# Endomorphisms of a monoidal functor

Let $$k$$ be a field and $$\mathcal{C}$$ be a finite $$k$$-linear abelian monoidal category, and let $$F:\mathcal{C}\to\text{Vec}$$ be a monoidal functor from $$\mathcal{C}$$ to the category of finite-dimensional vector spaces with the usual monoidal structure.

Let $$\operatorname{End}(F)$$ denote endomorphisms of $$F$$ as a natural transformation. Now everything I've read (I'm especially looking at Etingof's notes now) studies this when $$F$$ is faithful and exact, in which case $$\operatorname{End}(F)$$ is a bialgebra, and $$\mathcal{C}$$ is equivalent to the category of finite-dimensional $$H$$-modules.

However I happen to be interested in a functor which is neither faithful nor exact. My question is whether or not $$\operatorname{End}(F)$$ is still bialgebra?

It seems to me it would but I'm concerned I'm making a mistake. But my reasoning is that $$\operatorname{End}(F)$$ has a natural multiplication given by composition of natural transformations and that the coproduct is given (as in the case of a faithful exact functor) by composing the isomorphism $$\operatorname{End}(F\otimes F)\cong\operatorname{End}(F)\otimes_k\operatorname{End}(F)$$ with the map $$\alpha:\operatorname{End}(F)\to\operatorname{End}(F\otimes F)$$ by the following diagram:

$$\require{AMScd}$$ $$\begin{CD} F(X\otimes Y) @>{\eta_{X\otimes Y}}>> F(X\otimes Y)\\ @VVV @VVV\\ F(X)\otimes F(Y) @>{\alpha(\eta_{X\otimes Y})}>> F(X)\otimes F(Y) \end{CD}$$

where $$\eta\in\operatorname{End}(F)$$. Am I making a mistake somewhere? Or do people generally only care about the exact and faithful case because they want to construct an equivalence of categories?

If your category $$\mathcal{C}$$ has finitely many objects, then you have $$\mathsf{End}(F\otimes F)\cong \mathsf{End}(F)\otimes \mathsf{End}(F)$$ for sure and your argument is exactly the one described by Majid. However, in general you only have a morphism $$\mathsf{End}(F)\otimes \mathsf{End}(F) \to \mathsf{End}(F\otimes F)$$. This is one of the reasons why people work in the dual case of coalgebras: $$\mathsf{Coend}(F)$$ satisfies $$\mathsf{Coend}(F\otimes F)\cong \mathsf{Coend}(F)\otimes \mathsf{Coend}(F)$$ independently on te number of objects of your category and, in general, I believe it is much better behaved (see, for example: Schauenburg, Tannaka duality for arbitrary Hopf algebras; Pareigis, Lecture notes on Quantum Groups; Street, Quantum groups, a path to current algebra).