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This question has also been posted on MSE, 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.

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    $\begingroup$ Why can we assume $D$ is contained in the unit sphere? $\endgroup$ Jan 16 at 0:59
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    $\begingroup$ You should include a link to the question on MSE $\endgroup$ Jan 17 at 0:34
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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 question.

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  • $\begingroup$ Did you read the quoted article in Wikipedia (it seems to be wrong right now). $\endgroup$ Feb 7 at 5:11
  • $\begingroup$ @AntonPetrunin: No, I did not. Indeed, the article is suboptimal (should be "isometric action"); it is also missing a purely local form of the theorem as well as a modern textbook reference. Feel free to edit. $\endgroup$ Feb 7 at 10:42
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The answer is yes.

Since curvature is 1, there is an isometric immersion $\iota\colon D\looparrowright \mathbb{S}^2$. Note that the curve $\iota(\partial D)$ has constant curvature, therefore $\iota(\partial D)$ bounds a round disc $\Delta\subset\mathbb{S}^2$.

Acually we have two choices for $\Delta$, but for the right choice we get an isometry $D\to \Delta$. The latter can be done by using Morse-type argument for a function $f\colon\Delta\to\mathbb{R}$.

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  • $\begingroup$ What is $f$ here? $\endgroup$ Jan 16 at 3:26
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    $\begingroup$ @AntonPetrunin why does $\iota:D \to \mathbb{S}^2$ exist? $\endgroup$ Jan 16 at 18:09
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    $\begingroup$ I think this and the original question really belong in MSE. I’ve posted a different answer to the original question there. $\endgroup$
    – Deane Yang
    Jan 16 at 19:59
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    $\begingroup$ The existence of an isometric immersion of $D$ into $\mathbb{S}^2$ follows, for example, by Bonnet's Theorem, which is stated and proved in Do Carmo's book, Differential geometry of curves and surfaces. $\endgroup$
    – Deane Yang
    Jan 16 at 20:22
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    $\begingroup$ This is is same proof as Deane's argument and is similarly incomplete. You have to argue that $i$ is 1-1 (it would suffice to prove injectivity in the boundary). This is not har though. You can use the fact that the teardrop orbifold does not admit a spherical structure. $\endgroup$ Jan 17 at 3:39

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