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Let $(R, I)$ be a Henselian pair, with $I$ a finitely generated ideal.

We know that for any smooth $R/I$-algebra $A_0$, there exists a smooth $R$-algebra $A$ such that $A/I\simeq A_0$.

We also know that for any map $A_0\to B_0$ of smooth $R/I$-algebras, there exist $R$-smooth algebras $A$ and $B$, and a map $A\to B$ that lifts $A_0\to B_0$.

Suppose $A_0\to B_0$ is surjective. Can $A\to B$ be arranged to be surjective too?

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Yes. Let $i\colon Y=\mathbf{Spec}(B)\to \mathbf{Spec}(A)=X$ be the induced map of schemes, and let $K$ be the cokernel of $$ i^*\colon \mathcal{O}_X \to i_* \mathcal{O}_Y. $$ This is a coherent $\mathcal{O}_X$-module whose support does not meet $X_0$ by assumption. Thus after replacing $X$ with an affine open neighborhood of $X_0$ and $Y$ with its base change to this neighborhood we get $K=0$, so $i^*$ is becomes surjective.

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  • $\begingroup$ In the special case when $\text{Spec}(A)$ is an affine space over $R/I$, or a distinguished affine open of a projective space over $R/I$, can one lift the relations directly, to get a surjective lift $A\to B$ on the nose without replacing $\text{Spec}(A)$ with an open neighborhood of $\text{Spec}(A_0)$? I’d just like to be sure $\endgroup$
    – John P.
    Commented Apr 3, 2019 at 21:18
  • $\begingroup$ In general $i_*\mathcal{O}_Y$ is not coherent, only quasicoherent. Next, where do you use the assumptions that $(R,I)$ is henselian and $A$, $B$ are smooth? $\endgroup$
    – Laurent Moret-Bailly
    Commented Apr 4, 2019 at 6:30

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