This question has [also been posted on MSE](https://math.stackexchange.com/questions/3986936/characterization-of-disks-of-constant-curvature-and-whose-boundaries-have-consta), but maybe here is the right place to post it.


Is it true that if $D$ is a Riemannian $2$-disk having constant Gaussian curvature equal to $1$ and whose boundary has constant geodesic curvature, then $D$ is isometric to some geodesic ball of the unit sphere $\mathbb{S}^2 \subset \mathbb{R}^3$? I strongly suspect so, but I couldn't find a reasonable argument.



Constant Gaussian curvature disks
This is an addendum to the proofs by Anton and Deane, completing the missing part of the argument.


Lemma. Let $D$ be the closed unit disk and $f: D\to S^2$ an immersion such that $f(\partial D)$ is a circle $C$ in $S^2$. Then $f$ is 1-1.


Proof. Let $J: S^2\to S^2$ denote the reflection in $C$. Double $D$ cross its boundary to obtain the 2-sphere $\Sigma$ and let $j: \Sigma\to\Sigma$ denote the reflection in $\partial D$. Then extend $f$ to a local homeomorphism $F: \Sigma\to S^2$ by
$$
F(j(z))=J f(z). 
$$
Since $\Sigma$ is compact, $F$ is a covering map, hence, a homeomorphism. Thus, $f$ is 1-1. qed


Edit. As for the existence of the isometric immersion $\iota$ in Anton's answer (above, $\iota=f$), it is an application of Riemann's theorem (the local form of the Killing–Hopf theorem): It shows that for every $z\in D$ there exists neighborhood $U\subset D$ and an isometric embedding $U\to S^2$. Joe Wolf in his book attributes the local result to Riemann and he is probably right, but it is likely (since it is about surfaces) that Gauss already knew how to prove this.


Since $D$ is simply-connected, these local isometries can be combined to produce a globally-defined isometric immersion, see for instance, the answer to [this](https://mathoverflow.net/questions/322330/a-question-concerning-the-developing-map-of-g-x-manifolds) question.


