TITLE: The skew monoidal structure induced by a functor
QUESTION [6 upvotes]: $\require{AMScd}$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" holds in a generic 2-category.

REPLY [2 votes]: 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.