Is tensor product exact in abelian tensor categories with duals?

Question

Suppose we are in an abelian tensor category with duals, where all objects have finite length. Let $0 \to A \to B \to C \to 0$ be a short exact sequence and $Z$ an object of the category. Is $$0 \to Z \otimes A \to Z \otimes B \to Z \otimes C \to 0$$ exact?

Motivation: I am reading the proof of Proposition 5.7 in this paper of Deligne and trying to figure out why the lower sequence at the bottom of page 23 is exact. I believe $\mathcal{H}om(X,Y)$ here is $X^{\vee} \otimes Y$, although I have not actually found the point in the paper where he defines it. What he is trying to prove is that the corresponding sequence of external Hom's is exact, so he can't be using that fact.

There are, of course, tons of abelian tensor categories where $\otimes$ is not exact. For example, modules for any commutative ring $A$ which is not a field, with tensor product $\otimes_A$. But these don't usually have duals for all of their objects.

Answer 1

Yes, because if Z has a dual, then in particular Z ⊗ – has a left adjoint (Z^* ⊗ –) and hence commutes with limits (and similarly with colimits, but that's automatic if the category is closed monoidal).