Going from the unit-counit definition of the dual pair to the Hom-adjunction

Question

In "Categories and Sheaves" by Kashiwara and Schapira on the page 101, the definition of the dual pair in tensor category is given and in the next theorem it is shown that if $(X,Y)$ is a dual pair, then for every pair of objects $W, Z$ we have $$Hom(Z, W\otimes X)\cong Hom(Z\otimes Y, W)$$ He claims that the maps $$Hom(Z, W\otimes X)\ni u\mapsto \bigg(Z\otimes Y\xrightarrow{u\otimes Id_Y}W\otimes X\otimes Y\xrightarrow{Id_W\otimes \eta}W\otimes 1\cong W\bigg)\in Hom(Z\otimes Y, W)$$ and $$Hom(Z\otimes Y, W)\ni v\mapsto \bigg(Z\cong Z\otimes 1\xrightarrow{Id_Z\otimes\epsilon}Z\otimes Y\otimes X\xrightarrow{v\otimes Id_X}W\otimes X\bigg)\in Hom(Z, W\otimes X)$$ are mutually inverse. However, I was unable to show that (of course I know that unit-counit equations should somehow be involved, but I can't make any use of them) or find any other resource which tells about the proof of that fact something more than that "it is an easy diagram chase". For sure I am missing something obvious. Any help with that problem would be highly appreciated.

Answer 1

Here's a nice way to see this:

Let $C$ be a monoidal category. Then the functor $ad: C\to Fun(C,C)$ given by $X\mapsto X\otimes -$ has a canonical monoidal structure, where $Fun(C,C)$ is (strict) monoidal with composition.

In fact, the monoidal structure just comes from the associators: $ad(X)\circ ad(Y) = X\otimes (Y\otimes -) \cong (X\otimes Y)\otimes -$

The fact that this does in fact define a monoidal structure on $ad$ uses the pentagon identity (this is why it's there in a sense)

In particular, a dual pair $(X,Y)$ gives you functors $F,G: C\to C$ given by $X\otimes -,Y\otimes-$, with a unit $\eta: id \to GF$ (because $id\cong 1\otimes -$) and co-unit $\epsilon : FG\to id$ satisfying the triangle identities. In other words, this defines an adjunction.

Now the fact that adjunctions induce bijections as you suggest is well-known, this is e.g. proposition 1.4 on this nLab page

The point is simply the following: start with $f: Fa\to b$. Apply $G$ to that, to get $GFa \to Gb$, and precompose with $\eta_a$ to get $a\to Gb$. Now apply $F$ to that and postcompose with $\epsilon_b$: you want to prove that you get $f$ back.

By definition, the last map is the composite $Fa\to FGFa \overset{FGf}\to FGb \to b$. But $\epsilon$ is a natural transformation, so you have the following commutative diagram :

$$\require{AMScd}\begin{CD}Fa @>>> FGFa @>{FGf}>> FGb \\ & @V{\epsilon_a}VV @V{\epsilon_b}VV \\ & & Fa @>{f}>> b\end{CD}$$

Now, by definition, the new map you get is the composite "right-right-down". By commutativity of the diagram, this is "right-down-right". By the triangle identity, this is the same as just $f$, because "right-down" is $id_{Fa}$.

This proves that the composite $\hom(Fa,b)\to \hom(a,Gb)\to \hom(Fa,b)$ is the identity, and of course the dual version shows that the other composite is also the identity, so they are inverse bijections.