What is the adjunct of the evaluation morphism in a closed monodical category?

According to the nlab article about evaluation map if $$X, Y \in C$$, a closed monodical category then the adjunct to evaluation morphism $$[X, Y]\otimes X \rightarrow Y$$ is the identity morphism $$[X, Y] \rightarrow [X, Y]$$.

Is that correct and if yes why?
Because it seems to me that the adjunct of such has to be $$X \rightarrow [Y, (X \otimes Y)]$$

• nlab is already correct. The answer below is too as a textbook-like answer. But it would help us correct your misunderstanding the most if you explain WHY you think adjunct of $ev: [X,Y]\otimes X \rightarrow Y$ has type $X\rightarrow [Y,(X\otimes Y)]$ Oct 27, 2019 at 2:43
• @Apiwat. Thank you very much for your concern. The reason I think so is that according to nlab article for adjunct, $f: LX \rightarrow Y$ and $g: X \rightarrow RY$ are adjunct of each other if L and R are left and right adjoints respectively. When I draw the diagram the first adjoint matches what I have for $\epsilon$ that is $[X,Y] \otimes X \rightarrow Y$, but at the same time my $\eta$ shows as $X \rightarrow [Y, (X \otimes Y)]$ which I believe should be the other adjunct unless I'm wrong. Oct 27, 2019 at 16:43

2 Answers

Adjuncts are morphisms related by the natural isomorphism:

$$Hom(FA, B) \cong Hom(A, GB)$$

defining the adjunction. The adjuncts of (co)units are identities:

$$(ε_A : FGA → A) \longleftrightarrow (1_{GA} : GA → GA) \\ (η_A : A → GFA) \longleftrightarrow (1_{FA} : FA → FA)$$

This correspondence is basically what connects the $$Hom$$ definition of adjunctions with the (co)unit definition.

In this case, $$FA = A \otimes X$$ and $$GA = [X,A]$$.

Your question is not specific to the evaluation map.

Consider any morphism $$f : A \otimes B \to C$$ in a in a symmetric monoidal (not "monodical") category. There are two adjuncts because $$A \otimes B \cong B \otimes A$$:

1. We can take $$A$$ to the other side of the arrow to get $$f_1 : B \to [A, C]$$.
2. We can take $$B$$ to the other side of the arrow to get $$f_2 : A \to [B, C]$$.

Now let us apply this to the evaluation map $$e : [X, Y] \otimes X \to Y$$:

1. In the first case we get $$e_1 : X \to [[X,Y], Y]$$.
2. In the second case we get $$e_2 : [X,Y] \to [X,Y]$$. This is what the nLab page is talking about.

Now, if I had to guess where you made an error, I would hypothesize that you considered the first case, and additionally you confused which is the left and which is the right adjoint: while it is the case that $$[Y, [X, Y]] \cong [X \otimes Y, Y]$$, it is not the case that we can just switch things around and expect also that $$[[X, Y], Y] \cong [Y, X \otimes Y]$$. Internal hom-sets are not symmetric!