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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?

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    $\begingroup$ 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. $\endgroup$ – Moishe Kohan Feb 1 at 0:44
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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.

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  • $\begingroup$ Perfect counter-exemple. $\endgroup$ – Matheus Manzatto Feb 1 at 1:35

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