The Unapologetic Mathematician

Mathematics for the interested outsider

Chain Homotopies and Homology

Sorry about going AWOL yesterday, but I got bogged down in writing another exam.

Okay, so we’ve set out chain homotopies as our 2-morphisms in the category \mathbf{Kom}(\mathcal{C}) of chain complexes in an abelian category \mathcal{C}. We also know that each of these 2-morphisms is an isomorphism, so decategorifying amounts to saying that two chain maps are “the same” if they are chain-homotopic.

Many interesting properties of chain maps are invariant under chain homotopies, which means that they descend to properties of this decategorified version. Alternately, some properties are defined by 2-functors, which means that if we apply a chain homotopy we change our answer by a 2-morphism in the target 2-category, which must itself be an isomorphism. I like to call these “homotopy covariants”, rather than “invariants”. Anyway, then the decategorification of this property is an invariant, and what I said before applies.

The big one of these properties we’re going to be interested in is the induced map on homology. Let’s consider chain complexes A and B, chain maps f and g from A to B, and let’s say there’s a chain homotopy \alpha:f\Rightarrow g. The chain maps induce maps \widetilde{f}:H_\bullet(A)\rightarrow H_\bullet(B) and \widetilde{g}:H_\bullet(A)\rightarrow H_\bullet(B). I assert that \widetilde{f}=\widetilde{g}.

To see this, first notice that passing to the induced map is linear. That is, \widetilde{f}-\widetilde{g}=\widetilde{f-g}. So all we really need to show is that a null-homotopic map induces the zero map on homology. But if \alpha:f\Rightarrow0 makes f null-homotopic, then f_n=d^B_{n+1}\circ\alpha_n+\alpha_{n-1}\circ d^A_n. When we restrict f_n to the kernel of d^A_n, this just becomes d^B_{n+1}\circ\alpha_n, which clearly lands in the image of d^B_{n+1}, which is zero in H_n(B), as we wanted to show.

Now if we have chain maps f:A\rightarrow B and g:B\rightarrow A along with chain homotopies g\circ f\Rightarrow1_A and f\circ g\Rightarrow1_B, we say that A and B are “homotopy equivalent”. Then the induced maps on homology \widetilde{f}:H_\bullet(A)\rightarrow H_\bullet(B) and \widetilde{g}:H_\bullet(B)\rightarrow H_\bullet(A) are inverses of each other, and so the homologies of A and B are isomorphic.

This passage from covariance to invariance is the basis for why Khovanov homology works. We start with a 2-category \mathcal{T}ang of tangles (which I’ll eventually explain fore thoroughly). Then we pick a ring R and consider the 2-category \mathbf{Kom}(R\mathbf{-mod}) of chain complexes over the abelian category of R-modules. We construct a 2-functor K:\mathcal{T}ang\rightarrow\mathbf{Kom}(R\mathbf{-mod}) that picks a chain complex for each number of free ends, a chain map for each tangle, and a chain homotopy for each ambient isotopy of tangles. Then two isotopic tangles are assigned homotopic chain maps — the chain map is a “tangle covariant”. When we pass to homology, we get tangle invariants, which turn out to be related to well-known knot invariants.

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October 19, 2007 - Posted by | Category theory


  1. Why call this 2-category “Kom”? I thought the Germans lost World War II and math terminology was mainly English now.

    Anyway: if one wants still more fun, one can notice that there are chain homotopies between chain homotopies, and so on ad infinitum. So, Kom is really a strict ∞-category.

    And one can explain this by noticing that a chain complex in C (at least a “one-sided” chain complex) is itself a sort of strict ∞-category: precisely a strict ∞-category internal to C. From this viewpoint, chain maps are functors, chain homotopies are natural transformations, and so on.

    But I’m very glad you’re not baffling your readers with that sort of stuff, at least not yet.

    Comment by John Baez | October 19, 2007 | Reply

  2. For some reason, the references I’m using to refresh my memory say “Kom”. So I followed their lead for now.

    Besides, why not switch it up in the language department now and then? We still use “F” for sheaves, don’t we?

    Comment by John Armstrong | October 19, 2007 | Reply

  3. [...] is a homotopy, then the Poincaré lemma gives us a chain homotopy from to as chain maps, which tells us that the maps they induce on homology are identical. That is, passing to homology [...]

    Pingback by Homotopic Maps Induce Identical Maps On Homology « The Unapologetic Mathematician | December 6, 2011 | Reply

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