A closed symmetric monoidal category is enriched over itself.

Dear mathstackexchange,

I have for some hours been grappling with what should be an easy diagram chase but turns out to be a bit more involved than I would imagine. Let $\mathcal{V}$ be a closed symmetric monoidal category, with $\otimes$ the monoidal product so that we have a natural isomorphism $\mathcal{V}(A \otimes B,C) \cong \mathcal{V}(A, \underline{\mathcal{V}}(B,C))$ with $\underline{\mathcal{V}}(-,-)$ the adjoint to the monoidal product.

I want to show that $\mathcal{V}$ is enriched over itself, with the hom-objects between two objects $A,B \in \mathcal{V}$ defined to be $\underline{\mathcal{V}}(A,B)$. One gets a composition morphism $$\mu:\underline{\mathcal{V}}(B,C) \otimes \underline{\mathcal{V}}(A,B) \rightarrow \underline{\mathcal{V}}(A,C)$$ by requiring it to be adjoint to the morphism $$\underline{\mathcal{V}}(B,C) \otimes ( \underline{\mathcal{V}}(A,B) \otimes A) \xrightarrow{1 \otimes ev^A_B} \underline{\mathcal{V}}(B,C) \otimes B \xrightarrow{ev^B_C} C$$ where the evaluation morphisms are the adjoints to the identity $\underline{\mathcal{V}}(A,B) \rightarrow \underline{\mathcal{V}}(A,B)$. Now, I want to show that this composition operation is associative. I considered the first thing I could think of, using adjointness. So I wrote up a diagram:

$$\require{AMScd} \begin{CD} \mathcal{V}(\underline{\mathcal{V}}(C,D) \otimes \underline{\mathcal{V}}(B,C) \otimes \underline{\mathcal{V}}(A,B),\underline{\mathcal{V}}(A,D)) @<(1 \otimes \mu)^\ast<< \mathcal{V}(\underline{\mathcal{V}}(C,D)\otimes \underline{\mathcal{V}}(A,C),\underline{\mathcal{V}}(A,D)) \\ @VVV @VVV \\ \mathcal{V}(\underline{\mathcal{V}}(C,D) \otimes \underline{\mathcal{V}}(B,C) \otimes \underline{\mathcal{V}}(A,B) \otimes A , D) @<(1\otimes\mu \otimes 1 )^\ast << \mathcal{V}(\underline{\mathcal{V}}(C,D)\otimes \underline{\mathcal{V}}(A,C) \otimes A,D)\end{CD}$$

where the vertical arrows are isos and got that the morphism $\underline{\mathcal{V}}(C,D) \otimes ( \underline{\mathcal{V}}(B,C)\otimes \underline{\mathcal{V}}(A,B) ) \xrightarrow{1 \otimes \mu} \underline{\mathcal{V}}(C,D) \otimes \underline{\mathcal{V}}(A,C) \xrightarrow{\mu} \underline{\mathcal{V}}(A,D)$ is adjoint to the morphism $(ev^C_D(1 \otimes ev^A_C))(1 \otimes \mu \otimes 1).$ In the same way, I write up a diagram of the following form to see what $( \underline{\mathcal{V}}(C,D) \otimes \underline{\mathcal{V}}(B,C) )\otimes \underline{\mathcal{V}}(A,B) \xrightarrow{\mu \otimes 1} \underline{\mathcal{V}}(B,D) \otimes \underline{\mathcal{V}}(A,B) \xrightarrow{\mu} \underline{\mathcal{V}}(A,D)$ is adjoint to. The diagram is

$$\require{AMScd} \begin{CD} \mathcal{V}(\underline{\mathcal{V}}(C,D) \otimes \underline{\mathcal{V}}(B,C) \otimes \underline{\mathcal{V}}(A,B),\underline{\mathcal{V}}(A,D)) @<(\mu \otimes 1)^\ast<< \mathcal{V}(\underline{\mathcal{V}}(B,D)\otimes \underline{\mathcal{V}}(A,B),\underline{\mathcal{V}}(A,D)) \\ @VVV @VVV \\ \mathcal{V}(\underline{\mathcal{V}}(C,D) \otimes \underline{\mathcal{V}}(B,C) \otimes \underline{\mathcal{V}}(A,B) \otimes A , D) @<(\mu \otimes 1 \otimes 1 )^\ast << \mathcal{V}(\underline{\mathcal{V}}(B,D)\otimes \underline{\mathcal{V}}(A,B) \otimes A,D)\end{CD}$$ so that the above composition is adjoint to $(ev^B_D(1 \otimes ev^B_A))(\mu \otimes 1 \otimes 1)$. So what we now want to show is that, actually: $$(ev^B_D(1 \otimes ev^B_A))(\mu \otimes 1 \otimes 1) = (ev^C_D(1 \otimes ev^A_C))(1 \otimes \mu \otimes 1).$$

I am a bit unsure how to prove this to be honest - should I use some unit-counit to reduce it further? This should all be routine of course, but I am having some problems so anything would be more than greatly appreciated.

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@SanathDevalapurkar I don't understand your comment - isn't that what I have done? – user101036 Mar 30 '14 at 18:44

I think I see now what I was missing. One notes that under the counit-unit adjunctions, actually, $ev^A_C (\mu \otimes 1 )= ev^B_C \circ (1 \otimes ev^A_B)$ and similarily for the other expression. Thus, $(ev^C_D(1 \otimes ev^A_C))(1 \otimes \mu \otimes 1)= ev^C_D(1 \otimes (ev^B_C \circ(1 \otimes ev^A_B))).$ One can then see that $ev^B_D( \mu \otimes 1) = ev^C_D \circ (1 \otimes ev^B_C).$ Thus we get that $(ev^B_D(1 \otimes ev^B_A))(\mu \otimes 1 \otimes 1) = ev^C_D(1 \otimes (ev^B_C \circ (1 \otimes ev^A_B))).$ The diagram is by this commutative. I might have made some bracketing errors, but the idea should be clear, both sides are evaluation first from $A$ to $B$, then from $B$ to $C$ and then from $C$ to $D$.