Are all separable algebras Frobenius algebras? Let $\mathcal C$ be a [added later: semi-simple] tensor category, and let $A=(A,m:A\otimes A\to A,i:1\to A)$ be an algebra object in $\mathcal C$.
The algebra is...
Separable if there is an $A$-$A$-bimodule map $\Delta:A\to A\otimes A$ such that $m\circ\Delta=\mathrm{id}_A$
Frobenius if there is an $A$-$A$-bimodule map $\Delta:A\to A\otimes A$ that is coassociative and counital.

I'm wondering whether separable implies Frobenius.

I can show that separable implies coassociative but I suspect that separable does not imply counital. I'm having a hard time finding counterexamples.

PS: by a "tensor category", I mean a category that is monoidal (not necessarily symmetric) and linear over some field $k$.

PS2: In a previous version of this question, I had added the condition that $\mathcal C$ be rigid. I'd be actually more interested to have counterexamples where the ambient category $\mathcal C$ is semi-simple. I understand that there are many things one might mean by "semisimple" (e.g., is the category of all vector spaces over $k$ semisimple?), I'm deliberately keeping things vague to allow for more counterexamples.
 A: You should be able to get a (perhaps unnatural) counterexample by constructing a tensor category (or algebra object) with no maps to $1$. Let's try to make the most trivial such counterexample possible.
Let $C$ have two objects $A$ and $1$ with one non-identity morphism $i:1\to A$. Equip this with a structure where $1$ is the unit and $A\otimes A=A$. I believe these make $C$ unambiguously into a tensor category.
There are then unique maps $m:A\otimes A\to A$ and $\Delta:A\to A\otimes A$, both of which are the identity on $A$. Then $m$ and $i$ seem to make $A$ an algebra object in $C$ which trivially satisfies the separability condition. But this can't be counital because there are no maps $A\to 1$.
This is obviously a bit unnatural but the only "natural" tensor categories I know offhand where objects don't map to the unit are ones with disjoint union... But they can't supply any counterexamples because multiplication is forced to be the fold map but then you can't make a bimodule map $A\to A\sqcup A$ unless $A$ is empty.
