Monodromy theorem

Remarks

The idea of the proof is that the continuation along $\gamma_s$ depends continuously on $s$. For $s'$ close to $s$, the two paths stay close, and the continuations along them use overlapping discs, so they give the same germ at $b$. Compactness of $[0, 1]$ then passes from $s = 0$ to $s = 1$. An important consequence concerns a simply connected domain $D$. If $f$ can be continued along every path in $D$ that starts at $a$, then there is a holomorphic function $F$ on $D$ that equals $f$ near $a$. The homotopy hypothesis is needed. On $D = \C \setminus \set{0}$, a branch of the complex logarithm near $a = 1$ continues along every path, but continuation along the unit circle returns to $1$ with the value increased by $2 \pi i$. The unit circle is not homotopic to a constant path in $D$, so there is no contradiction.

Going deeper

For a student

The theorem is the standard way to construct functions on simply connected domains, such as logarithms and square roots of zero-free holomorphic functions. It is the analytic form of a topological fact: a covering space over a simply connected space is trivial. Seen in this way, the continuations of a germ form a Riemann surface spread over the plane. The theorem says that this surface has only one sheet over a simply connected region. The monodromy group measures the failure over regions that are not simply connected.