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In this question: Constructing a cochain complex out of a chain complex , palio asked how to construct a co-chain complex when given a chain complex as well as how to go in the opposite direction, i.e., given a cochain complex, how to construct its dual chain complex. I am a bit confused about some of the details in the answer, and have some additional questions:

i) Do we just apply $M\rightarrow Hom(X,-)$ , to go in the opposite direction? If so, how do we get $\partial$ from the coboundary $\delta$; do we just use the fact that the two are adjoints as linear maps?

ii)Are we using the fact that $Hom(X,-)$ and $Hom(-, X)$ are contravariant and covariant (respectively) and right-exact (i.e., both preserve exactness of the respective long-exactsequences(co)homology (co)chain complexes )?

I am specifically interested in recovering the boundary operator in De Rham cohomology, given the exterior derivative on forms as the coboundary operator, so that the forms are the cochains ; in this case $\partial_{i+1}$ should satisfy the relation $\delta^i:f\rightarrow f\circ\partial_{i+1}$ , for any $f: C_i \rightarrow R$ , where $C_i$ is the graded subspace of i-forms (so that $\delta_i$ is the linear adjoint to $\partial _{i+1}$ ). Is this the right way of doing this?

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user99680
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In general, given a chain complex, you can take $Hom(X, -)$ to get a chain complex in which the maps go "in the other direction", which gets called a co-chain complex. If you apply $Hom(X, -)$ to that cochain complex, you again reverse the maps, and get a chain complex. Will it be (in some sense) the same as the one you started with? That depends on whether "double dual" is "the same as" identity in your category. For vector spaces, it is: $V^{**}$ is isomorphic to $V$ in general. For other categories...maybe not. Does that help any?

John Hughes
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  • I assume you mean $V$ finite dimensional. Otherwise $V^{**}\neq V$. – Cheerful Parsnip Jun 12 '14 at 00:20
  • @John: But are both Hom(X,_) and Hom(-,X) of the same variance, i.e., both covariant? And, is the dual map (be it boundary or coboundary) given by the linear adjointness relation? – user99680 Jun 12 '14 at 00:30
  • For $V$ finite dimensional, $V^{}$ is isomorphic to $V$, but not for the infinite-dimensional case. But the double dual...for that, there's the map $V \to V^{*} : v \mapsto (f \mapsto f(v))$ which I thought was an isomorphism. Could be wrong; it's been a long time. – John Hughes Jun 12 '14 at 03:01
  • I believe that they're are not of the same variance (see the commentary in the question you cited!). I don't understand the second part of your question -- I just don't know the words. Sorry. – John Hughes Jun 12 '14 at 03:02