Existence of a simple homeomorphism Definitions:
Define $Q=[0,1] \times [0,1]$ with the product topology and $C=\{(s,t) \in Q:s=0\} \cup \{(s,t) \in Q:t=1\} \cup \{(s,t) \in Q:s=1\}$.
Define $Q/C$ the quotient space by the relation:
$a,b \in Q$ satisfies $a \mathscr{R} b$ if and only if $a=b$ or $a,b \in C$.
with the quotient topology.
Define $D^2=\{z \in \mathbb{C}: ||z|| \leq 1\}$.
Problem: There exists an homeomorphism $f : Q/C \to D^2$ such that for all $s \in [0,1]$ holds that $f(s,0)=e^{2\pi i s}$.
It is clear to me that the statement above is true but I would like to prove it formally.
I do not understand if this problem can be solved with elementary tools or not. I really don't know how to start to prove it.
Furthermore, we can find such homeomorphism explicitly? Or we can just prove that it exists?

We can find such an $f$ explicitly by constructing a map $g \colon Q \to D^2$ that factors through $Q/C$ and induces the desired homeomorphism.
Such a $g$ must map every horizontal line segment $t = \operatorname{const}$ to a closed Jordan curve with base point $1$, and these loops must sweep the whole disk and be disjoint except for the base point. The most easily seen way to do it is
$$g(s,t) = t + (1-t)\cdot e^{2\pi i s}\,.$$
It is clear that $g$ is a continuous map from $Q$ to $D^2$ that factors through the quotient $Q/C$. It is also clear that $g(s,t) = 1 \iff (s,t) \in C$. That any two of the loops $s \mapsto g(s,t)$ for fixed $t$ have only the base point in common is easily seen, and that every point of $D^2$ lies in the image of $g$ also isn't difficult to check.
Thus $f \colon Q/C \to D^2$ induced by $g$ is a continuous bijective map, by compactness it follows that $f$ is a homeomorphism. The condition $f(s,0) = e^{2\pi i s}$ also holds by construction.
