I have a general question about techniques used in [@Emerton's proof](https://mathoverflow.net/q/62405), sketched below, in the answer to [$\mathbb{P}^n$ is simply connected](https://mathoverflow.net/q/62282).


Given a finite étale map $\pi: Y \to \mathbb P^n$ (we regard all involved schemes as $k$-schemes for some fixed base field $k$), a base point $x \in \mathbb P^n$ and a point $y \in Y$ lying over $x$. Then naïvely/on the level of sets we construct a map $ \mathbb P^n \setminus \{x\} \to Y \setminus \{y\}$, $x' \mapsto y'$ by taking a unique line $L$ joining $x'$ with $x$, choosing the unique component $L'$ of the preimage $ \pi^{-1}(L)$ which contains $y$ and letting $y' \in Y'$ be the point lying over $x'$.


Now the funny thing is that $x' \mapsto y'$ is *algebraic*, i.e., a morphism in the category of schemes and that sounds quite surprising to me at first glance.


Emerton's argument was to observe that this map is realized as the composition of three maps, which can presumably be recognized as morphisms on their own:


1. $x' \mapsto \pi^{-1}(L)$, which is a map from $\mathbb P^n \setminus \{x\}$ to the Hilbert scheme of $Y$
2. picking out of $L' \subset \pi^{-1}(L)$ the component containing $y$, which should be a morphism from a locally closed subset of the Hilbert scheme of $Y$ to the Hilbert scheme of $Y$ itself and finally
3. mapping $L'$ to $L' \cap \pi^{-1}(x')$.


My naïve question is why the steps 1–3 are all morphisms in sense of algebraic geometry?


There are two standard ways in algebraic geometry to specify a morphism $f: X \to Y$ between schemes: ‘old school style’ by writing down an explicit polynomial equations with variables living in projective spaces which implicitly contain $X $ and $Y$.


The more modern functorial approach is to constuct a natural transformation
between the functors $F\_X= \operatorname{Hom}(-,X)$ and $F\_Y= \operatorname{Hom}(-,Y)$ which the schemes $X$, $Y$ represent. That amounts to associating a family
a $(f\_S: F\_X(S) \to F\_Y(S)\_{S \in (\mathrm{Sch}/k)}$ of maps indexed by the class $(\mathrm{Sch}/k)$ of $k$-schemes which commute with morphisms $m: S \to T$ in compatible way (= natural transform).


Seemingly morphisms 1–3 above were specified via the second approach. But the aspect which irritates me is that seemingly all three maps were only specified on the level of $k$-points,  

for example the first map was seemingly specified just as a map $(\mathbb P^n \setminus \{x\})(k) \to \operatorname{Hilb}\_Y(k)$, but in the spirit of the functorial construction above it is expected to specify this map as a *family of maps* $(\mathbb P^n \setminus \{x\})(S) \to \operatorname{Hilb}\_Y(S)$ for all $k$-schemes $S$.


Or is it in this special situation sufficient to specify everything on level of $k$-points only? E.g., naïvely one could be tempted to interpret $ x'$ not as a $ k $-point, but an $ S $-point etc. But on the other hand it seems to make no sense to talk about a line between a $ k $-point (= $x$) and an $ S $-point, so I'm a bit skeptical if the construction can be ‘prolonged’ naturally to $ S $-points….


Update: as Marsault Chabat [pointed out](https://mathoverflow.net/questions/437038/construct-morphisms-of-schemes-on-level-of-associated-functors#comment1126407_437038) the constructions go through fine as long as $ S $ are spectra of fields, but I am not sure why these should work for arbitrary $ S $. Clearly it's wrong that a morphism is already determined by what it does on finite field extensions of ground field (“geometric points”).



Construct morphisms of schemes on level of associated functors
I can't speak for Matt Emerton specifically, but my understanding is that it is conventional to describe maps in terms of $k$ points in such a way that there is a clear extension of the definition to $S$ points. This is perhaps less rigorous, but if you know how to fill in the details, it removes clutter and leaves the essential geometric idea intact.


A partial justification for this practice is that a map between varieties over an algebraically closed field is determined by its value on $\overline k$ points, so in "nice" cases this description is not too lossy.


In your specific case, note that the $k$ point $x$ yields a constant $S$ point $S \to {\rm Spec} k \to \mathbb P^n$. There is a map that takes a pair of disjoint points of $\mathbb P^n$ to a line in $\mathbb P^n$. One way of describing it on $S$ points is as follows. Let $\mathcal O\_S^{\oplus n+1} \to L\_1$ and $\mathcal O\_S^{\oplus n + 1} \to L\_2$ be the surjections corresponding to two $S$ valued points $x\_1, x\_2$. Then dualizing, we obtain an injection $L\_1^{\vee} \oplus L\_2^{\vee} \to \mathcal O\_S^{\oplus n+1}$ (this follows from Nakayama and the fact that $x\_1, x\_2$ are disjoint). Then take the subscheme of $\mathbb P^n\_S$ defined by the vanishing of the homogeneous ideal of ${\rm Sym}(\mathcal O\_S^{\oplus n+1})$ generated by $L\_1^{\vee} \oplus L\_2^{\vee}$.
This yields the map $(\mathbb P^n-x)(S) \to {\rm Hilb}(\mathbb P^n)(S)$. Then pullback the subscheme along the flat map $Y \to \mathbb P^n$ to get the first map in Emerton's answer.


As you can see, this description more verbose. In my opinion, it obscures the key point: given two disjoint lines you can take their span and get a plane!


(One final nitpick: I don't think your dichotomy "old school" v.s. "modern" is very accurate. Classical algebraic geometers were comfortable with defining maps without always writing down explicit equations.)


