://mathoverflow.net/feeds/question/10974 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence Does homology detect chain homotopy equivalence? Stephen Bigelow 2010-01-06T22:05:33Z 2010-01-06T23:50:39Z <p>Is the following true: If two chain complexes of free abelian groups have isomorphic homology modules then they are chain homotopy equivalent.</p> http://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence/10983#10983 Answer by Tyler Lawson for Does homology detect chain homotopy equivalence? Tyler Lawson 2010-01-06T23:22:57Z 2010-01-06T23:22:57Z <p>Yes, this is true. Suppose <code>$C_*$</code> is such a chain complex of free abelian groups.</p> <p>For each $n$, choose a splitting of the boundary map <code>$C_n \to B_{n-1}$</code>, so that <code>$C_n \cong Z_n \oplus B_{n-1}$</code>. (You can do this because <code>$B_{n-1}$</code>, as a subgroup of a free group, is free.) For all $n$, you then have a sub-chain-complex <code>$\cdots \to 0 \to B_n \to Z_n \to 0 \to \cdots$</code> concentrated in degrees $n$ and $n+1$, and <code>$C_*$</code> is the direct sum of these chain complexes.</p> <p>Given two such chain complexes <code>$C_*$</code> and <code>$D_*$</code>, you get a direct sum decomposition of each, and so it suffices to show that any two complexes <code>$\cdots \to 0 \to R_i \to F_i \to 0 \to \cdots$</code>, concentrated in degrees $n$ and $n+1$, which are resolutions of the same module $M$ are chain homotopy equivalent; but this is some variant of the fundamental theorem of homological algebra.</p> <p>This is special to abelian groups and is false for modules over a general ring.</p> http://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence/10984#10984 Answer by Mariano Suárez-Alvarez for Does homology detect chain homotopy equivalence? Mariano Suárez-Alvarez 2010-01-06T23:31:07Z 2010-01-06T23:31:07Z <p>The natural functor $K^b(\mathbb Z\mathrm{-free})\to D^b(\mathbb Z)$ from the homotopy category of bounded complexes of finitely generated free abelian groups to the derived category of bounded complexes of finitely generated abelian groups is an equivalence. This means that a map of bounded complexes of finitely generated free abelian groups which induces an isomorphism in homology is an homotopy equivalence. </p> <p>This and the fact that one can always lift a morphism $f:H_\bullet(X)\to H_\bullet(Y)$ between the homologies of two complexes of free abelian groups to a morphism $\tilde f:X\to Y$ of complexes which induces $f$ give an affirmative answer to your question.</p> http://mathoverflow.net/questions/10974/does-homology-detect-chain-homotopy-equivalence/10987#10987 Answer by Leonid Positselski for Does homology detect chain homotopy equivalence? Leonid Positselski 2010-01-06T23:50:39Z 2010-01-06T23:50:39Z <p>Yes, this is true, and it does not matter whether the complexes are bounded from any side (nor of course does it matter whether the homology is finitely generated). This is so because:</p> <ol> <li>The homotopy category of free abelian groups is equivalent to the derived category of abelian groups. This holds even for unbounded complexes, since the category of abelian groups has a finite homological dimension.</li> <li>Any complex of abelian groups is quasi-isomorphic to its homology, since the category of abelian groups has homological dimension 1.</li> </ol>