Show homeomorphism between two quotient topologies Consider two subspaces of $\Bbb R^2$ in the usual topology:
Square $I^2$ := $\{(x, y) ∈ \Bbb R^2: 0 \leq x, y \leq 1\}$
Annulus $A$:= $\{(x, y) \in \Bbb R^2: 1 \leq x^2 + y^2 \leq 4\}$,
Define the following equivalence relations $∼$ and $≈$ on $I^2$ and $A$ respectively by
$(x, y) ∼ (x, y)\ \forall (x, y) \in I^2$, $(0, y) ∼ (1, y)$ and $(x, 0) ∼ (x, 1)$ if $0 \leq x, y \leq 1$
$(x, y) ≈ (x, y)\ \forall(x, y) \in A,\ (x, y) ≈ (2x, 2y)$ if $x^2 + y^2 = 1$.
I wish to prove that in the respective quotient topologies of $[I^2]_∼$ and $[A]_≈$ are homeomorphic . It is perhaps helpful to view quotient spaces as homeomorphic to the torus; so intuitively they should be homeomorphic. So, we can show show each quotient space is homeomorphic to the torus. But does there exist an explicit homeomorphism for this or something? Is there a proof that need not necessarily involve the torus or by pictures, if possible hmm~
*Just an additional remark my classmate and I were working on an exercise in a new topology course where we encountered the above problem and just realise that he had a similar post at Show two topological spaces are homeomorphic but I will be eligible and happy to offer a bounty hopefully to an answer that is helpful for us, thank you!

Identifying $\mathbb{R}^2$ with $\mathbb{C}$, we have $A = \{ z \in \mathbb{C} \mid 1 \le \lvert z \rvert \le 2 \}$. Define
$$q : I^2 \to A, q(x,y) = 2(x+1)e^{2\pi iy} .$$
This is a well-defined continuous map because $\lvert q(x,y) \rvert = 2(x+1) \in [1,2]$. It is a closed map because $I^2$ is compact and $A$ is Hausdorff. Moreover, $q$ is surjective: If $z \in A$, then $z = \lvert z \rvert e^{2\pi it}$ for some $t \in [0,1]$. Hence $(\frac{\lvert z \rvert -1}{2},t) \in I^2$ and $q(\frac{\lvert z \rvert -1}{2},t) = z$.
So $q$ is a continuous closed surjection, hence a quotient map. Obviuosly we have $q(x,y) = q(x',y')$ if and only if $x = x'$ and either $y = y'$ or $\{y, y' \} = \{0, 1 \}$. Note that the latter implies $(x,y) \sim (x',y')$.
Let $p : A \to T = A/\approx$ denote the quotient map. Then $r = p \circ q : I^2 \to T$ is a quotient map. We claim that $r(x,y) = r(x',y') \Leftrightarrow (x,y) \sim (x',y')$.
"$\Leftarrow$" We have $q(x,0) = q(x,1)$, hence $r(x,0) = r(x,1)$, and we have $q(0,y) = e^{2\pi iy}, q(1,y) = 2e^{2\pi iy} = 2q(0,y)$, hence $r(0,y) = r(1,y)$.
"$\Rightarrow$" In case $q(x,y) = q(x',y')$ we are done. So let $q(x,y) \ne q(x',y')$. Hence w.l.o.g. (1) $\lvert q(x,y) \rvert = 1$ and  (2) $q(x',y') = 2q(x,y)$. But (2) implies (3) $\lvert q(x',y') \rvert = 2$ and (4) $y = y'$ or $\{y, y' \} = \{0, 1 \}$. From (1) and (3) we conclude $x = 0$ and $x' = 1$. Then (4) shows that $(x,y) \sim (x',y')$.
Our above claim proves that $r$ induces a homeomorphism $\hat{r} : I^2 /\sim \phantom{.} \to A/\approx$. See also When is a space homeomorphic to a quotient space?.
