Definition of sphere and disk are following
\begin{align} S^n =\{ (x_1 , \cdots x_{n+1}) \in \mathbb{R}^{n+1} | \sum x_i^2 =1 \} \end{align} \begin{align} D^n =\{ (x_1 , \cdots x_{n}) \in \mathbb{R}^{n} | \sum x_i^2 \leq 1 \} \end{align}
In Algebraic topology class, Prof, says there is a following homemorphic relation \begin{align} D^n / S^{n-1} \simeq S^n \end{align} Here $\simeq$ means homemorphic. Is it true?(Or i mistake the note?) In the above definition $D^n$ has $n$ dimension and $S^n$ is $n+1$ dimension, So in terms of dimension i felt uncomfortable.
Can anyone give some derivation or hint for above relation?
Also i want to know the $D^2 \simeq S^1$ is valid expression.
Further, i want to find some restriction such that
\begin{align} S^2 \rightarrow S^1 \end{align}
Here is some intuition as to why $D^n/S^{n-1} \simeq S^n$ is true.
Observe that the boundary of $D^n$ is homeomorphic to $S^{n-1}$, so what this relationship is expressing is that if you glue the boundary of an $n$-dimensional ball to a point, what you get is the $n$-dimensional sphere. This can be nicely visualized in the case of $n=1$ and $n=2$. In the first case, you are simply gluing together the endpoints of an interval.
A relevant invariant you can use to see that $D^2$ and $S^1$ are not homeomorphic is to look at how loops can move around them. In $D^2$, any loop can be shrunk continuously to a point, but the same is not true in $S^1$. This argument is not rigorous just yet, but no doubt you have or will see it in the future.