Choose a big $\mathit{fppf}$site $(\mathbf{Sch})_{\mathit{fppf}}$ and let $S$ be a scheme in that site.
Let $\{\mathcal{X}_i\mid i\in I\}$ be a family of stacks in groupoids over $S$ and let $\mathcal{Y}\to\mathcal{Z}$ be a morphism of stacks in groupoids over $S$.
Let $\mathcal{X}\colon\!\!=\coprod_{i\in I}\mathcal{X}_i$ be the disjoint union as described in [Champs algébriques, G.Laumon/ L.MoretBailly, (3.3)].
Let $\times$ denote the $2$fibre product of stacks in groupoids over $S$.
Is there a canonical morphism of stacks
$$\coprod_{i\in I}(\mathcal{X}_i\times_{\mathcal{Z}}\mathcal{Y})\to\Big(\coprod_{i\in I}\mathcal{X}_i\Big)\times_{\mathcal{Z}}\mathcal{Y}\quad$$ an isomorphism?

5$\begingroup$ Have you tried writing down suitable functors between the two sides? $\endgroup$ – S. Carnahan♦ Mar 1 at 14:08

2$\begingroup$ Of course $\mathbb{Z}$ can be replaced by any index set. $\endgroup$ – Martin Brandenburg Mar 1 at 21:31

1$\begingroup$ This is just universality of colimits in a higher topos. $\endgroup$ – user147129 Mar 7 at 22:46

1$\begingroup$ @S.Carnahan The definition of the coproduct of categories fibred in groupoids that sdigr is indeed correct. The point is that the (2,1)Yoneda embedding doesn't preserve colimits (nor does the (1,1)Yoneda embedding!). The stackified (2,1)Yoneda embedding, however, does commute with coproducts. $\endgroup$ – Harry Gindi Mar 8 at 1:46

1$\begingroup$ @sdigr My mistake  your guess is correct. $\endgroup$ – S. Carnahan♦ Mar 10 at 12:55
I would split this problem up into two parts (here, 'sheaf (of groupoids)' is used instead of stack in order to disambiguate between Algebraic stacks (geometric objects) and mere (pseudo)functors satisfying descent). :
1.) Show that the inclusion of algebraic stacks into the category of fppf sheaves of groupoids on Sch preserves coproducts. This follows immediately from the fact that algebraic stacks are a full (2,1)subcategory of fppf sheaves and that if $F,G$ are two algebraic stacks, their sheafy coproduct is representable by an algebraic stack. If D is a diagram landing in a full subcategory whose limit or colimit exists in the ambient category and is in the full subcategory, this is also a limit or colimit of the diagram landing in the full subcategory without reference to the ambient category. I think the proof here is immediate by taking a disjoint union of the atlases.
2.) Show that colimits are universal in (2,1)stack topoi. This follows from the leftexactness of the stackification (2,1)functor together with the altogether more obvious version of this fact for (2,1)topoi of groupoid fibrations (also called (2,1)presheaf topoi), where one can immediately reduce to proving the statement for groupoids pointwise.
An object over a scheme $T$ on the left is given by a decomposition of $T$ into a parametrized disjoint union $T_i$ of schemes, and a parametrized family of triples $(x_i, y_i, \phi_i)$, where $x_i$ is an object of $X_i$ over $T_i$, $y_i$ is an object of $Y$ over $T_i$, and $\phi_i$ is an isomorphism $\rho_{X_i}(x_i) \to \rho_Y(y_i)$ in $Z$ over $T_i$. A morphism over $id_T$ is a parametrized family of pairs of maps $(f_i: x_i \to x'_i, g_i: y_i \to y'_i)$ that satisfy suitable commutative square relations. In particular, if two objects come from unequal decompositions of $T$, then there are no morphisms between them. Let us omit discussion of other morphisms, and pretend the "fibered category" property takes care of them.
An object over a scheme $T$ on the right is given by a decomposition of $T$ into a parametrized disjoint union $T_i$ of schemes, and a tuple $((x_i), y, \phi)$, where $x_i$ is an object of $X_i$ over $T_i$, $y$ is an object of $Y$ over $T$, and $\phi$ is an isomorphism $\rho_{\coprod X_i}((x_i)) \to \rho_Y(y)$ in $Z$ over $T$. A morphism over $id_T$ is a pair $((f_i: x_i \to x'_i), g:y \to y')$ that satisfies conditions that I won't describe.
In order to match these data, we need to identify $y$ with the parametrized family $(y_i)$, and $\phi$ with $(\phi_i)$ for objects, and $g$ with $(g_i)$ for morphisms. This is just using the fibered category property: pulling back along the isomorphism $\coprod T_i \to T$ yields a decomposition of $y$ that is unique up to unique isomorphism. It might be helpful to check that the object $\rho_{\coprod X_i}((x_i))$ in $Z$ over $T$ is identified with the tuple $(\rho_{X_i}(x_i))$ of objects in $Z$ over $T_i$.

$\begingroup$ Thanks for the explanation. I understand, that one can pullback $y$ along $\coprod_i T_i\to T$ because of the fibre product property (after a choice of pullbacks was made in the very beginning). But given a family $(y_i)$ of $y_i\in \mathcal{Y}_{T_i}$ as on the left. How can one obtain $y$? I guess that more is needed than only the property that pullbacks exist uniquely. I guess that we have to keep track of the descent data coming from the stackification process aswell, do we not need this? $\endgroup$ – sdigr Mar 7 at 21:50

$\begingroup$ I don't understand the description of the $T$valued points of a disjoint union $\coprod_{i\in I}\mathcal{X}_i$ of stacks as the choice of a decomposition $T=\coprod_{i\in I}T_i$ and $(x_i)_{i\in I}, x_i\in {\mathcal{X}_i}_{T_i}$, because if I take the construction of stackification as in Lemma 02ZN literally, there does not seem to be an obvious reason why one can choose the covering indexed by the same index set $I$ and it is also not clear to me, why the descent data is not mentioned in this description. I think this causes extra confusion to me. $\endgroup$ – sdigr Mar 7 at 22:06

1$\begingroup$ @sdigr To get $y$ from $y_i$, you pull back along the isomorphism $T \to \coprod_i T_i$. As far as stackification is concerned, if you have a finer covering, then we are concerned with gluing objects over $T_i$ in $X_i$, and descent for these objects is handled by the stack property of $X_i$. $\endgroup$ – S. Carnahan♦ Mar 10 at 13:01