Lemma 73.9.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$. If $f$ is flat at $x$ and $X$ is geometrically unibranch at $x$, then $Y$ is geometrically unibranch at $f(x)$.

Proof. Consider the map of étale local rings $\mathcal{O}_{Y, f(\overline{x})} \to \mathcal{O}_{X, \overline{x}}$. By Morphisms of Spaces, Lemma 66.30.8 this is flat. Hence if $\mathcal{O}_{X, \overline{x}}$ has a unique minimal prime, so does $\mathcal{O}_{Y, f(\overline{x})}$ (by going down, see Algebra, Lemma 10.39.19). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).