I couldn't add to the well-written $n$Lab page about relative adjoints, so let me start taking that definition for granted.

I have a few questions about how the classical results on adjoints remain true, even in this fairly asymmetric setting.

I would like to understand better the meaning of this structure (e.g. when, how and why does it arise "in practice"). I collect here a few questions I have no clue how to attack.

Let's assume that there are relative adjunctions $f\dashv_j g$ and $f' \,{}_{j'}\!\!\dashv g'$; we can then arrange a pair of commutative triangles $$ \begin{array}{ccc} \bullet & \overset{f}\to & \bullet\\ \hskip-4mm g\uparrow & \nearrow \\ \bullet && \end{array} \qquad \epsilon : fg \Rightarrow j $$ $$ \begin{array}{ccc} &&\bullet \\ &\nearrow& \hskip5mm \uparrow g'\\ \bullet &\underset{f'}\to& \bullet \end{array} \qquad \eta : j' \Rightarrow g'f' $$ assume, now, that there exists a natural transformation $j\Rightarrow j'$; under which assumptions the resulting pasting square is filled by an invertible 2-cell?

Assume that $$ g_1 \dashv_{j_1} f \;{}_{j_2}\!\!\dashv g_2 $$ are relative adjoints; in this particular case it is possible to build a 2-cell $j_2 g_1\Rightarrow g_2 j_1$ pasting the unit of $f \;{}_{j_2}\!\!\dashv g_2$ with the counit of $g_1 \dashv_{j_1} f$. Under which conditions is this 2-cell an iso? In the same situation, the composition $j_1 j_2$ is parallel to $f$; is there a "comparison 2-cell" to/from $f$, and if there is, under which conditions it is invertible?

The $n$Lab says that $f \; {}_j\!\!\dashv g$ if $f\cong \text{LIFT}_gj$ (absolute lift); is this an if

**f**? This relation means that $g$ uniquely determines $f$; the converse is not true in general: does it mean that there can be two non-isomorphic $g,g'$ such that $f \; {}_j\!\!\dashv g$*and*$f \; {}_j\!\!\dashv g'$? Is there an instructive example of this? What is the structure, if any, of the class $\{g\mid f \; {}_j\!\!\dashv g\}$?If $j$ is an isomorphism and $f \;{}_{j}\!\!\dashv g$, then this means that $f\dashv j^{-1}g$; is this condition any different from the particular case $f\dashv g = f \;{}_1\!\!\dashv g$? Is there a nontrivial, instructive example of this situation? I am confused by the loss of uniqueness outlined in the previous point...

- One of the reasons why relative adjunctions are useful is that they model weighted co/limits: writing the weight in a suitable form, one has that $j\otimes f \dashv_j \hom(f,1)$, which is, well, quite obvious if you write $$ {\bf hom}(j\otimes f,1) \cong {\bf hom}(j, \hom(f,1)) $$ how far can this analogy be pushed? (write $j\otimes(j'\otimes f) \cong \dots$: what does it mean for the associated relative adjunction?)
- What is the meaning of relations like $1_A \dashv_j g$ for suitable functors? such a situation gives an isomorphism of profunctors $\hom \cong \hom(j,g)$, so in some sense finding "all $j$-right adjoints of the identity" means finding all factorizations of the identity profunctor $\hom_A$. How far can this equivalence be pushed? What is the formal meaning of this analogy in terms of profunctor theory?
- What does the invertibilty of the relative unit $\eta : j\Rightarrow gf$ imply for $g,f$ ($f$ is "relatively fully faithful"?!). Dual question for the counit.