## 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 of chain complexes in an abelian category . 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 and , chain maps and from to , and let’s say there’s a chain homotopy . The chain maps induce maps and . I assert that .

To see this, first notice that passing to the induced map is linear. That is, . So all we really need to show is that a null-homotopic map induces the zero map on homology. But if makes null-homotopic, then . When we restrict to the kernel of , this just becomes , which clearly lands in the image of , which is zero in , as we wanted to show.

Now if we have chain maps and along with chain homotopies and , we say that and are “homotopy equivalent”. Then the induced maps on homology and are inverses of each other, and so the homologies of and are isomorphic.

This passage from covariance to invariance is the basis for why Khovanov homology works. We start with a 2-category of tangles (which I’ll eventually explain fore thoroughly). Then we pick a ring and consider the 2-category of chain complexes over the abelian category of -modules. We construct a 2-functor 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.

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

betweenchain 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

thatsort of stuff, at least not yet.Comment by John Baez | October 19, 2007 |

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 |

[…] 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 |