# Existence of a special homeomorphism on $\mathbb{T}^2$.

Let $$A, B$$ be closed topological subspaces of $$\mathbb{T}^2$$. Suppose that $$A$$ and $$B$$ are homeomorphic as topological spaces.

My Question: Is it possible to construct a homeomorphism $$h: \mathbb{T}^2 \to \mathbb{T}^2$$, such that $$h(A) = B?$$

If necessary, we can assume $$A$$ and $$B$$ as $$\mathcal{C}^0$$- manifold with boundary (topological manifold with boundary).

I've been stuck in this problem for a long time; everything I tried did not come close to achieving the result. Can anyone help me?

• It is impossible in general, when $A, B$ are topological circles. When each of them is a disjoint union of compact arcs, such a homeomorphism does exist, but it is not completely trivial. One needs Schoenflies theorem for topological arcs. – Moishe Kohan Feb 1 at 0:44

## 1 Answer

In general it's not possible. Consider A = circle that you can colapse in a point and B = circle that "cuts" the torus(One generator of the fundamental group). So $$A^c$$ retracts to a 8 shapped figure and $$B^c$$ is a cilinder.

• Perfect counter-exemple. – Matheus Manzatto Feb 1 at 1:35