# Proof of existence of maximal analytic continuation of a holomorphic germ

The following is from Lectures on Riemann Surfaces by O. Forster:

7.8. Theorem. Suppose $$X$$ is a Riemann surface, $$a\in X$$ and $$\varphi\in\mathcal{O}_a$$ is a holomorphic function germ at the point $$a$$. Then there exists a maximal analytic continuation $$(Y, p,f, b)$$ of $$\varphi$$.

Pʀᴏᴏғ. Let $$Y$$ be the connected component of $$|\mathcal{O}|$$ containing $$\varphi$$. Let $$p$$ also denote the restriction of the mapping $$p: |\mathcal{O}| \to X$$ to $$Y$$. Then $$p: Y \to X$$ is a local homeomorphism. By Theorem (4.6) there is a complex structure on $$Y$$ so that it becomes a Riemann surface and the mapping $$p: Y \to X$$ is holomorphic. Now define a holomorphic function $$f: Y \to \mathbb{C}$$ as follows. By definition every $$\eta\in Y$$ is a function germ at the point $$p(\eta)$$. Set $$f(\eta) := \eta(p(\eta))$$. One easily sees that $$f$$ is holomorphic and $$p_*(p_\eta(f)) = \eta$$ for every $$\eta\in y$$. Thus if one lets $$b := \varphi$$, then $$(Y, p,f, b)$$ is an analytic continuation of $$\varphi$$.

Now we will show that $$(Y, p,f, b)$$ is a maximal analytic continuation of $$\varphi$$. Suppose $$(Z, q, g, c)$$ is another analytic continuation of $$\varphi$$. Define the map $$F: Z \to Y$$ as follows. Suppose $$\zeta \in Z$$ and $$q(\zeta) =: x$$. By Lemma (7.7) the function germ $$q_*(p_\zeta(g))\in\mathcal{O}_x$$ arises by analytic continuation along a curve from $$a$$ to $$x$$ from the function germ $$\varphi$$. By Lemma (7.2) $$Y$$ consists of all function germs which are obtained by the analytic continuation of $$\varphi$$ along curves. Hence there exists exactly one $$\eta\in Y$$ such that $$q_*(p_\zeta(g))=\eta$$. Let $$F(\zeta) = \eta$$. It is easy to check that $$F: Z \to Y$$ is a fiber-preserving holomorphic map such that $$F(c)=b$$ and $$F^*(f)=g$$. $$\square$$

I don't understand the second paragraph of the proof. Why is the map $$F:Z\to Y$$ well-defined? If we choose a different curve from $$a$$ to $$x$$, we might get a different $$\eta$$...

For the relevant definitions:

Suppose $$X$$ and $$Y$$ are Riemann surfaces and $$\mathcal{O}_X$$ and $$\mathcal{O}_Y$$ are the sheaves of holomorphic functions on them. Suppose $$p: Y \to X$$ is an unbranched holomorphic map. Since $$p$$ is locally biholomorphic, for each $$y\in Y$$ it induces an isomorphism $$p^*: \mathcal{O}_{X, p(y)} \to \mathcal{O}_{Y, y}$$. Let

$$p_*: \mathcal{O}_{Y, y} \to \mathcal{O}_{X, p(y)}$$

be the inverse of $$p^*$$.

7.6. Definition. Suppose $$X$$ is a Riemann surface, $$a\in X$$ is a point and $$\varphi\in\mathcal{O}_a$$ is a function germ. A quadrupel $$(Y, p,f, b)$$ is called an analytic continuation of $$\varphi$$ if:

1. $$Y$$ is a Riemann surface and $$p:Y \to X$$ is an unbranched holomorphic map.
2. $$f$$ is a holomorphic function on $$Y$$.
3. $$b$$ is a point of $$Y$$ such that $$p(b) = a$$ and $$p_*(p_b(f)) = \varphi.$$

An analytic continuation $$(Y, p,f, b)$$ of $$\varphi$$ is said to be maximal if it has the following universal property. If $$(Z, q, g, c)$$ is any other analytic continuation of $$\varphi$$, then there exists a fiber-preserving holomorphic mapping $$F: Z\to Y$$ such that $$F(c) = b$$ and $$F^*(f) = g$$.

• That's the whole point of defining $Y$ so different (homotopically) curves in $X$ lead to different points in $Y$ or conversely given a path in $X$ there is a unique point in $Y$ corresponding to the homotopy class of the path in $X$ (and the continuation of $\varphi$ on that path) – Conrad Apr 21 at 17:11
• How to make simple things very abstract. For a function $f$ analytic on a small disk around $a$ you are considering the set of curves $\gamma : \gamma(0)=a \to \gamma(1)$ such that $f$ can be analytically continued along $\gamma$, which gives a new function $f_\gamma$ analytic around $\gamma(1)$ (a power series), the full analytic continuation of $f$ is the set of those $(\gamma(1),f_\gamma)$ which is a Riemann surface $Y$. You can define homotopy of two such curves in term of if $f_{\gamma_r}$ is well-defined, which is the same as homotopy on $Y$. – reuns Apr 21 at 17:43
• @Conrad In general it is not the homotopy in $X$, to do so you need to assume for any $b \in Y$ and simply connected set $U \ni f(b)\subset X$ then $U = f(V)$ with $V\subset Y$ simply connected. – reuns Apr 21 at 18:50
• @Conrad Sorry for the late response. I think the reasoning is this: For $\zeta\in Z$ find curve $\gamma:[0,1]\to Z$ such that $\gamma(0)=c$, $\gamma(1)=\zeta$. Then $q_*(\rho_\zeta(g))$ is the analytic continuation of the function germ $\varphi$ along $q\circ\gamma$. Now there is a lifting $\Gamma:[0,1]\to Y\subset|\mathcal{O}|$ of $q\circ\gamma$ along $p$. Then we define $F(\zeta)=\Gamma(1)$. Why is it independent of $\gamma$? – Colescu May 2 at 13:28
• Hmmm I seem to understand it now. So $F(\zeta)$ is just $q_*(\rho_\zeta(g))$ and the lifting $\Gamma$ is only used to show that $q_*(\rho_\zeta(g))$ lies in the connected component $Y$? Am I right? – Colescu May 2 at 13:35