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Let $V$ and $V^\prime$ - complexes of modules over ring $A$, and $f, g$ - homomorphisms $V\rightarrow V^\prime$.

I am interested in various conditions on $A, V, V^\prime$: ($f$ and $g$ are homological) $\Rightarrow$ ($f$ and $g$ are homotopic).

(I knew one example: $A$ - hereditary algebra and $V, V^\prime$ - complexes of projective modules, bounded from the right. But recently I understood that in this case it's not true that ($f$ and $g$ are homological) $\Rightarrow$ ($f$ and $g$ are homotopic))

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I assume that by "homological" you mean "homologous" –  Igor Rivin Dec 21 '11 at 21:12
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I mean that maps $H_n(f)=H_n(g)$ for all $n$. The term "homological" is used in the book Y.A. Drozd, V.V Kirichenko "Finite-Dimensional Algebras". –  One_math_boy Dec 21 '11 at 21:22
    
Some of the discussion at mathoverflow.net/questions/8974/chain-homotopy-why-duud-and-not-duvd/ is relevant. –  David Speyer Dec 21 '11 at 22:53

1 Answer 1

This question relates to a very complicated problem known as Freyd's generating hypothesis. The problem was first posed for the stable homotopy category but it can be extended to more general triangulated categories, such as the derived category $D(R)$ of a ring $R$. In this context the hypothesis (which may or may not be satisfied, depending on $R$) says that your $\Rightarrow$ is satisfied whenever the complexes are (quasi-isomorphic to) bounded complexes of f.g. projectives. Keir H. Lockridge proved (JPAA, 2007) that for $R$ commutative this is true if and only if $R$ is von Neumann regular. You can look at this problem in other contexts (e.g. modular representation theory) to get more examples and counterexamples.

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Sorry, but can you give some simple examples? –  One_math_boy Dec 22 '11 at 19:36
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Don't you find simple enough what I've said? –  Fernando Muro Dec 25 '11 at 20:56
    
Well, it would be nice to see an example of a pair of complexes over a non-vnr ring (e.g. over Z) and a pair of maps between them that are homologous but not homotopic. –  Neil Epstein Dec 9 '12 at 18:02
    
Over $\mathbb Z$ it is very easy, take two abelian groups $A$ and $B$ with non-vanishing $\operatorname{Ext}^1(A,B)$. This abelian group classifies homotopy classes $A\rightarrow B[1]$, and all these maps obviously induce the trivial morphism in homology by dimension reasons. I took this as being "non research level". –  Fernando Muro Dec 9 '12 at 23:19

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