# Under what conditions two spaces that are homeomorphic with a point removed are homeomorphic

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.

• Have you considered one point compactification? en.wikipedia.org/wiki/Alexandroff_extension Apr 2, 2019 at 9:28
• Have you not just written down an explicit homeomorphism using the stereographic projection maps? Apr 2, 2019 at 9:37
• @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
– Dac0
Apr 2, 2019 at 9:40
• 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? Apr 2, 2019 at 9:42
• @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
– Dac0
Apr 2, 2019 at 9:49

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.