Tensor functor between rigid tensor categories preserves $\text{Hom}$-objects

Question

I was looking at these notes on Tannakian categories. Let me briefly recall the notion of tensor functors: Let $(\mathcal{C},\otimes)$ and $(\mathcal{C'},\otimes')\DeclareMathOperator{\uphom}{\underline{Hom}} \DeclareMathOperator{\hom}{Hom}$ be two rigid tensor categories. (He defines the notion of rigid tensor categories in page 2 in his notes). A tensor functor $(\mathcal{C},\otimes)\rightarrow (\mathcal{C'},\otimes')$ is a pair $(F,c)$ where $F:\mathcal{C}\rightarrow \mathcal{C'}$ is a functor and $c$ is a collection of isomorphisms $c_{X,Y} : F(X)\otimes F(Y)\rightarrow F(X\otimes Y)$, functorial in $X$ and $Y$, that “commute with the associativity and commutativity constraints” in the obvious way, and send identity objects to identity objects.

Next, the author states the following fact, which is not clear to me:

Claim: One checks that if $\mathcal{C}$ and $\mathcal{C'}$ are rigid, then the axioms force the induced map $F(\uphom(X,Y))\rightarrow \uphom(FX,FY)$ is an isomorphism, where $\uphom(X,Y)$ is the object that represents the functor $\mathcal{C}^{op}\rightarrow\operatorname{Sets}$ $$T\mapsto \hom(T\otimes X, Y)\,.$$

Could someone please explain/ suggest some reference where I can find a proof of this fact?

Thanks in advance!

Answer 1

Your $F$ preserves dual objects because it preserves the duality situations: quadruples $(X,Y,\alpha : X\otimes Y \rightarrow I, \beta: I \rightarrow Y \otimes X)$.

It remains to recall the formula for the internal hom: $$ \underline{\mathrm{Hom}} (X,Y) = X^\ast \otimes Y. $$

Answer 2

$\DeclareMathOperator{\uphom}{\underline{Hom}}$

Looking at the notes, I can see that $\uphom(X,Y)$ is the internal hom of the category.

This has the universal property that that maps into it $ Z\to \uphom(X,Y)$ are in bijection with maps $ Z\otimes X \to Y$.

A suitable proof of your statement might go like this:

The map $$ F(\uphom(X,Y))\to \uphom(FX,FY)$$ corresponds to a map (from property above) $$ F(\uphom(X,Y))\otimes FX\to FY$$ which in turn corresponds to a map (from $F$ being a tensor functor) $$ F(\uphom(X,Y)\otimes X)\to FY$$

If the counit of the adjunction $-\otimes X \dashv \uphom(X,-)$ is called $ev:\uphom(X,Y)\otimes X\to Y$, then one may check that the map above is $F(ev)$. And since the isomorphism between homs defined by the adjunction takes (naturally) its counit to the identity morphim on $\uphom(X,Y)$. Then the functor $F$ must take $ev$ to an isomorphism $F(ev)$.

--- edit ---

Here are the corners of a relevant square depicting the chase of $ev$. Where square brackets denote the external hom sets. $$ [\uphom(X,Y)\otimes X, Y] \cong [\uphom(X,Y),\uphom(X,Y)] $$ $$ [F(\uphom(X,Y)\otimes X), F(Y)] \cong [F(\uphom(X,Y)),F(\uphom(X,Y))] $$ There should be vertical arrows with labels $F$ which I'll won't draw now. :)