# Interpolation of scheme-theoretic endomorphisms of closed fibers

Let $$S$$ be a scheme and $$f : X\to S$$ be an $$S$$-scheme. This question asks for examples of maps of sets $$X(S) \to X(S)$$ that do not come from an $$S$$-scheme endomorphism of $$X$$, but that, roughly, specialize to maps $$X_s(\kappa(s))\to X_s(\kappa(s))$$ that do come from a $$\kappa(s)$$-scheme endomorphism of the fiber $$X_s$$ of $$f$$ at each closed point $$s\in S$$, with $$\kappa(s)$$ the residue field of $$S$$ at $$s$$.

More precisely, let $$R$$ be a strictly henselian complete discrete valuation ring and $$s$$ the closed (geometric) point in $$S=\text{Spec}(R)$$ and assume $$f$$ is proper and smooth.

Suppose there are:

• a set-theoretic self map $$a : X(R)\to X(R)$$
• a set-theoretic self map $$a_0 : X_s(\kappa(s))\to X_s(\kappa(s))$$
• an endomorphism of $$\kappa(s)$$-schemes $$\alpha_0 : X_s\to X_s$$ such that $$a_0$$ is induced by $$\alpha_0$$ on $$\kappa(s)$$-points
• calling $$\pi$$ the natural map $$X(R) \to X_s(\kappa(s))$$, $$a$$ and $$a_0$$ satisfy the condition $$a_0\circ\pi = \pi\circ a$$

Q1: Is there an endomorphism of $$S$$-schemes $$\alpha : X\to X$$ such that $$a$$ is induced by $$\alpha$$ on $$S$$-points and $$\alpha_0 = \alpha\times_S\text{Spec}(\kappa(s))$$?

To give a sense of what the question asks, for general $$S$$ this would mean whether and when we can interpolate scheme theoretic endomorphisms of the closed fibers of $$f$$ to an $$S$$-scheme endomorphism of $$f$$, and whether the condition that we can do so set-theoretically on $$S$$-points is enough.

I'd expect the answer to be no. I'm asking for a couple of concrete counterexamples, if any. In other words

Q2: Is there an explicit example of such an $$f$$ together with the data described before Q1 and with no such $$\alpha$$?

• When you say "let $S$ be a strictly complete henselian discrete valuation ring", that should be $R$, right? Jun 22, 2021 at 19:58
• The condition $\pi\circ a=a_0\circ \pi$ (where $\pi:X(R)\to X(\kappa(s))$ is the natural map) is necessary. Did you mean to assume about $a$ and $a_0$? Jun 22, 2021 at 20:44
• We can take $X$ to be a curve of genus $\geq 2$, $a_0$ the identity map, $\alpha_0$ the identity, and $a$ any nontrivial permutation of the points reducing to a given one. Because there are infinitely many rational points, $\alpha_0$ is uniquely determined by $a_0$, and then $\alpha_0$ has a unique lift, which is the identity, so any non-identity $a$ works. Jun 22, 2021 at 22:22
• Thanks, this already settles it
– user290895
Jun 24, 2021 at 2:50

Choose a smooth projective $$X/R$$ of positive dimension, and pick a set-theoretic splitting $$\varphi:X(\kappa(s))\to X(R)$$ of the reduction map $$\pi:X(R)\to X(\kappa(s))$$. Take $$a=\varphi\circ \pi$$, and let $$a_0$$ and $$\alpha_0$$ be the identity. Then $$a$$ is constant on residue disks, so is locally constant for the analytic topology. If $$a$$ were induced by an algebraic morphism $$\alpha:X\to X$$, then $$\alpha$$ would have to be locally constant for the Zariski topology (I'm using the fact that the Zariski closure of an analytic neighborhood contains a Zariski neighbohood). This would imply the image of $$\alpha$$ is finite, which is impossible because $$\#Im(a)=\# X(\kappa(s))=\infty$$ since $$X$$ is positive dimensional and $$\kappa(s)$$ is separably closed.