Left split subobject in a $2$-category

Question

Let $\mathcal{K}$ be a $2$-category. Keeping in mind the Cat-intuition on $\mathcal{K}$ I say that:

Def : A $1$-cell $A \stackrel{f}{\to} B$ is a left split subobject if $f$ is a right adjoint and the counit is invertible.

In Cat this condition is equivalent to be fully faithful (and a right adjoint) for $f$.

In Lemma 2.3 of his 2-categories companion Steve Lack points out that:

In a 2-category, when the counit is invertible then the right adjoint is representably fully faithful.

Thus one implication remains true. Is it possible to have sort of a converse for this statement? Even adding somme additional structure on $\mathcal{K}$.

I was hoping to have something like the following:

A $1$-cell is a left split subobject if and only if $f$ is a right adjoint and a monomorphism in $\mathcal{K}$,

which - I understand - is a vain expectation. I was hoping to use the left cancellation in the triangle equality somehow. Unfortunately it is not working as I expected.

Answer 1

This is true if by "monomorphism" you mean "representably fully faithful" as in the result of Lack: just argue representably. If $f:A\to B$ has a left adjoint $\ell:B\to A$ in $\mathcal K$, then $\mathcal K(X,f): \mathcal K(X,A) \to \mathcal K(X,B)$ has a left adjoint $\mathcal K(X,\ell)$ in $\mathrm{Cat}$ for any $X\in \mathcal K$. If $f$ is representably fully faithful, then by definition $\mathcal K(X,f)$ is fully faithful. Thus, the counit of the adjunction $\mathcal K(X,\ell)\dashv \mathcal K(X,f)$ is an isomorphism in $\mathrm{Cat}$. This counit has components that are 2-cells $\ell f g \to g$ in $\mathcal K$ for $g:X\to A$, obtained by whiskering with the counit $\epsilon : \ell f \to 1_A$. So just take $X=A$ and $g = 1_A$ to conclude that $\epsilon$ is an isomorphism.