Proving continuity of a map from $D^2$ onto $S^2.$

Question

I am following the lecture note in topology given by our instructor. There I found a proof of the fact that $D^2/S^1$ is homeomorphic to $S^2.$ The proof of this result goes along the following lines $:$

Let us denote the inverse of the stereographic projection from $S^2 - N$ onto $\Bbb R^2$ by $\mathscr I.$ Then define a function $f : D^2 \longrightarrow S^2$ as follows $:$

$$ f : x \longmapsto \begin{cases} \mathscr {I} \left (\frac {x} {1 - |x|} \right ) & x \in D^2 \setminus S^1 \\ N & x \in S^1 \end{cases} $$

This map is continuous and surjective and hence by universal property of quotient topology we have a bijective continuous map $\overline {f} : D^2/S^1 \longrightarrow S^2.$ Now since $D^2/S^1$ is compact (being the image of the compact space $D^2$ under the quotient map) and $S^2$ is Hausdorff it follows that $\overline {f}$ is a homeomorphism and this proves the result.

Here I don't understand why $f$ is continuous. It is clear that $f$ is a continuous bijection from $D^2 \setminus S^1$ onto $S^2 - N.$ But how do I show that $f$ is continuous $S^1\ $? Any help or suggestion will be appreciated.

Thanks in advance.

Answer 1

On $S^1$, $f$ is constant. We must show that if $x_n\to x\in S^1$, where $x_n\in D^2$, then $f(x_n)\to N$. That seems clear. As $|x|\to1$, $\left\lvert\frac x{1-|x|}\right\rvert\to\infty$ and $$\mathscr I\left(\frac x{1-|x|}\right)\to N$$

Answer 2

Another proof (less explicit but topologically correct): $D^2\setminus S^1$ is just the open unit disk which is homeomorphic to $\Bbb R^2$. So when we form $D^2{/}S^1$ we're left with in essence $\Bbb R^2$ unioned with the identified boundary, which is just a point in the quotient space, and together the form a compact Hausdorff space. So $D^2{/}S^1$ is just a one-point compactification of $\Bbb R^2$ and by unicity of that space it's homeomorphic to $S^2$, another one-point compactification of $\Bbb R^2$.