On the join of simplicial sets as a dependent product Prop. 3.5 of Joyal notes in quasicategories gives a description of $X\star Y$ as $i_*(X,Y)$, where $i^*\dashv i_*$ is the adjunction 
$$
i^*\colon \mathbf{sSet}/\Delta[1] \leftrightarrows \mathbf{sSet}/\partial\Delta[1]
$$
where $i^*$ is "pullback along $\partial\Delta[1]\to \Delta[1]$ and $i_*$ is the associated dependent product, which being $\partial\Delta[1]$ discrete should morally be a mere product; this baffles me a lot, since $X\star Y$ seems much more complicated than what I get if I blind myself and I try to write what $i_*(A,B)$ is for $A,B\in \bf sSet$ (confusing ${\bf sSet}/\Delta[1]$ with ${\bf sSet}^{\Delta[1]}$ it chooses $A\times B\to B$).
Here are the words of Joyal:

any clue?
I have some observations to make: somewhere there must be a typo, since either Joyal's symbol for adjointness is reversed, or he uses the left adjoint of pullback, $i_!$, which simply composes $A\to \partial\Delta[1]$ with $i$ (but this is pretty strange too, since then I am dealing with the mere coproduct map $X\sqcup Y\to \partial\Delta[1]\to\Delta[1]$). 
I have absolutely no clue about what he does in the proof so I tried something different: my alternative proof strategy was to simply prove that the unit $X\star Y\to i_*(X,Y)$ is a monic (only in this component); it's always a split epi since $i^*$ is a full functor. And, alas, I'm not able to prove this either!

By the Yoneda lemma for the category $\mathbf{\Delta}/[1]$, using the canonical equivalence $(\mathbf{\Delta}/[1])^\wedge = \mathbf{\Delta}^\wedge/\Delta^1$, it suffices to show that the morphism in question induces a bijection
  $$ Hom((\Delta^n,f), X \star Y) \to Hom((\Delta^n,f), i_*(X \sqcup Y)) $$
for each $n$, and each $f : \Delta^n \to \Delta^1$.
By adjunction, it suffices to compute
  $$ Hom(i^*(\Delta^n,f), X \sqcup Y) $$
and compare the result with the formula in 3.2.
For each morphism $f$ (there are $n+2$ choices), it is easy to compute $i^*(\Delta^n,f)$ by hand; the possible values are listed in the formula Joyal writes.
