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The definition of cohomology of a complex is based on the following:

We have a complex (of appropriate objects) $$0\leftarrow C_0\leftarrow C_1\leftarrow C_2\ldots \leftarrow C_n\ldots$$ Then for an abelian group $G$, we consider the dual complex $$0\rightarrow Hom(C_0,G)\rightarrow Hom(C_1,G)\rightarrow\ldots Hom(C_n,G)\ldots$$ This complex generates the cohomology associated to the initial complex $C_n$.

Now, for an appropriate object $G$, let's consider the following complex:

$$0\leftarrow Hom(G,C_0)\leftarrow Hom(G,C_1)\ldots \leftarrow Hom(G,C_n)\ldots$$

Is there any terminology for this construction? Is there an analogue of the universal coefficient theorem in order to give a relation between the homology of the latter complex and the homology of the initial complex for $C_n$? Are there any applications of this construction which show that the latter homology is more useful or more convenient than the initial one?Can one give some nontrivial examples in algebraic topology?

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    $\begingroup$ If $C$ is the chain complex of a finite CW-complex $X$, then the $C_n$-s are simply finite direct sums of copies of $\mathbb{Z}$. So, this boils down to $\mathrm{Hom}(G,\mathbb{Z})$. If $G$ is a finitely generated abelian group, then your complex discards any torsion of $G$, and you get a complex of finite direct sums of $\mathbb{Z}$. In particular, your complex does not care about coefficients like $\mathbb{Z} / p \mathbb{Z)$ :-) $\endgroup$
    – M.G.
    Commented Nov 9, 2020 at 21:31
  • $\begingroup$ In fact, the complex is also boring if $G$ is the additive group of any field regardless if its char. $\endgroup$
    – M.G.
    Commented Nov 9, 2020 at 21:49
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    $\begingroup$ This is not good because it doesn't identify quasi-isomorphic complexes. Maybe you want something like $\mathbb{R}\operatorname{Hom}(G,C_*(X))$? It can be computed by taking a dg-injective replacement for $C_*(X)$ so it's less explicit. $\endgroup$
    – Denis Nardin
    Commented Nov 10, 2020 at 10:24
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    $\begingroup$ @AliTaghavi $\mathrm{R} \mathrm{Hom}(-, -)$ is the derived functor of $\mathrm{Hom}$. See Ch 3 of Methods of Homological Algebras by Gelfand and Manin for a nice textbook on homological algebra. $\endgroup$
    – Wille Liu
    Commented Nov 11, 2020 at 0:33
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    $\begingroup$ The point is that for $\mathrm{Hom}(G, C_*(X))\cong \mathrm{RHom}(G, C_*(X))$ to hold, you need to require that $\mathrm{Ext}^{> 0}(G, C_k) = 0$ for all $k$. Otherwise, the most we can say is that there is a spectral sequence connecting $\mathrm{Hom}(G, C_*(X))$ and $\mathrm{RHom}(G, C_*(X))$. $\endgroup$
    – Wille Liu
    Commented Nov 11, 2020 at 0:39

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