# The Unapologetic Mathematician

## Chain Homotopies

We’ve defined chain complexes in an abelian category, and chain maps between them, to form an $\mathbf{Ab}$-category $\mathbf{Kom}(\mathcal{C})$. Today, we define chain homotopies between chain maps, which gives us a 2-category.

First, we say that a chain map $f:A\rightarrow B$ given by $f_n:A_n\rightarrow B_n$ is “null-homotopic” if we have arrows $h_n:A_n\rightarrow B_{n+1}$ such that $f_n=d^B_{n+1}\circ h_n+h_{n-1}\circ d^A_n$. Here’s the picture:

In particular, the zero chain map with $f_n=0$ for all $n$ is null-homotopic — just pick $h_n=0$.

Now we say that chain maps $f$ and $g$ are homotopic if $f-g$ is null-homotopic. That is, $f_n-g_n=d^B_{n+1}\circ h_n+h_{n-1}\circ d^A_n$. We call the collection $h=\{h_n\}$ a chain homotopy from $f$ to $g$. Then a chain map is null-homotopic if and only if it is homotopic to the zero chain map. We can easily check that this is an equivalence relation. Any chain map is homotopic to itself because $f_n-f_n=0$ and the zero chain map is null-homotopic. If $f$ and $g$ are homotopic by a chain homotopy $h$, then $-h$ is a chain homotopy from $g$ to $f$. Finally, if $f-f'$ is null-homotopic by $h$ and $f'-f''$ is null-homotopic by $h'$, then $f-f''=(f-f')+(f'-f'')$ is null-homotopic by $h+h'$.

Another way to look at this is to note that we have an abelian group $\hom_{\mathbf{Kom}(\mathcal{C})}(A,B)$ of chain maps from $A$ to $B$, and the null-homotopic maps form a subgroup. Then two chain maps are homotopic if and only if they differ by a null-homotopic chain map, which leads us to consider the quotient of $\hom_{\mathbf{Kom}(\mathcal{C})}(A,B)$ by this subgroup. We will be interested in properties of chain maps which are invariant under chain homotopies — properties that only depend on this quotient group.

In the language of category theory, the homotopies $h:f\Rightarrow g$ are 2-morphisms. Given 1-morphisms (chain maps) $f$, $f'$, and $f''$ from $A$ to $B$, and homotopies $h:f\Rightarrow f'$ and $h':f'\Rightarrow f''$, we compose them by simply adding the corresponding components to get $h+h':f\Rightarrow f''$.

On the other hand, if we have 1-morphisms $f$ and $g$ from $A$ to $B$, 1-morphisms $f'$ and $g'$ from $B$ to $C$, and 2-morphisms $h:f\Rightarrow g$ and $h':f'\Rightarrow g'$, then we can “horizontally” compose these chain homotopies to get $h'\circ h:f'\circ f\Rightarrow g'\circ g$ with components $(h'\circ h)_n=h'_n\circ f_n+g'_{n+1}\circ h_n$. Indeed, we calculate
$d^C_{n+1}\circ(h'_n\circ f_n+g'_{n+1}\circ h_n)+(h'_{n-1}\circ f_{n-1}+g'_n\circ h_{n-1})\circ d^A_n=$
$d^C_{n+1}\circ h'_n\circ f_n+d^C_{n+1}\circ g'_{n+1}\circ h_n+h'_{n-1}\circ f_{n-1}\circ d^A_n+g'_n\circ h_{n-1}\circ d^A_n=$
$d^C_{n+1}\circ h'_n\circ f_n+h'_{n-1}\circ d^B_n\circ f_n+g'_n\circ d^B_{n+1}\circ h_n+g'_n\circ h_{n-1}\circ d^A_n=$
$(d^C_{n+1}\circ h'_n+h'_{n-1}\circ d^B_n)\circ f_n+g'_n\circ(d^B_{n+1}\circ h_n+h_{n-1}\circ d^A_n)=$
$(f'_n-g'_n)\circ f_n+g'_n\circ(f_n-g_n)=$
$f'_n\circ f_n-g'_n\circ g_n$

We could also have used $h_n\circ f'_n+g_{n+1}\circ h'_n$ and done a similar calculation. In fact, it turns out that $h'_n\circ f_n+g'_{n+1}\circ h_n$ and $h_n\circ f'_n+g_{n+1}\circ h'_n$ are themselves homotopic in a sense, and so we consider them to be equivalent. If we pay attention to this homotopy between homotopies, we get a structure analogous to the tensorator. I’ll leave you to verify the exchange identity on your own, which will establish the 2-categorical structure of $\mathbf{Kom}(\mathcal{C})$.

One thing about this structure that’s important to note is that every 2-morphism is an isomorphism. That is, if two chain maps are homotopic, they are isomorphic as 1-morphisms. Thus if we decategorify this structure by replacing 1-morphisms by isomorphism classes of 1-morphisms, we are just passing from chain maps to homotopy classes of chain maps. In other words, we pass from the abelian group $\hom_{\mathbf{Kom}(\mathcal{C})}(A,B)$ of chain maps to its quotient by the null-homotopic subgroup.

October 17, 2007 - Posted by | Category theory

## 5 Comments »

1. And now we’re up to the point where all the things I’ve spent my last month in Sydney doing are explicable. Enjoying your blog more and more every day!

In other news, finding homotopies is now about to be a part of Magma.

Comment by Mikael Johansson | October 18, 2007 | Reply

2. […] 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 […]

Pingback by Chain Homotopies and Homology « The Unapologetic Mathematician | October 19, 2007 | Reply

3. […] comes in handy here as well, for it’s exactly what we need to say that the Lie derivative is null-homotopic. And like any null-homotopic map, it defines the zero map on cohomology. That is, if we take some […]

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4. […] now I say that a homotopy between two maps induces a chain homotopy between the two chain maps and . And, indeed, if the homotopy is given by a smooth map then we […]

Pingback by The Poincaré Lemma (setup) « The Unapologetic Mathematician | December 2, 2011 | Reply

5. […] can now prove the Poincaré lemma by proving its core assertion: there is a chain homotopy between the two chain maps and induced by the inclusions of into either end of the homotopy […]

Pingback by The Poincaré Lemma (proof) « The Unapologetic Mathematician | December 3, 2011 | Reply