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 $$a\_0\circ\pi = \pi\circ a$$



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> **Q1:** Is there an endomorphism of $S$-schemes $\alpha : X\to X$ such that $a$ is induced by $\alpha$ on $S$-points and $\alpha\_0 = \alpha\times\_S\text{Spec}(\kappa(s))$?
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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



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> **Q2:** Is there an explicit example of such an $f$ together with the data described before **Q1** and with no such $\alpha$?
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Interpolation of scheme-theoretic endomorphisms of closed fibers
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.


