The Stacks project

Lemma 74.9.5. Let $f : X \to Y$ be a smooth morphism of algebraic spaces. If $Y$ is reduced, then $X$ is reduced. If $f$ is surjective and $X$ is reduced, then $Y$ is reduced.

Proof. Choose a commutative diagram

\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]

where $U$ and $V$ are schemes, the vertical arrows are surjective and ├ętale, and $U \to X \times _ Y V$ is surjective ├ętale. Observe that $X$ is a reduced algebraic space if and only if $U$ is a reduced scheme by our definition of reduced algebraic spaces in Properties of Spaces, Section 66.7. Similarly for $Y$ and $V$. The morphism $U \to V$ is a smooth morphism of schemes, see Morphisms of Spaces, Lemma 67.37.4. Since being reduced is local for the smooth topology for schemes (Descent, Lemma 35.18.1) we see that $U$ is reduced if $V$ is reduced. On the other hand, if $X \to Y$ is surjective, then $U \to V$ is surjective and in this case if $U$ is reduced, then $V$ is reduced. $\square$

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