$\require{AMScd}$Let $\cal K$ be a 2-category, and $j : A\to B$ one of its 1-cells. Assume that the induced map
$$
j^\* : {\cal K}(B,B)\to {\cal K}(A,B)
$$ precomposing with $j$ has a left adjoint $j\_!$, the *left extension along $j$*.


Given 1-cells $x,y\in {\cal K}(A,B)$, it is possible to obtain canonical maps


* $\gamma\_{xyz} : j\_! (j\_!x\circ y) \circ z \to j\_!x\circ j\_!y\circ z$, a map that in turn is determined by a map $\bar\gamma : j\_!(j\_!x\circ y) \to j\_!x\circ j\_! y$ via the universal property of $j\_!$, so that $\gamma = \bar\gamma \* z$; in ${\cal K}=\bf Cat$, a sufficient condition for this to be invertible is that left extensions along $j$ are absolute.
* $\sigma : j\_!(j)\to 1\_B$ the counit of the density comonad of $j$; in ${\cal K}=\bf Cat$, this is invertible if and only if $j$ is dense.
* $\eta : x \to j\_!(x)\circ j$, the $x$-component of the unit of $j\_!\dashv j^\*$; in ${\cal K}=\bf Cat$, this is invertible if and only if $j$ is fully faithful.


I want to prove that the following two diagrams of 2-cells are commutative:
$$
\begin{CD}
j\_!(j\_!x\circ y)\circ j @>\gamma\_{xyj}>> j\_!x\circ j\_!y\circ j\\
@A\eta\_{j\_!x\circ y}AA @| \\
j\_!(x)\circ y @>>j\_!x \* \eta\_y> j\_!x\circ j\_!y\circ j
\end{CD}
\begin{CD}
j\_!(j\_!j\circ x)\circ y @>j\_!\sigma \* y>> j\_!x\circ y \\
@| @AA\sigma \* j\_!x\circ y A \\
j\_!(j\_!j\circ x)\circ y @>>\gamma\_{jxy}> j\_!j\circ j\_!x \circ y
\end{CD}
$$
But, as I expected when I began, this is going to be quite painful. Any advice to make the proof simpler and clearer?


A conceptual proof in ${\cal K}=\bf Cat$, that can be adapted to a generic $\cal K$, is welcome!


**Edit** (Jul 23, 2018). What I am trying to prove is that the claim that appears as Theorem 3.1 in "[Monads need not be endofunctors](https://arxiv.org/pdf/1412.7148.pdf)" holds in a generic 2-category.



The skew monoidal structure induced by a functor
There is a fast and down to earth proof. We use the letters $R$ and $L$ to indicate the right and left adjoints of morphisms.


For the first one, just call $a=j\_!x\circ y$ and $b=j\_!x\circ j\_!y$, we have one morphism $\phi:a\to j^\*b$ with left adjoint $L\phi:j\_!a\to b$. Then your diagram becomes
$$\begin{CD}
j^\*j\_!a @>j^\*L\phi>> j^\*b\\
@A\eta AA@|\\
a@>\phi>>j^\*b
\end{CD}$$
which is an easy $1$ categorical fact about adjunction. It's just the right adjoint diagram of $L\phi\circ\text{id}=\text{id}\circ L\phi$, with $\eta=R\text{id}$ and $\phi=RL\phi$.


For the second one, observe that you can prove commutativity before composing with $y$: every arrow has the form $\phi\ast y$ for some $\phi$. Now just use adjunction of $j^\*$ and $j\_!$: your diagram becomes
$$\begin{CD}
j\_!j\circ x @>Rj\_!(\sigma\ast\text{id})>> j^\*j\_!x\\
@|@A\sigma\ast\text{id} AA\\
j\_!j\circ x@>\text{id}\ast\eta>>j\_!j\circ j^\*j\_!x
\end{CD}$$


We can now unpack
$$Rj\_!(\sigma\ast\text{id}):j\_!j\circ x\to j^\*j\_!x$$
as
$$j\_!j\circ x\xrightarrow{\sigma\ast\text{id}} x\xrightarrow{\eta} j^\*j\_!x$$
To check that this unpacking is right, just observe that $R\eta=\text{id}$. Hence the diagram above becomes


$$\begin{CD}
x @>\eta>> j^\*j\_!x\\
@A\sigma\ast\text{id} AA@A\sigma\ast\text{id} AA\\
j\_!j\circ x@>\sigma\ast\eta>>j\_!j\circ j^\*j\_!x
\end{CD}$$
which is obvious.


