Let $A$ be a square and $B$ be a triangle on the plane. They are homeomorphic.
Given $\phi:\partial A\rightarrow \partial B$ homeomorphism, does there exist a map $\phi':A\rightarrow B$ homeomorphism such that restriction of $\phi'$ on $\partial A$ is $\phi$?
I guess this is an elementary theorem or exercise in some topic, but I don't know it. Any reference will help a lot.
Thanks.
Observe that $A = \partial A * \{p\}$, where $p \in A^o$. Similarly. $B = \partial B * \{q\}$. Then $\phi: \partial A \to \partial B$ extends to a homeomorphism between $A$ and $B$ by defining $\phi(tx + sp) = t \phi(x) + sq$.
Note it in general, if two homeomorphic sets have homeomorphic boundaries, the homeomorphism between the boundaries may not extend to a homeomorphism between the sets themselves. eg. Let W be the unit circle with the horizontal diameter, X be the union of W with the solid upper semicircle and Y be the union of W with the solid lower semicircle. Then X and Y are homeomorphic with boundary W. Then the identity on W does not extend to a homeomorphism between X and Y.