Intuition for the valuative criterion for properness of morphisms? I've always been told that the intuition for the valuative criterion for properness is something like this: a morphism $X\rightarrow Y$ is proper if, given a map of a small disk $D$ into $Y$ and a lifting $\alpha$ of $D\setminus\{p\}$ to $X$ for some point $p\in D$, there exists a unique lifting of $D$ to $X$ extending $\alpha$.
In the valuative criterion, we can think of the discrete valuation ring $R$ as representing a germ of the curve around a point, or $D$ in the above picture. Its field of fractions $K$ is the localization at its generic point. 
Questions:


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*How can I think of $K$ geometrically as a "punctured germ'' or "punctured disk"? It seems that $K$ is actually a "point'' of $D$, so that the criterion is saying something like, given a lift of the generic point of a small piece of a curve, we can lift to the entirety of the small piece. This makes less geometric sense to me, though. Perhaps one could view the generic point as a non-closed "fuzzy" thing whose closure includes the single closed point $\mathfrak m$ in the DVR and preserve the analogy that way.

*It seems reasonable to expect the valuative criterion to be able to prove the following: if $X$ and $Y$ are projective (and hence proper) curves, and $p\in X$ is a regular point, then any morphism $X\setminus \{p\}\rightarrow Y$ extends uniquely to a morphism $X\rightarrow Y$. We can interpret this in terms of extending a lift of a small punctured disk (the image of some small punctured disk around $p$ mapped to $Y$). Can the valuative criterion be used to prove this, and if so, how?
I think something like the following works. We have morphisms
$$FF(\mathcal O_{X,p})\rightarrow Y \rightarrow \operatorname{Spec} \mathbb C$$
$$FF(\mathcal O_{X,p}) \rightarrow \mathcal O_{X,p}\rightarrow \operatorname{Spec} \mathbb C$$
that commute. The map $$FF(\mathcal O_{X,p})\rightarrow Y$$ is given by the restriction of the given map $X\rightarrow Y$ to the generic point, since $FF(\mathcal O_{X,p})$ is the function field for $X$ (right?). Then the valuative criterion produces a map $f:\mathcal O_{X,p}\rightarrow Y$.
It seems like this map should be the desired extension, but I'm not sure how to say that precisely.
 A: Lemma. Let $X$ be a variety over $k$. Then $X$ is proper if and only if for every smooth proper curve $C$ and every $U \subseteq C$ open, any morphism $f \colon U \to X$ can be extended to a map $C \to X$.
Proof. We extend one point at a time. Suppose $P \in C\setminus U$, and let $V = \operatorname{Spec} B$ be an affine open neighbourhood of $P$. Let $\eta \in C$ be the generic point; note that it is contained in any open. Restricting $f$ to the generic point gives a map
$$\operatorname{Spec} \kappa(\eta) \to U \to X.$$
By the valuative criterion, we can extend uniquely to a map $g \colon \operatorname{Spec} \mathcal O_{C,P} \to X$. Let $W = \operatorname{Spec} A$ be an affine open neighbourhood of $g(P)$ in $X$. Then $g(\eta) \in W$ as well, since open sets are stable under generisation. Thus, $g$ is given by a ring homomorphism
$$A \to \mathcal O_{C,P} = B_{\mathfrak m_P}.$$
Since $A$ is of finite type, we can collect denominators to get a map $A \to B_f$ for some $f \in B$; that is, we get a morphism
$$h \colon D(f) \to X.$$
For every $Q \in U \cap V$, the valuative criterion of separatedness shows that the maps
$$h, f \colon \operatorname{Spec} \mathcal O_{C,Q} \to X$$
agree (since their restrictions to $\operatorname{Spec} \kappa(\eta)$ agree by definition of $g$). Then a simple algebra argument shows that $f = h$ on $U \cap V$. Thus, they glue to a well-defined map on $C \cup \{P\}$. 
Conversely, if $X$ is proper, and $f \colon U \to X$ is a map, let $C'$ be the scheme-theoretic image. If $C'$ is a point, then $f$ is constant, so we can trivially extend. Otherwise, $C'$ is a proper curve (being closed inside a proper). Then the normalisation of $C'$ is isomorphic to $C$, since there is only one smooth proper curve with a given function field. Thus, the normalisation map $C \to C'$ extends $f$. $\square$
Remark. For a slightly more high-brow version of the same proof, see this blog post.
Remark. To see the analogy with the punctured disk, note that if we identify all nonzero points of the complex unit disk $\Delta$, we get a topological space with two points: one closed point (the image of the origin), and one non-closed point (the image of the punctured disk $\Delta^*$). Thus, removing the origin from this quotient space also gives a one-point space.
Of course, topological spaces behave very differently from ringed spaces, so to some extent the analogy ends there.
In another direction, if we equip $\Delta$ with the sheaf of rings of holomorphic functions, then the local ring $\mathcal O_{\Delta, 0}$ is the ring of convergent power series $\mathbb C\{x\}$. This looks a lot like the localisation $\mathbb C[x]_{(x)}$; for example the completion of both rings is the ring of power series $\mathbb C[[x]]$. This is the beginning of Serre's GAGA principle.
The analogy between the punctured disk and the function field of a curve does not end there. For example, the fundamental group of $\Delta^*$ is $\mathbb Z$, with covers given by $\Delta^* \to \Delta^*$, $x \mapsto x^n$. The absolute Galois group of $\mathbb C((x))$ is the profinite completion $\hat{\mathbb Z}$ of $\mathbb Z$, with extensions given by $\mathbb C ((x^{\frac{1}{n}}))$. This is no coincidence (the key word here is étale fundamental group).
