$\newcommand{\ul}[1]{\underline{#1}}$I'm stuck on a detail when trying to prove centralizers/normalizers/transporters of algebraic groups over a separably closed field are representable.
Let $k = k_s$ be a separably closed field and $G$ a $k$-group of finite type acting on a finite type $k$-scheme $X$. Let $W \subset X$ a (geometrically) reduced separated $G$-stable closed subscheme and $\alpha_w : G \to W$ be the orbit map for each $w \in W(k)$. We would like to show that the functorial centralizer $$ \ul{Z}_G(W) : S \longmapsto \{g \in G(S) : g.w = w \;\; \forall w \in W(S'),\,S'/S\} $$ is represented by the candidate subgroup $$ Z'_G(W) := \bigcap_{w \in W(k)} \alpha_w^{-1}(w) : S \longmapsto \{g \in G(S) : g.w_S = w_S \;\; \forall w\in W(k)\}. $$
Clearly (1) we have a subfunctor $\ul{Z}_G(W) \subseteq Z'_G(W)$, and (2) they are equal on $k$-points: since $W$ is generically smooth $k$-points are dense, and since it is separated the morphisms $g$ and $\operatorname{id}_W : W \to W$ which agree on a Zariski-dense subset are equal.
How do I show that $\ul{Z}_G(W)(S) = Z'_G(W)(S)$ for an arbitrary $S$? For general $S$, the argument above gives two morphisms $g, \operatorname{id}_{W_S} : W_S \to W_S$ which agree on those $S$-points which come from $k$-points of $W$, but I don't see how to carry over the density argument.
It would also be enough to show $\ul{Z}_G$ is a sheaf for an appropriate topology, and then show sheaf-surjectivity. I don't see how this makes things any easier.