Let $X\to Y$ be a map between algebraic stacks, and $U\to Y$ a smooth atlas of $Y$. Suppose we know that $X\times\_Y U$ is a scheme, can we show that $X\to Y$ is schematic?



Can being schematic be checked on an atlas?
No. Consider the morphism $pt \to BG$ and the smooth atlas $pt \to BG$, for a smooth group scheme $G$. Then $pt \times\_{BG} pt = G$ is a scheme. However, $pt \to BG$ is in general only representable by algebraic spaces, and may not be schematic. See [this answer](https://mathoverflow.net/a/316738) for an example.


However, in your situation, one can always conclude that $X \to Y$ will be representable (by algebraic spaces, not necessarily by schemes). This follows from the fact that algebraic spaces satisfy étale descent, and even fppf descent, see [Stacks 04SJ](https://stacks.math.columbia.edu/tag/04SJ). Strictly speaking to apply the cited reference you need to reduce to the case where $Y$ is a scheme first. In more details:


Let $T \to Y$ be a morphism with $T$ a scheme and let $S = T \times\_Y X$. We need to show that $S$ is an algebraic space. By assumption there exists a smooth atlas $Y\_0 \to Y$ such that the base change $X\_0 = Y\_0 \times\_Y X \to X$ is a smooth atlas with $X\_0$ also a scheme. Base change everything along $Y\_0 \to Y$ so that you get a cartesian square $$\require{AMScd}\begin{CD}
U @>>> V\\
@VVV @VVV\\
S @>>> T
\end{CD}$$
where $V = T \times\_Y Y\_0$, $U = S \times\_X X\_0$, and the vertical maps are smooth surjections. Also, $U$ and $V$ are algebraic spaces; that's because $X\_0 \to X$ and $Y\_0 \to Y$ are representable (automatic from the definition of algebraic stack). In particular, the upper map is representable. In other words, $S \to T$ is representable "smooth-locally on the target" (i.e. becomes representable after base change along a smooth atlas). Now the Stacks Project reference applies and implies that $S \to T$ itself is representable, hence $S$ is an algebraic space.


