Let $H$ be a bialgebra in a braided (if necessary symmetric) monoidal category with unit $\eta$, multiplication $\mu$, counit $\epsilon$ and comultiplication $\Delta$. My question is:
If we assume that the "shear map" $\sigma \colon H \otimes H \xrightarrow{H \otimes \Delta} H \otimes H \otimes H \xrightarrow{\mu \otimes H} H \otimes H$ is an isomorphism, does it follow that $H$ admits an antipode, i.e. a morphism $S \colon H \to H$ such that $\mu (\operatorname{Id}_H \otimes S) \Delta = \eta \epsilon = \mu (S \otimes \operatorname{Id}_H) \Delta$?
I think the converse is true: If $S \colon H \to H$ is an antipode, then
$H \otimes H \xrightarrow{H \otimes \Delta} H \otimes H \otimes H \xrightarrow{\operatorname{Id}_H \otimes S \otimes \operatorname{Id}_H} H \otimes H \otimes H \xrightarrow{\mu \otimes H} H \otimes H$
is an inverse of $\sigma$.
Moreover, I think the statement holds if the monoidal structure is given by the categorical product: Consider $\operatorname{Hom}(H , H)$ as a monoid via the convolution product $\star$. Under the isomorphism $\operatorname{Hom}(H, H \times H) \cong \operatorname{Hom}(H, H) \times \operatorname{Hom}(H , H)$, $\operatorname{Hom}(H , \sigma)$ corresponds to $(f, g) \mapsto (f \star g, g)$. As $\sigma$ is an isomorphism, we can construct an inverse for every element of $f \colon H \to H$ with respect to convolution by considering a preimage of $(\eta \epsilon, f)$. In particular, there exists an inverse of $\operatorname{Id}_H \in \operatorname{Hom}(H , H)$, i.e. an antipode. (A dual argument should work if the tensor product is a coproduct.)
However, "splitting $H \times H$ into two factors" seems to be very crucial in this argument, so I can imagine that the statement is not true in general monoidal categories.