If "tensor" has an adjoint, is it automatically an "internal Hom"?

2010-04-14T19:55:24Z

Let $\mathcal C,\otimes$ be a monoidal category, i.e. $\otimes : \mathcal C \times \mathcal C \to \mathcal C$ is a functor, and there's a bit more structure and properties. Suppose that for each $X \in \mathcal C$ , the functor $X \otimes - : \mathcal C \to \mathcal C$ has a right adjoint. I will call this adjoint (unique up to canonical isomorphism of functors) $\underline{\rm Hom}(X,-) : \mathcal C \to \mathcal C$ . By general abstract nonsense, $\underline{\rm Hom}(X,-)$ is contravariant in $X$ , and so defines a functor $\underline{\rm Hom}: \mathcal C^{\rm op} \times \mathcal C \to \mathcal C$ . If $1 \in \mathcal C$ is the monoidal unit, then $\underline{\rm Hom}(1,-)$ is (naturally isomorphic to) the identity functor.

Then there are canonically defined "evaluation" and "internal composition" maps, both of which I will denote by $\bullet$ . Indeed, we define "evaluation" $\bullet_{X,Y}: X\otimes \underline{\rm Hom}(X,Y) \to Y$ to be the map that corresponds to ${\rm id}: \underline{\rm Hom}(X,Y) \to \underline{\rm Hom}(X,Y)$ under the adjuntion. Then we define "composition" $\bullet_{X,Y,Z}: \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to \underline{\rm Hom}(X,Z)$ to be the map that corresponds under the adjunction to $\bullet_{Y,Z} \circ (\bullet_{X,Y} \otimes {\rm id}) : X \otimes \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to Z$ . (I have supressed all associators.)

Question: Is $\bullet$ an associative multiplication? I.e. do we have necessarily equality of morphisms $\bullet_{W,Y,Z} \circ (\bullet_{W,X,Y} \otimes {\rm id}) \overset ? = \bullet_{W,X,Z} \circ ({\rm id}\otimes \bullet_{X,Y,Z})$ of maps $\underline{\rm Hom}(W,X) \otimes \underline{\rm Hom}(X,Y) \otimes \underline{\rm Hom}(Y,Z) \to \underline{\rm Hom}(X,Z)$ ? If not, what extra conditions on $\otimes$ are necessary/sufficient?

Answer by Evan Jenkins for If "tensor" has an adjoint, is it automatically an "internal Hom"?

2010-04-14T20:33:38Z

It is associative. Consider the evaluation cube drawn here. Four of the faces commute by definition of the composition map, and one by functoriality of the tensor product. The commutativity of these five faces implies that any of the maps $W \otimes \operatorname{Hom}(W, X) \otimes \operatorname{Hom}(X, Y) \otimes \operatorname{Hom}(Y, Z) \to Z$ are equal, so by adjunction, the two composites of compositions are equal.

Answer by Buschi Sergio for If "tensor" has an adjoint, is it automatically an "internal Hom"?

2010-05-25T21:46:52Z

In S. Eilenberg and G. M. Kelly Closed categories, in Proc. C. O. C. A.. (La Jolla, 1965),

These is a comprehensive study about Monoidal and Close structure on a category, and the relation and equivalence between these.

If "tensor" has an adjoint, is it automatically an "internal Hom"? - MathOverflow

If "tensor" has an adjoint, is it automatically an "internal Hom"?

Answer by Evan Jenkins for If "tensor" has an adjoint, is it automatically an "internal Hom"?

Answer by Buschi Sergio for If "tensor" has an adjoint, is it automatically an "internal Hom"?