Exactness of functors in a $C^*$-tensor category I'm reading the book "Compact quantum groups and their representation categories" by Neshveyev-Tuset. In section 2.3 "Fiber functors and reconstruction theorems*, the following definition is given for a $C^*$-tensor category $\mathscr{C}$:
Definition: A tensor functor $F: \mathscr{C}\to \text{Hilb}_f$ is called a fiber functor if it is faithful and exact.
Question: What does exactness of a functor mean in this context? I know what it means in the context of (semi)abelian categories where it means that the functor $F$ preserves short exact sequences, but I believe in a general $C^*$-tensor category (without dual objects) it is possible that kernels and cokernels do not exist, so the aforementioned notion of exactness doesn't seem to apply in this situation.
 A: I agree, and wanted to follow up on @Noah's comment.
Following the definition of $\mathcal C^*$ tensor categories, there are only two assumptions that allow us to construct new objects:

(vi) $\mathcal C$ has finite direct sums in the sense that for any objects $U$ and $V$ there exist
an object $W$ and isometries $u\in\mathsf{Mor}(U, W )$ and $v\in\mathsf{Mor}(V, W )$ (that is, $u^∗u = 1$
and $v^∗v = 1$) such that $uu^∗ + vv^∗ = 1$;
(vii) $\mathcal C$ has subobjects in the sense that for every projection $p\in\mathsf{End}(U)$ there exists
an object $V$ and an isometry $v\in\mathsf{Mor}(V, U)$ such that $vv^∗ = p$.

In particular, only condition (vii) creates subobjects.  If a surjection $f:U\twoheadrightarrow V$ doesn't split, we have no way of guaranteeing that the kernel of $f$ exists.  Thus, you are right in suspecting that these categories need not be abelian.  However buried in the lower half of page 28 is the sentence:

Furthermore, we will mainly deal
with semisimple categories, meaning that every object is a direct sum of simple
ones.

... which implies that the category is abelian.  I think it's safe to assume that the author is not attempting to deal with the situation where $\mathcal C$ is not abelian.  The discussion immediately following the definition of a fiber functor supports this conclusion as well.
(This bit is just speculation) If I were attempting to formulate a version in the nonabelian setting, I would probably assume that $\otimes$ and all functors are right-exact (cocontinuous) in each argument, in addition to being linear.  Without this assumption you would need to work very hard in order to deal with exceptional examples, which would probably best be dealt with using ad hoc techniques.
