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I'm trying to state crystal clear that the extended complex plane $\widehat{\mathbb C}$ is homeomorphic to the sphere $S^2$ through the stereographic projection. Of course in this case it easy to see that the two spaces $S^2-(0,0,1)$ and $\mathbb C$ are homeomorphic. But what is the theorem that states that then $S^2$ and $\widehat{\mathbb C}$ are homeomorphic. I know it's a well known theorem but I'd like to have the complete details.

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    $\begingroup$ Have you considered one point compactification? en.wikipedia.org/wiki/Alexandroff_extension $\endgroup$ Apr 2, 2019 at 9:28
  • $\begingroup$ Have you not just written down an explicit homeomorphism using the stereographic projection maps? $\endgroup$
    – Tyrone
    Apr 2, 2019 at 9:37
  • $\begingroup$ @Tyrone yes I did, I don't have problem there. My point is that the homeomorphism is between the two spaces removing one point (namely the north pole in $s^2$ and infinity in the complex extended palne. I want to know how can I prove that adding these two point the space are necessarely homeomorphic $\endgroup$
    – Dac0
    Apr 2, 2019 at 9:40
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    $\begingroup$ I mean write down an explicit homeomorphism between the two spaces without removing a point. It seems like you have correctly defined the maps on all but one point. Can you extend your defintion over the final point? $\endgroup$
    – Tyrone
    Apr 2, 2019 at 9:42
  • $\begingroup$ @Tyrone You are right, but then I have to be sure of the continuity of the map in that point... so if there was already a theorem it would be already done $\endgroup$
    – Dac0
    Apr 2, 2019 at 9:49

1 Answer 1

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The extended complex plane $\widehat{\mathbb C}$ is defined by adjoining to $\mathbb C $ an additional point at infinity, that is $\widehat{\mathbb C} = \mathbb C \cup \{ \infty \}$. Algebraic operations are defined in the obvious way. However, this does not automatically provide a topology on $\widehat{\mathbb C}$.

The standard topological model of the extended complex plane $\widehat{\mathbb C}$ is the Riemann sphere $S^2 \subset \mathbb{R}^3$. In fact, many authors use the stereographic projection $p : S^2 \setminus \{ (0,0,1) \} \to \mathbb C$ to introduce $\widehat{\mathbb C}$. This map is a homeomorphism, and $i = p^{-1}$ embeds $\mathbb C$ as an open subset into $S^2$. Clearly $i$ extends to a bijection $h : \widehat{\mathbb C} \to S^2$ by defining $h(\infty) = (0,0,1)$. Then $h$ induces a unique topology on $\widehat{\mathbb C}$ making $h$ a homeomorphism. With this topology $\widehat{\mathbb C}$ is a compact metrizable space and the subspace $\mathbb C$ receives its original topology.

If you use this construction as the definition of $\widehat{\mathbb C}$ as a topological space, then nothing remains to be shown.

On the other hand, it suggests itself to define the space $\widehat{\mathbb C}$ as the Alexandroff compactification of $\mathbb C$. Open neighborhoods of $\infty$ are the complements of compact subsets of $\mathbb C$.

Here are some well-known facts.

(1) The Alexandroff compactification of a space $X$ is a compact Hausdorff space if and only if $X$ is a locally compact Hausdorff space.

(2) For any two embeddings $i_1: X \to C_1, i_2: X \to C_2$ of a locally compact Hausdorff space $X$ into compact Hausdorff spaces $C_k$ such that $C_k \setminus i_k(X)$ is a one-point set, there exists a unique homeomorphism $g : C_1 \to C_2$ such that $gi_1 = i_2$.

If we apply this to $i : \mathbb C \to S^2$ and $\mathbb C \hookrightarrow \widehat{\mathbb C}$, we get the desired homeomorphism.

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