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Working in the category of modules over a fixed ring $A$ $\operatorname{Mod}_A$, we have the adjunction: $$F(X) = X\otimes_A M \dashv G(X) = \operatorname{Hom}_A(M,X)$$ for a fixed module $M$. Is there a version where we do not fix $M$. That is, is there a right adjoint to the following functor: $$H: \operatorname{Mod}_A\times \operatorname{Mod}_A \to \operatorname{Mod}_A: H(X,Y) = X\otimes_A Y.$$ I am also interested in a version where the ring is not fixed either. I am not sure what category to work from in this case though.

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Multivariable adjunctions do not work like that. For instance, if $(X, Y) \mapsto X \otimes_A Y$ had a right adjoint, then it would have to preserve direct sums, i.e. the comparison $$(X_0 \otimes_A Y_0) \oplus (X_1 \otimes_A Y_1) \to (X_0 \oplus X_1) \otimes_A (Y_0 \oplus Y_1)$$ would have to be an isomorphism. But, in fact, $$(X_0 \oplus X_1) \otimes_A (Y_0 \oplus Y_1) \cong (X_0 \otimes_A Y_0) \oplus (X_0 \otimes_A Y_1) \oplus (X_1 \otimes_A Y_0) \oplus (X_1 \otimes_A Y_1)$$ so $(X, Y) \mapsto X \otimes_A Y$ cannot have a right adjoint (assuming $A \ncong \{ 0 \}$).

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  • $\begingroup$ Thanks. What do you mean by they do not work like that? I assume this is implicitly saying you can make them work some other way? $\endgroup$ – Asvin Mar 1 '16 at 13:45
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    $\begingroup$ The other way is exactly to have all the functors given by fixing one coordinate have adjoints, as occurs in this case. $\endgroup$ – Kevin Carlson Mar 1 '16 at 15:46

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