## 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.

[...] 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 [...]

Pingback by Pullbacks on Cohomology « The Unapologetic Mathematician | July 21, 2011 |

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

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

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 |

Exactly; the duality functor is contravariant, so it sends the differential to the codifferential .

Comment by John Armstrong | June 24, 2014 |