Existence hypothesis of the monodromy theorem

The monodromy theorem gives equality of continuations that exist under its hypothesis. It does not assert that every holomorphic function admits analytic continuation everywhere, even within a simply connected domain.

For an explicit obstruction, take $f(z)=1/(1-z)$ near $a=0$ and attempt to apply the theorem on $D=\mathbb{C}$. The path $\eta (t)=t$ ends at $1$, where no terminal germ of a holomorphic function exists. If a holomorphic representative $G$ near $1$ continued $f$, then the identity theorem would extend $(1-z)G(z)=1$ to that neighbourhood. Substituting $z=1$ would give $0=1$. Thus the pole prevents the existence hypothesis, although $\mathbb{C}$ is a simply connected domain. The analytic continuation article explains why the same rational function does continue throughout $\mathbb{C}\setminus \{1\}$.

Existence at an endpoint by some route is also weaker than existence along every path from the base point. Along the real segment from $1$ towards $0$, a holomorphic square root of $z$ tends to $0$, yet cannot continue holomorphically through $0$: a local representative $G$ would satisfy $G^2=z$ by the identity theorem, and differentiating at $0$ would give $2G(0)G^{\prime }(0)=1$ with $G(0)=0$. A finite limiting value therefore does not remove a branch point. More severe obstructions appear at a natural boundary of a holomorphic function.