TITLE: Estimate of number of boundary components of a compact Riemannian 2-surface
QUESTION [7 upvotes]: Let $X$ be a compact smooth 2-dimensional Riemannian manifold with boundary. Assume that the Gauss curvature of $X$ is at least $-1$ and the diameter is at most $D$. Assume that near the boundary the surface is locally geodesically convex.
Is it true that the number of connected components of the boundary is bounded above by a constant depending on $D$ only?
Remark. I am not a specialist, but if I understand correctly the answer should be positive and follow from very general results in the Alexandrov geometry. Indeed due to local convexity, $X$ is an Alexandrov space of curvature at least $-1$. Each boundary component is an extremal subset in the sense of Perelman-Petrunin. In general the number of extremal subsets of a compact $n$-dimensional Alexandrov space of curvature at least $-1$ and diameter at most $D$ is bounded above by a constant depending on $n,D$ only. Is this argument correct? Anyway I would prefer to have a more direct argument in my case.

REPLY [7 votes]: I think it follows from Gauss-Bonnet.  Suppose $X$ has genus $g$ and $n$ boundary components.  Gauss-Bonnet says that
$$\int_X K\;dA+\int_{\partial X}k\;ds=2\pi\chi(X)=2\pi(2-2g-n),$$
where $K$ is sectional curvature, $k$ is the curvature of the boundary.  The local convexity implies that the boundary is positively curved, so $\int_{\partial M}k\;ds\ge 0$ and 
$$-\mathop{\text{area}}X \le \int_X K\;dA\le 2\pi(2-2g-n),$$
i.e., $2\pi(2g+n-2)\le \mathop{\text{area}}X$.
The curvature and diameter bound imply that the area of $X$ is at most the area of a ball of radius $D$ in the hyperbolic plane, so $g$ and $n$ are bounded.