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 $X \rightarrow [Y, (X \otimes Y)]$

  • 1
    $\begingroup$ 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 $X\rightarrow [Y,(X\otimes Y)]$ $\endgroup$ – Apiwat Chantawibul Oct 27 '19 at 2:43
  • $\begingroup$ @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. $\endgroup$ – al pal Oct 27 '19 at 16:43

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!


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