Extenting a homeomorphism on subsurface to the entire surface Suppose you have surface and a subsurface. The complement of subsurface is union of open discs and once punctured open discs. Can all homeomorphisms of the subsurface be extended to homeomorphisms of the surface? There is a way of doing this by assuming the identity on the complement. Is this way unique?
 A: Turning my comment into an answer:  No, you cannot extend all homeomorphisms to the entire surface.
For a counterexample, let $S$ be the surface of the unit sphere in $\mathbb{R}^3$ with the point $(0,1,0)$ removed.  Let $R \subset S$ be the closed annulus consisting of points $(x,y,z)$ of $S$ with $\lvert y \rvert \leq \frac{1}{2}$.  Then the complement of $R$ is an open disc on the bottom and a punctured open disc on the top, so it fulfills the requirements you've stated.
Now consider the map $R\rightarrow R$ that takes $(x,y,z)$ to $(x, -y, z)$.  This is a homeomorphism on the annulus, but it switches the boundary components.  In order to extend this to a homeomorphism of $S$, you would have to homeomorphically map the disc to the punctured disc and vice versa, which is impossible.
If you strengthen your assumptions, for example by assuming the homeomorphism of the subsurface preserves boundary components, then the answer becomes yes, and you can prove it using the Alexander trick as janmarqz suggests.  However, these extensions will not be unique, unless you start adding qualifiers like "up to isotopy".
