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

Question

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 1

It is associative. Consider the following evaluation cube:

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 2

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.

EDIT (I explain better):

Let given a monoidal category $\mathscr{C}, \otimes , I, l, r)$ if for each $B, C\in \mathscr{C}$ I have natural isomorphism $\pi: (A \otimes B, C) \cong (A, [B, C])$, we can get (by Yoneda lemma) naturally a functor $[-, ?]: \mathscr{C}^{op}\times \mathscr{C} \to \mathscr{C} $ and a parametric adjunction described by the above isomorphism.

From [EK] subsection 3, p. 477 (with a condition about a functor $V: \mathscr{C}\to Set$ that is not essential, you can chose $V= (I, -)$ with little modification to the [EK] esposition), we have that $\mathscr{C}, \otimes , I, l, r)$ is a monoidal closed category (see [EK] p. 475). From [EK] T.5.2 p.445, $\mathscr{C},$ is a enriched category enriched on itself considered as closed category. From [EK] T. 6.4 p. 468, $\mathscr{C}$ is also enriched on itself considered considered as monoidal category and in particular you get the associativity condition VC3' of [EK] p. 496.