The sheaf $\mathfrak{S}$ of germs of analytic functions over $D$ is a topological group (Ahlfors)

Question

In Ahlfors' complex analysis text, page 286 he gives the following definition:

Definition 1. A sheaf over $D$ is a topological space $\mathfrak S$ and a mapping $\pi:\mathfrak S \to D$ with the following properties: (i) The mapping $\pi$ is a local homeomorphism; this shall mean that each $s \in \mathfrak S$ has an open neighborhood $\Delta$ such that $\pi(\Delta)$ is open and the restriction of $\pi$ to $\Delta$ is a homeomorphism. (ii) For each $\zeta \in D$ the stalk $\pi^{-1}(\zeta)=\mathfrak S_\zeta$ has the structure of an abelian group. (iii) The group operations are continuous in the topology of $\mathfrak S$.

In the next page (287) he attempts to prove (iii) by proving that subtraction is continuous. I'm having trouble with his proof. Firstly he introduces the symbols $\Delta_0, \Delta_0',\Delta$ without precisely defining them. Also he uses the equality $\pi(s-s')=\pi(s)-\pi(s')$ which I believe is false.

Could anyone help me review this proof? Thanks.

Answer 1

The $\pi(s-s') = \pi(s) - \pi(s')$ is nonsense, of course. However, it is almost certainly a trivial typo and was meant to read $\pi(s-s') = \pi(s) = \pi(s')$. That is correct and makes sense.

In the last paragraph on page 286, he writes

With $s_0$ and $(f,\Omega)$ as above, let $\Delta$ be the set of all germs $\mathbf{f}_\zeta$ determined by $(f,\Omega)$.

In view of that, and since it's the interpretation that makes sense, as the two germs $s_0$ and $s_0'$ were determined by $(f,\Omega)$ and $(g,\Omega)$ in $\zeta_0$ repsectively,

• $\Delta_0$ is the set of germs determined by $(f,\Omega)$, $\Delta_0 = \{\mathbf{f}_\zeta : \zeta \in \Omega\}$,
• $\Delta_0'$ is the set of germs determined by $(g,\Omega)$, $\Delta_0' = \{\mathbf{g}_\zeta : \zeta \in \Omega\}$, and
• $\Delta$ is the set of germs determined by $(f-g,\Omega)$, $\Delta = \{ \mathbf{(f-g)}_\zeta : \zeta \in \Omega\}$.

The proof of continuity then simply shows that the stalkwise subtraction maps

$$\{ (s,t) \in \Delta_0 \times \Delta_0' : \pi(s) = \pi(t)\}$$

into $\Delta$ and ends with the remark that $\Delta_0$ and $\Delta_0'$ can be shrunk to have the image contained in any prescribed neighbourhood of $s_0-s_0'$. That is by the definition of a germ, any representative of the germ $s_0 - s_0'$ coincides with $f-g$ in some neighbourhood $W$ of $\zeta_0$. Here, since we are dealing with germs of analytic functions, we have the identity theorem that yields something stronger, any two representatives of a germ coincide in the entire component of the intersection of their domains containing $\zeta_0$.