# The Unapologetic Mathematician

## Chain maps

As promised, something lighter.

Okay, a couple weeks ago I defined a chain complex to be a sequence $\cdots\rightarrow C_{i+1}\stackrel{d_{i+1}}{\rightarrow}C_i\stackrel{d_i}{\rightarrow}C_{i-1}\rightarrow\cdots$ with the property that $d_{i-1}\circ d_i=0$. The maps $d_i$ are called the “differentials” of the sequence. As usual, these are the objects of a category, and we now need to define the morphisms.

Consider chain complexes $C=\cdots\rightarrow C_{i+1}\rightarrow C_i\rightarrow C_{i-1}\rightarrow\cdots$ and $D=\cdots\rightarrow D_{i+1}\rightarrow D_i\rightarrow D_{i-1}\rightarrow\cdots$. We will write the differentials on $C$ as $d^C_i$ and those on $D$ as $d^D_i$. A chain map $f:C\rightarrow D$ is a collection of arrows $f_i:C_i\rightarrow D_i$ that commute with the differentials. That is, $f_{i-1}\circ d^C_i=d^D_i\circ f_i$. That these form the morphisms of an $\mathbf{Ab}$-category should be clear.

Given two chain complexes with zero differentials — like those arising as homologies — any collection of maps will constitute a chain map. These trivial complexes form a full $\mathbf{Ab}$-subcategory of the category of all chain complexes.

We already know how the operation of “taking homology” acts on a chain complex. It turns out to have a nice action on chain maps as well. Let’s write $Z_i(C)$ for the kernel of $d^C_i$ and $B_i(C)$ for the image of $d^C_{i+1}$, and similarly for $D$. Now if we take a member (in the sense of our diagram chasing rules) $x\in_mC_i$ so that $d^C_i\circ x=0$, then clearly $d^D_i\circ(f_i\circ x)=f_{i-1}\circ d^C_i\circ x=0$. That is, if we restrict $f_i$ to $Z_i(C)$, it factors through $Z_i(D)$. Similarly, if there is a $y\in_mC_{i+1}$ with $d^C_{i+1}\circ y=x$, then $f_i\circ x=d^D_{i+1}(\circ f_{i+1}\circ y)$, and thus the restriction of $f_i$ to $B_i(C)$ factors through $B_i(D)$.

So we can restrict $f_i$ to get an arrow $f_i:Z_i(C)\rightarrow Z_i(D)$ which sends the whole subobject $B_i(C)$ into the subobject $B_i(D)$. Thus we can pass to the homology objects to get arrows $\widetilde{f}_i:H_i(C)\rightarrow H_i(D)$. That is, we have a chain map from $H(C)$ to $H(D)$. Further, it’s straightforward to show that this construction is $\mathbf{Ab}$-functorial — it preserves addition and composition of chain maps, along with zero maps and identity maps.

October 16, 2007 - Posted by | Category theory

1. […] this fact that pullbacks commute with the exterior derivative is that it makes pullbacks into a chain map between the chains of the and . And then immediately we get homomorphisms , which we also write as […]

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2. […] complex. Since pullbacks of differential forms commute with the exterior derivative, they define a chain map between two chain […]

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3. Hi, me again , excellent resource for reviewing ( and learning many things I did not get the first time around): so we can get a cochain complex from a given chain complex, by tensoring each term with Hom(-, R), to get, object-wise, Hom(C_i ,R) , which are the cochain groups by definition, where R is the coefficient ring we are working with. But how do we get the codifferential this way ? I imagine we use some adjointness relation (as linear maps, between differential and codifferential), i.e., if d: C_n –>C_{n-1} is the differential, then d*: (C_{n-1})* –> (C_n)* is defined by d*((c_n)*):= (c_n)*(d(c_n) . Is this it?
Sorry if this is too simple, and keep up the great site!

Comment by Larry | June 23, 2014 | Reply

4. Exactly; the duality functor $\left(\underline{\hphantom{X}}\right)^*=\mathrm{Hom}_R(\underline{\hphantom{X}}, R)$ is contravariant, so it sends the differential $d_n:C_n\to C_{n-1}$ to the codifferential $d_n^*:C_{n-1}^*\to C_n^*$.

Comment by John Armstrong | June 24, 2014 | Reply