# 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 | 4 Comments