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Let $\textbf C$ be a chain complex of right $R$-modules, let $A$ be a left $R$-module and let $B$ be an abelian group.

How to make sense of the $\mathrm{Hom}(\textbf {C}\otimes_R A,B)$ and $\mathrm{Hom}_R(\textbf {C},\mathrm{Hom}(A,B))$?

Madara Uchiha
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    Can you provide some more context? – Ruben Jul 23 '19 at 14:02
  • Actually, $\text{Hom}(A,B)$ seems to be an $R$-module by defining $r \cdot f(x) := f(rx)$, so that part seems correct. – Ruben Jul 23 '19 at 14:15
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    @Ruben $\operatorname{Hom}(A,B)$ has a canonical right $R$-module structure induced by the left $R$-module structure on $A$: $(\xi r)(x)=\xi(rx)$ for $r\in R$, $x\in A$ and $\xi:A\to B$. – Fabio Lucchini Jul 23 '19 at 14:16
  • From your notation I would assume $\text{Hom}_R$ to mean Hom sets of $R$-modules, which would imply that we consider the Hom sets of individual objects from the complex. This certainly exists. Otherwise we could take it to mean Hom sets of complexes of $R$-modules, with $B$ for example taken as the complex with $B$ at zero and $0$ everywhere else. This comes down to morphisms $f : C_0 \to B$ such that the kernel contains the image of the map $C_1 \to C_0$ I think. There may be a third interpretation. – Ruben Jul 23 '19 at 14:21
  • @FabioLucchini Exactly, thank you. – Ruben Jul 23 '19 at 14:22

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Usually, if $\mathbf C$ denote a complex $$\cdots\to C_n\xrightarrow{\gamma_n}C_{n+1}\to\cdots$$ then $\DeclareMathOperator\Hom{Hom}\Hom(\mathbf C\otimes_RA,B)$ denote the complex $$\cdots\leftarrow\Hom(C_n\otimes_RA,B)\xleftarrow{\Hom(\gamma_n\otimes_R1_A,1_B)}\Hom(C_{n+1}\otimes_RA,B)\leftarrow\cdots$$ while $\Hom(\mathbf C,\Hom(A,B))$ denote the complex: $$\cdots\leftarrow\Hom_R(C_n,\Hom(A,B))\xleftarrow{\Hom_R(\gamma_n,\Hom(1_A,1_B))}\Hom_R(C_{n+1},\Hom(A,B))\leftarrow\cdots$$

Fabio Lucchini
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