First-time here... I hope my question isn't silly or anything... anyway...

Consider the category of chain complexes and chain maps. We can also define chain homotopies between chain maps. Does this form a 2-category? I am able to construct vertical and horizontal composition of chain homotopies but am unable to prove that the horizontal composition is associative and that the interchange law holds.

I like to include a lot of details in case the reader isn't familiar with everything, so the following is the problem in detail.

To be a bit more concrete, let's let $(C_n, \partial_n), (C'_n, \partial'_n),$ and $(C''_n, \partial''_n)$ be chain complexes (sorry, my differential is going up in degree) and let $f_n, g_n, h_n: C_n \to C'_n$ and $f'_n, g'_n, h'_n : C'_n \to C''_n$ be chain maps. Suppose that $\sigma : f \Rightarrow g,$ $\tau : g \Rightarrow h,$ $\sigma' : f' \Rightarrow g',$ and $\tau' : g' \Rightarrow h'$ are chain homotopies, where $\sigma_n, \tau_n : C_n \to C'_{n-1}$ and similarly for $\sigma'$ and $\tau'.$ Recall that these definitions say

$f_n \circ \partial_{n-1} = \partial'_{n-1} \circ f_{n-1}$

and similarly for primes, $g$'s, $h$'s, and

$\sigma_{n+1} \circ \partial_{n} + \partial'_{n-1} \circ \sigma_{n} = f_{n} - g_{n},$

$\tau_{n+1} \circ \partial_{n} + \partial'_{n-1} \circ \tau_{n} = g_{n} - h_{n},$

and similarly for primes.

We can define the vertical composition of $\sigma : f \Rightarrow g$ with $\tau : g \Rightarrow h$ by

$(\tau \diamond \sigma)_{n} := \tau_n + \sigma_n$

(I denote vertical composition with a diamond, $\diamond$).

We can also define the horizontal composition of $\sigma : f \Rightarrow g$ with $\sigma' : f' \Rightarrow g'$ by

$(\sigma' \circ \sigma)_n := \partial''_{n-2} \circ \sigma'_{n-1} \circ \sigma_{n} - \sigma'_{n} \circ \sigma_{n+1} \circ \partial_{n} + \sigma'_{n} \circ g_{n} + f'_{n-1} \circ \sigma_{n}.$

Now we just verify that these are indeed chain homotopies. The vertical composition is easy:

$(\tau \diamond \sigma)_{n+1} \circ \partial_{n} + \partial'_{n-1} \circ (\tau \diamond \sigma)_{n} = f_n - g_n + g_n - h_n = f_n - h_n$

and the horizontal composition is a bit more challenging but doable (I won't include the derivation here).

The identity for vertical composition is the zero map and similarly for the horizontal composition (note that for the vertical composition, the zero map is a chain homotopy between any two chain maps provided they are the same while for the horizontal composition, the chain maps are both the identities). The vertical composition is easily seen to be associative. However, the horizontal composition satisfies

$( \sigma'' \circ ( \sigma' \circ \sigma ) )_{n} - ( ( \sigma'' \circ \sigma' ) \circ \sigma )_n$

$= ( f''_{n-1} - g''_{n-1} ) \circ ( f'_{n-1} - g'_{n-1} ) \circ \sigma_{n} - \sigma''_{n} \circ ( f'_{n} - g'_{n} ) \circ ( f_n - g_n ).$

The interchange law doesn't hold either and the difference is given by

$( ( \tau' \diamond \sigma' ) \circ ( \tau \diamond \sigma ) )_{n} - ( ( \tau' \circ \tau ) \diamond ( \sigma' \circ \sigma ) )_{n}$

$= ( f'_{n-1} - g'_{n-1} ) \circ \tau_{n} + (g'_{n-1} - h'_{n-1} ) \circ \sigma_{n} - \tau'_{n} \circ ( f_{n} - g_{n} ) - \sigma'_{n} \circ (g_{n} - h_{n} )$

So I'm not sure if chain complexes, chain maps, and chain homotopies are supposed to form a 2-category, but I would've liked this result to be true. Does anyone know what the correct categorical structure of this category is? I have not yet considered chain homotopies of chain homotopies, so if the answer requires all such higher morphisms, then an answer in that direction would also be acceptable.

Any thoughts? Thanks in advance.

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    $\begingroup$ I think you need to use homotopy classes of chain homotopies. $\endgroup$ – Qiaochu Yuan Jan 28 '12 at 23:08
  • $\begingroup$ Do you mean "... and $(C''_n, \partial''_n)$ be ..." in the fourth paragraph ? $\endgroup$ – Ralph Jan 29 '12 at 13:51

Part of the problem is that you're not using a convenient definition of chain homotopy. I'll use a differential going down in degree. Let $I$ be the interval object in the category of chain complexes; that is, $I$ is $\text{span}(0, 1)$ in degree $0$ and $\text{span}(e)$ in degree $1$ ($k$ the underlying commutative ring), and the differential sends $e$ to $1 - 0$. If $C, D$ are two complexes, I claim that a chain homotopy between two chain maps $f, g : C \to D$ is precisely a chain map $$H : I \to \text{Hom}(C, D)$$

such that the restriction of the map to $0$ is $f$ and the restriction of the map to $1$ is $g$, where $\text{Hom}$ is the hom chain complex. With this definition you can work guided by analogy to the topological situation (there the 2-category is the 2-category of topological spaces, continuous functions, and homotopy classes of homotopies between such functions); all of the maps you need have obvious topological definitions, although I admit I have never worked out the details. I think everything reduces to working with fairly concrete maps between some chain complexes constructed from $I$.

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    $\begingroup$ Equivalently, a homotopy is a map $H: C \otimes I \rightarrow D$, and again you can be guided by the topology :). And $I$ didn't come from nowhere: it is the simplicial chain complex associated to the unit interval. It is also equivalent to what you get when you take the normalized complex of the simplicial abelian group $\mathbb{Z}\Delta^1$... etc. $\endgroup$ – Dylan Wilson Jan 29 '12 at 1:04
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    $\begingroup$ Yeah, I completely forgot to input the homotopy condition. Thanks for the reminder and also the quick response. Perhaps this viewpoint will allow me to find this condition as well as the conditions for higher chain homotopies. I can see that it works for the 1-category description (namely, such an $H$ produces the $f,g,$ and $\sigma$ by considering the restriction to 1,0, and $\mathrm{span}(e)$ respectively). This should take care of those leftover terms. For the higher homotopies the idea is very similar--restrict to the appropriate corners, boundaries, and faces. Awesome! $\endgroup$ – Arthur Jan 29 '12 at 1:08
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    $\begingroup$ Yeah, this is pretty beautiful. You immediately get the algebraic condition for when two $(n-1)$-chain homotopies are homotopy equivalent by applying a chain map $I \otimes \cdots \otimes I \otimes C \to D$ ($n$ interval objects) to the ``face'' of the $n$-cube and viewing that as an $n$-chain homotopy between the two. This gives a very elegant algebraic expression that looks just like the usual chain homotopy condition for the first level. The differences between the two chain homotopies I wrote above satisfy this requirement, so this solves the problem! Thanks again! :) $\endgroup$ – Arthur Jan 29 '12 at 3:49
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    $\begingroup$ Strictly speaking, what you get this way isn't a bicategory, it is a $(\infty,1)$-category, as they call it in nLab. Your multiplication of 2-cells is defined and associative only up to a coherent action of 3-cells. And multiplication of 3-cells is ok only up to 4-cells. This is just the same in topology: if you have a homotopy from $f$ to $g$ and from $g$ to $h$, you don't have a uniquely defined homotopy from $f$ to $h$! There are numerous ways to contract $[0;2]$-interval into a $[0;1]$-interval. $\endgroup$ – Anton Fetisov Jan 29 '12 at 22:38
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    $\begingroup$ @Anton: isn't the point of taking homotopy classes of homotopies to quotient out by the action of those 3-cells? I am pretty sure we get an honest bicategory, perhaps even a (strict) 2-category, this way. $\endgroup$ – Qiaochu Yuan Jan 29 '12 at 23:20

Viewing homotopy via interval objects analogously to topological homotopy as done in Qiaochu Yuan's answer is certainly the best way to consider the problem. Nevertheless it's also possible to get along directly with your definition of homotopy.

In this point of view the problem is that your horizontal composition isn't appropriate: First note that we have $\sigma: f \Rightarrow g,\; \sigma': f' \Rightarrow g'$. Thus (I just write $f'f$ for $f'\circ f$) $$f'f - g'g = f'f - f'g + f'g - g'g = ...=\partial^{''}(f'\sigma+\sigma'g) + (f'\sigma+\sigma'g)\partial. $$

Therefore $\sigma' \circ \sigma:= f'\sigma+\sigma'g: f'f \Rightarrow g'g$ is a homotopy.

Let $\sigma'': f'' \Rightarrow g''$ be another homotopy. By setting in the definition of the former homotopy, one finds that $$(\sigma'' \circ \sigma') \circ \sigma = f''f'\sigma + f'' \sigma' g + \sigma'' g'g = \sigma'' \circ (\sigma' \circ \sigma)$$ is a homotopy $(f'' f')f = f''(f'f) \Rightarrow g''(g'g) = (g''g')g$.

Hence the horizontal composition is associative.

  • $\begingroup$ Thanks. And yes, I believe your definition of horizontal composition is the same as mine under equivalence of higher chain homotopies now that I know what that condition is. $\endgroup$ – Arthur Jan 29 '12 at 23:43

I think the right context for this question is that of monoidal closed categories with a unit interval object, and probably the first place to explore this is the ncat lab.

I mention that we set up a number of monoidal closed categories in our book

R. Brown, P.J. Higgins, R. Sivera, Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids, EMS Tracts in Mathematics Vol. 15, 703 pages. (August 2011). http://www.bangor.ac.uk/r.brown/nonab-a-t.html pdf available there.

for example:

crossed complexes, chain complexes with a groupoid of operators,

and explore the relations between them. As explained in the above answer and comments, the cubical approach has some advantages when dealing with homotopies and higher homotopies, basically because of the formula $I^m \otimes I^n \cong I^{m+n}$.

Added as edit: the other point about the cubical formulation is that there is a notion of cubical set with compositions (and also extra structure such as connections) but we do not have a similar notion simplicially, or not so easily. Globular notions have compositions, though multiple compositions are awkward, and tensor products are not so easy.


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