Let be $\mathcal{D}^{2}=\{z\in\mathbb{C}:|z|\leq 1\}$ and $\mathcal{D}^{2}(\frac{1}{2})=\{z\in\mathbb{C}:|z|\leq \frac{1}{2}\}$. I'm asked to identify who is the collapse $\mathcal{D}^{2}/\mathcal{D}^{2}(\frac{1}{2})$. My intuition is that this quotient space is homeomorphic to the torus, but really I'm not able to show it.
The thing that I know is the following: let be $X$ a topological space, and $A$ a topological subspace, then the collapse of $A$ in $X$ is given by $\mathcal{R}$ where the classes are given by $[x]=\begin{cases}x, x\notin A\\ A, x\in A\end{cases}$.
Any hint to formalize this intuition is appreciated!
The quotient is homeomorphic to the original closed disk.
As the inner disk that is collasped to a point p, is closed it is possible to place p on the real plane as a not isolated point.
The open sets of the quotient that contain p correspond to the open sets of the original disk that contain the inner disk and those that do not contain p correspond to the open sets of the original disk that are disjoint to the inner disk.
Were the inner disk an open disk, then the quotient would be homeomorphic to a closed annulus with an isolated point. It makes no difference topologically if the point is inside or outside the ring.