The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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October 16, 2007 - Posted by | Category theory

4 Comments »

  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 [...]

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

  2. [...] 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 | Reply

  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


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