# Lifts of smooth algebras

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?

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.
• 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 – John P. Apr 3 '19 at 21:18
• 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? – Laurent Moret-Bailly Apr 4 '19 at 6:30