I am trying to understand the proof adjunction described in page 244-245, of Joyals Theory of Quasicategories.
Background: $$i^*:S/I \rightarrow S/\partial I = S \times S$$
- $S$ is category of simplicial sets.
- $I$ is the simplicial set $1 \star 1$, where $1$ is terminal object $\star$ is join operation (explained also in the notes). $\partial I= 1 \sqcup 1$.
- $S/I$ is over category. Objects are $X \rightarrow I$, $X \in S$.
- The construction is as follows, we given $X \rightarrow I$ in $S/I$, we $i^*X$, is the pullback of $X$ along inclusion $\partial I \rightarrow I$.
In proof Prop. 3.5, pg 245, the author assumes the existence of right adjoint $i_*$. How is this true?
The category $S$ is locally cartesian closed, as is any presheaf category.
Indeed, for any $A\in S$, the category $S/A$ is equivalent to the category of presheaves over $\Delta/A$ (the category of elements of $A$). Any map $f:A\to B$ induces a functor $\Delta/f: \Delta / A \to \Delta / B$ and you can check that the restriction functor $$ {-} \circ (\Delta/f) : \widehat{\Delta/B} \to \widehat{\Delta/A}$$ fits in a commutative square as follows: $$\require{AMScd} \begin{CD} \widehat{\Delta/B} @>{{-} \circ (\Delta/f)}>> \widehat{\Delta/A} \\ @V\simeq VV @VV\simeq V \\ S/B @>>{f^\ast}> S/A \end{CD} $$ where the two verticals maps are the equivalence mentioned above. As the top map admits a right adjoint (the right Kan extension functor), so does the bottom map. All the constructions are explicit, so you can derive a formula for the right adjoint $f_\ast$ if needed.
Alternatively you can follow the recipe of showing that exponential object exist in $S$ (this is after all the special case of $B=1$ above) and adapt it to directly describe $f_\ast$.