Preamble: I am trying to show that $D^n/\partial D^n \approx S^n$, where $\partial D^n = S^{n-1}$. I've defined a mapping $f:D^n \to S^n$ as $f(x) = \begin{cases} s_1(s_2(x)) &: |x| < 1\\ N &: |x| = 1 \end{cases}$, where $N = (\underbrace{0,\dots,0}_{=n},1)$, $s_1(x) = \frac{1}{1 + |x|^2}\left(2x + (|x|^2 - 1)N\right)$, $s_2(x) = \frac{x}{1 - |x|^2}$ are mappings $s_1:\mathbb{R}^n \to S^{n} - N, s_2:D^n \to \mathbb{R}^n$. I've also already argued why the quotient space $D^n/R_f, xR_f y \Longleftrightarrow f(x) = f(y)$ is equivalent to $D^n/\partial D^n$.
Question context: If I can show that $f$ is an identification mapping, I'm done by a result in my reading material. I.e. I have to argue why $f$ is surjective and it coinduces $S^n$'s original topology. Surjectivity follows easily, as $s_1, s_2$ are bijective, and the North Pole is handled as a special case. But, consequently, I don't know how to argue about the "coninduces original topology" exactly because the North Pole is handled as a special case: If $V \subset S^n$ is open and $N \not\in V$, then $f^{-1}[V]$ is open in $D^n$ since $s_1\circ s_2$ is continuous. Moreover, if $U \subset D^n$ s.t. $\forall x \in U:|x| < 1$ and $U$ is open, then also $f[U] \subset S^n$ is open, since $s_1$ and $s_2$ have continuous inverses. But what about the special case?
Question: How can I show that $f$ coinduces the original topology $\mathcal{T}_{S^n}$ of $S^n$ to $S^n$, i.e. $\{V \subset S^n\mid f^{-1}[V] \in \mathcal{T}_{D^n}\} = \mathcal{T}_{S^n}$?
See the Closed Map Lemma. You have defined a continuous surjection $f$ from a compact space $\mathbb{D}^n$ to a Hausdorff space $\mathbb{S}^n$. The lemma implies that it is a quotient map. Insofar as $f$ identifies the points on $\partial \mathbb{D}^n$, the spaces $\mathbb{D}^n/\partial \mathbb{D}^n$ and $\mathbb{S}^n$ are homeomorphic.