Prove that $\frac{{S^1} \times [0,1]}{S^1\times \{1\}}$ is homeomorphic to $\mathbb{D}^2?$
My attempt :I know that ${S^1} \times [0,1]$ is regular cylinder and $S^1\times \{1\} \cong S^1$
We know that $\mathbb{D}^2$ is a contractible so there exist a map $f:\mathbb{D^2} \to {x_0}$ and $g:{x_0} \to \mathbb{D}^2$ such that
$fg\cong id_{x_0}$ and $gf\cong id_{\mathbb{D}^2} \implies f $ is homeomorphism.
Therefore ,i think $$\frac{{S^1} \times [0,1]}{S^1\times \{1\}} \cong \{x_0\}$$
If $X$ is a topological space, and $A$ is a subspace of $X$, then $X/A$ is homeomorphic to a space $Y$ if there is a quotient map $f : X \to Y$ such that $$(\forall x_1,x_2 \in X) \quad f(x_1) = f(x_2) \iff x_1 = x_2 \textrm{ or } x_1,x_2 \in A.$$ In your case, let $X = S^1 \times [0,1]$, $A = S^1 \times \{1\}$, $Y = D^2$, and consider $f : X \to Y$ such that $f(z,t) = (1-t)z$ for all $z \in S^1$ and $t \in [0,1]$.