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?
First of all, this answer could be of interest for you. Moreover, you may find an (in)formal treatment of the general case in Majid, Foundations of Quantum Group Theory, section 9.4.
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).