Lifting of meromorphic function along a finite morphism I am currently reading the book "Geometry of algebraic curves II", by Arbarello, Cornalba and Griffiths, and I am having some difficulties understanding a passage p.105. 
The setting is the following: we are working with separated schemes of finite type over $\mathbb{C}$, and we are given a finite surjective morphism $Z \rightarrow \bar{M_g}$, where $Z$ is normal, and a meromophic map $\dot{\Delta} \rightarrow \bar{M_g}$ (where $\dot{\Delta}$ denotes a punctured disc).
The authors then claim the following: "after a harmless base change, we can lift any such map to a map $f$ from $\dot{\Delta}$ to Z".

My questions are the following : First, I am not certain what they mean by a meromorphic function to some general scheme, but I expect it to be a morphism $ \operatorname{Spec}( \mathbb{C}\{X\}[X^{-1}]) \rightarrow  \bar{M_g}$ ?
If this is correct, How do we get the lifting after base change they are talking about ?
I was thinking about the following: let $K = \mathbb{C}\{X\}[X^{-1}] $, $x\in \bar{M_g}$ be the image of the given morphism, and $z\in Z$ be any point mapping to $x$. We have 2 extensions of $\kappa(x)$, namely $\kappa(z)$ and $K$, and we would like to embed $\kappa(z)$ in $K$ modulo finite extension. Suppose that $\kappa(x)\rightarrow \kappa(z)$ is a finite extension of degree $d$, could we find $d'\geq d$ such that an extension of this form would do the job ? 
$$\begin{align} K& \rightarrow K \\ 
X & \mapsto  X^{d'}
\end{align}$$
Or should I look for some ramification index of $Z \rightarrow \bar{M_g}$ at $z$ ?
Any help would be greatly appreciated, thanks in advance.
 A: Let's dig out the relevant text for those who might not have access to the book. We also do this because it makes resolving your questions easier. Here goes:

On the other hand, what turns out to be true is that there exists a normal algebraic variety $Z$, a finite morphism $$ m:Z\to \overline{M}_g \qquad (4.1)$$ and a family of stable curves of genus $g$ $$\eta:\mathcal{X}\to Z \qquad (4.2)$$ such that, for every $z\in Z$, $[\eta^{-1}(z)]=m(z)$.
One of the main results of the first five chapters of this volume, to be proved in Chapter XIV, is that $\overline{M}_g$ is a projective variety. A basic step in the proof will be to show that $\overline{M}_g$ is complete. This will be checked by using the valuative criterion for completeness.
Informally speaking, the valuative criterion for completeness asserts that an algebraic variety $X$ is complete whenever any holomorphic map from a punctured disc to $X$ which is meromorphic at the origin can be extended across the puncture. Clearly this criterion can be somewhat softened by only requiring that the extension be possible after a base change $z=\zeta^k$ on the punctured disc.
In applying the valuative criterion of completeness to the moduli space $\overline{M}_g$, we are thus confronted with maps from a punctured disc $\dot{\Delta}=\{z\in\Bbb C:0<|z|<\epsilon\}$ to $\overline{M}_g$ which are meromorphic at the origin. In view of the existence of the finite map (4.1), after a harmless base change, we can lift any such map to a map $f$ from $\dot{\Delta}$ to $Z$.

The first thing to note is that they discuss the valuative criterion in infomal terms. They are using an analogy instead of dealing with the usual statement (available here) which talks about morphisms from DVRs and their fraction fields. The intuition is that the spectrum of a DVR looks like a little disc around a particular closed point inside a smooth curve, and the spectrum of it's field of fractions is the same scheme but without the closed point, thus intuitively corresponding to a little punctured disc inside a curve. What the valuative criterion intuitively means is that we can always fill in the puncture, or that "limits exist" along curves.
As to the question about meromorphicness, note that they say "meromorphic at the origin" - what they're really talking about is a map from the punctured disc which is holomorphic everywhere except possibly at the origin. It's a little confusing because what we actually do here is show that there's always a holomorphic extension of such a map across the puncture, so we never really have a map which is meromorphic and can't be extended across the possible pole.
The usual way to talk about the qualities of a meromorphic map in algebraic geometry when dealing with arbitrary schemes is by using the language of rational maps. A rational map $X\dashrightarrow Y$ is an equivalence class $(f,U)$ where $U$ is an open subscheme of $X$ and $f:U\to Y$ is an honest morphism. One can apply this in our case here by thinking about the map from the spectrum of the field of fractions of our DVR as being a rational map from the spectrum of our DVR, and then both the DVR and punctured disc side of things line up in terms of what they're doing.

Now for the lifting/base change argument. There's a problem with trying to lift a map from the punctured disc to $\overline{M}_g$ to a map from the the punctured disc to a finite cover of $\overline{M}_g$: think about the example where all the spaces are $\Bbb C^\times$, the covering map is the $n$-sheeted covering $\Bbb C^\times\to \Bbb C^\times$ by $z\mapsto z^n$ and the map we want to lift is the identity. By looking at the map on fundamental groups we see this is impossible: the map on fundamental groups corresponding to the covering is multiplication by $n$, which doesn't have $1$ in it's image like the map we're trying to lift. The solution is to allow us to take finite covers of the punctured disc: if we now have $d$ points from the disc mapping to each point in $\overline{M}_g$ and $d$ points from $Z$ mapping to each point in $\overline{M}_g$, we then have a chance to really make a lift. (We might not always need all $d$ points: consider a covering that's just a disjoint union - then we don't need any base change at all.)
The algebraic-geometry side of this translates as the following. Given a diagram
$$\require{AMScd}
\begin{CD}
? @>{}>> Z\\
@VVV @VVV \\
\operatorname{Spec} K @>>> X
\end{CD}$$
with $K$ the field of fractions of a DVR and $Z\to D$ finite of degree $d$, then we want to find a finite extension $K'$ of $K$ so that the diagram commutes when we plug in $\operatorname{Spec} K'$ for $?$. Now let's turn this all in to an algebra problem.
Write $x$ for the image of $\operatorname{Spec} K$ in $X$ and $z$ for a point above $x$ in $Z$. What we're looking for in our diagram of schemes above now turns in to the following diagram of local rings:
$$\require{AMScd}
\begin{CD}
K' @<<< \mathcal{O}_{Z,z} \\
@AAA @AAA \\
K @<<< \mathcal{O}_{X,x}
\end{CD}$$
where the horizontal maps can be factored as $K\leftarrow \kappa(x)\leftarrow \mathcal{O}_{X,x}$ and $K'\leftarrow \kappa(z)\leftarrow \mathcal{O}_{Z,z}$ by the condition that they're local ring homomorphisms. By assumption, $\kappa(z)$ is algebraic of degree at most $d$ over $\kappa(x)$, and since we are in characteristic zero, the extension is separable and by the theorem of the primitive element, we may find some $\alpha\in \kappa(z)$ so that $\kappa(x)(\alpha)=\kappa(z)$. Let $p(t)$ be the minimal polynomial of $\alpha$. If $p(t)$ has a root in $K$, then we can take $K'=K$ and have no problem. If not, then take $K'=K[t]/p(t)$ and we're finished.
