Let $I^2$ be the square $\{(x, y) ∈ \Bbb R^2: 0 \leq x, y \leq 1\}$
$C$ be the circle $\{(x, y) \in \Bbb R^2: 1 \leq x^2 + y^2 \leq 4\}$, regarded as subspaces of $\Bbb R^2$ in the usual topology.
Let us have equivalence relations $∼$ and $≈$ on $I^2$ and $C$ respectively by
$(x, y) ∼ (x, y)\ \forall (x, y) \in I^2$, $(0, y) ∼ (1, y)$ and $(x, 0) ∼ (x, 1)$ if $0 \leq x, y \leq 1$
$(x, y) ≈ (x, y)\ \forall(x, y) \in C,\ (x, y) ≈ (2x, 2y)$ if $x^2 + y^2 = 1$.
How to show that $[S^2]_∼$ and $[C]_≈$ are homeomorphic in their respective quotient topologies?
I am trying to visualise a picture proof i.e. view quotient spaces as homeomorphic to the torus; but not sure if that helps
I am not writing an answer but the comment section is not appropriate for this, you may take it as an answer.
Using the fact in the answer here, it is sufficient to show that the Disk and the Solid square are homeomorphic. See here for proof. For visualization of how square is homeomorphic to circle here is an image. Hope this works.