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 $X\rightarrow [Y,(X\otimes Y)]$ $\endgroup$ Oct 27, 2019 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, 2019 at 16:43

2 Answers 2


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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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