Chain maps
As promised, something lighter.
Okay, a couple weeks ago I defined a chain complex to be a sequence 
with the property that 
. The maps 
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 
and 
. We will write the differentials on 
as 
and those on 
as 
. A chain map 
is a collection of arrows 
that commute with the differentials. That is, 
. That these form the morphisms of an 
-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 -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 
for the kernel of and 
for the image of 
, and similarly for . Now if we take a member (in the sense of our diagram chasing rules) 
so that 
, then clearly 
. That is, if we restrict 
to , it factors through 
. Similarly, if there is a 
with 
, then 
, and thus the restriction of to factors through 
.
So we can restrict to get an arrow 
which sends the whole subobject into the subobject . Thus we can pass to the homology objects to get arrows 
. That is, we have a chain map from 
to 
. Further, it’s straightforward to show that this construction is -functorial — it preserves addition and composition of chain maps, along with zero maps and identity maps.
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October 16, 2007 -
Posted by John Armstrong |
Category theory
The Unapologetic Mathematician 
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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!
Exactly; the duality functor
is contravariant, so it sends the differential 
to the codifferential 
.