Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
@SanathDevalapurkar I don't understand your comment - isn't that what I have done? – user101036 Mar 30 '14 at 18:44
up vote 0 down vote accepted

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 $D$.

share|cite|improve this answer
I'm working at the same problem at the momento, so would you be so kind to explain how that last equality allows you to conclude? – Marco Vergura Apr 10 '14 at 11:31
@MarcoVergura Sure, but I might post first tomorrow if this is okay for you! – user101036 Apr 10 '14 at 13:17
If it is ok for you, it is for me ;) I'll try to work it out by myself in the meanwhile! – Marco Vergura Apr 10 '14 at 13:42
@MarcoVergura Now I have added more details. – user101036 Apr 11 '14 at 14:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.