Suppose we have a commutative square
$\require{AMScd}$
\begin{CD}
A @>{f}>> B\\
@V{h}VV @V{k}VV \\
C @>{g}>> D
\end{CD}
of quasi-categories, such that $f,g,h,k$ are cofibrations in the Joyal model structure (i.e. these are monomorphisms of simplicial sets). Suppose further that $A,B,C,D$ have all finite coproducts, and that $f,g,h,k$ preserve these finite coproducts. Let $E$ be the homotopy pushout of $f$ along $h$ in the Joyal model structure. Assume that $E$ is a quasi-category. Does $E$ have all finite coproducts, and do the functors $B\to E, C\to E, E\to D$ preserve these?


I think the answer should be yes. A model for $E$ could be the following. We know that $B\cup C \subset D$ is the ordinary pushout, hence a homotopy pushout, since the Joyal model structure is left proper. Then we can take $E$ to be the smallest sub-simplicial set of $D$ such that all squares
\begin{CD}
\Lambda^k[n] @>{}>> E\\
@V{}VV @V{}VV \\
\Delta[n] @>{}>> D
\end{CD}
for $0<k<n$ have a diagonal filler $\Delta[n] \to E$. I think an $n$-simplex $\sigma:\Delta[n] \to D$ is then in $E$ if and only if there is some surjection $\delta:[m] \to [n]$ such that all edges of the spine of $\delta^\*\sigma:\Delta[m] \to D$ are in $B\cup C$.


We have an initial object $0\in A$ which is also initial in $B,C$ and $D$. This should give the initial object of $E$, but I have a hard time proving that $E(0,X)$ is contractible, even for $X \in B \setminus C$. It seems to get quite messy to show this by hand, and it would be nice if there is either a slick argument or a reference I could use.


**Edit** Simon's example convinced me the answer to the question is probably no, and that at least my construction of $E$ is wrong as stated. I'd like to add the condition $B \cap C = A$, to more closely reflect the situation I am interested in.



Pushout of quasi-categories with finite coproducts
A pushout $B \coprod\_A C$ will almost never have all coproduct. the problem is that objects in $B \coprod\_A C$ are all either objects of $B$ or objects of $C$, so if $B \coprod\_A C$ has coproduct, it means that every time you take the coproduct of $b \in B$ with $c \in C$ it would have to be either in $B$ or in $C$.


To give a concrete example. Take $D$ to be the quasi-category of pairs of spaces, so $D=\mathcal{S} \times \mathcal{S}$.


Take $B$ to be the full subcategory of pairs whose first component is empty, and $C$ the full subcategory of objects whose second component is empty. $B$ and $C$ are equivalent to the quasi-category $\mathcal{S}$ of spaces, their intersection $A$ is equivalent to the terminal category $\Delta[0]$ (it only contains the empty space).


The homotopy pushout $B \coprod\_A C$ can be shown to identify with the full subcategory of $D = \mathcal{S} \times \mathcal{S}$ of pairs of spaces $X \times Y$ where either $X$ or $Y$ is empty (this is not completely trivial). And it is not closed under coproducts in $D$.


