Let $(C,\partial)$ be a chain complex where $C_i$ is an $R$-module ($R$ is a given ring) , we can always construct a cochain complex out of the chain complex $(C,\partial)$ in the following way: We construct the $i$th $R$-module of the cochain complex as $C^i=Hom_R(C_i,R)$ this is the $R$-module of $R$-module homomorphisms from $C_i$ to $R$. The $R$-module homomorphism $\delta_i:C^i\rightarrow C^{i+1}$ sends a morphism $f:C_i\rightarrow R$ to the morphism $f\circ \partial_{i+1}:C^{i+1}\rightarrow R$.

Question 1 : What if we take another functor other than $Hom_R(-, R)$ and get another cochain complex out of the original chain complex, do we obtain isomorphic quotients $ker/Im$ in the two cochain complexes?

Question 2 : Given a cochain complex $(C,\delta)$, how can we construct a chain complex out of $(C,\delta)$? I think we can still use the $Hom(-,R)$ functor to the cochain complex to get a chain complex, Is this correct? and why this question seems to be not interesting as i can't find anything about this converse construction, it seems like we always need to get a cochain complex out of a chain complex but not a chain complex out of a cochain complex?


1 Answer 1


For Question 1, the answer is no. The simplest counterexample is to start with any chain complex whose homology is not trivial and to take the functor sending all modules to the zero module. A less trivial example is to take the functor $M\mapsto \text{Hom}_R(-,X)$ for some $R$-module $X$ other than $R$.

For Question 2, the answer is yes, you can go from cochain to chain complexes the same way. The reason people usually go from chain to cochain complexes is probably that in the original applications to topology, it was easy and natural to define the desired chain complexes directly (singular, simplicial, and cell complexes, for example) and then one needed a Hom construction to get cochain complexes. But in other situations, one can define a cochain complex directly (e.g., the deRham complex for a smooth manifold), and then one could use Hom to get a chain complex.

  • $\begingroup$ Thank you Andreas! I also want to know why when dualizing we always use $Hom_R(-, R)$ and not $Hom_R(R,-)$ ? $\endgroup$
    – palio
    Nov 9, 2013 at 20:22
  • 2
    $\begingroup$ @palio $\text{Hom}_R(R,-)$ is (naturally isomorphic to) the identity functor. More generally, $\text{Hom}_R(X,-)$ is a covariant functor, so it will convert chain complexes to chain complexes (and cochain complexes to cochain complexes), so this would not be dualizing. $\endgroup$ Nov 9, 2013 at 20:24
  • $\begingroup$ @AndreasBlass: Sorry for the necroquestion; re the issue of going from a cochain complex "in the same way", do you mean that we would dualize in the opposite direction by using $M\rightarrow Hom_R (X, -)$? And , do we use the same equation of linear map adjoints to recover $\partial $ from the coboundary operator? $\endgroup$
    – user99680
    Jun 11, 2014 at 22:15
  • $\begingroup$ @user99680 The dualization that I intended would just apply the contravariant functor $\text{Hom}_R(-,R)$. One could get other chain complexes from a cochain complex by applying $\text{Hom}_R(-,X)$ for other $R$-modules $X$. The functor in your comment is covariant, so it would send cochain complexes to cochain (not chain) complexes. $\endgroup$ Jun 12, 2014 at 6:18
  • $\begingroup$ @AndreasBlass: Thanks for the explanation, and for replying to my necroquestion. $\endgroup$
    – user99680
    Jun 12, 2014 at 6:20

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