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Let $S$ be a two dimensional surface and $D$ a closed disk contained in $S$. Is it true that if $f$ is a homeomorphism from a closed ball $B$ in the plane onto $D$, then $f$ extends to a homeomorphism from a neighbourhood of $B$ in the plane onto a neighbourhood of $D$ in the surface?

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Assuming that you don't mean manifold with boundary (in which case it's not true, since $D$ might already be the whole space!) then yes, it's true. One way to do it is with the Tubular Neighborhood Theorem by switching to the smooth category.

Let $J$ be the boundary curve of your disc in $S$. Then it has a tubular neighborhood, which is diffeomorphic to a neighborhood of $J$ in its normal tangent bundle. For a simple closed curve $J$ that's just an open annulus. This neighborhood is separated into two components by $J$, one of which will meet the interior of your disc. Keep the other one and glue your annulus homeomorphism and disc homeomorphism together along $J$.

If it's annoying having to go into the smooth category, just be aware that these sorts of theorems get VERY hard without appealing to smooth, algebraic/PL or complex-analytic techniques.

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    I agree your argument works in the smooth category, but I don't see why you can move to the smooth category to begin with. In higher dimensions, you can't in general. For example, if you embed a ball $B^3$ in $\mathbb{R}^3$ as the interior and boundary of the Alexander Horned sphere, then this embedding of $B^3$ into $\mathbb{R}^3$ does not extend to a homeomorphism of a neighborhood of $B^3$. – Jason DeVito - on hiatus Mar 01 '21 at 18:31
  • Is it a possible to give a reference? – aaa acb May 20 '25 at 23:01
  • Curious if there is any updates/follow-ups to @JasonDeVito-onhiatus’s question? I agree one can always impose a smooth structure on the surface. But the embedding of $\partial B$ into the surface may not been smooth, hence I failed to see why one can pass to the smooth category. If I’m missing something obvious please let me know. – aaa acb May 23 '25 at 13:04