One-point compactification of $R^n$

Question

I need to prove that there exists a homeomorphism between the one-point compactification of $\mathbb{R}^n$, that is, $\mathbb{R}^n\cup\lbrace\infty\rbrace$, and the $n$-dimensional sphere $S^n$. Let's denote the north pole of $S^n$ by $N$.

Let's remember that if $\tau$ is the usual topology of $\mathbb{R}^n$, then there exists a unique topology $\varphi$ that makes $\mathbb{R}^n\cup\lbrace\infty\rbrace$ Hausdorff and compact. And this topology is defined as $\varphi = \tau\cup\lbrace V|\infty\in V; \mathbb{R}^n\setminus V$ compact$\rbrace$.

We know that the stereographic projection $f\colon\mathbb{R}^{n}\longrightarrow S^n\setminus \lbrace N\rbrace$ given by $f(x_1,\cdots , x_n) = \left(\dfrac{2x_1}{x_1^2+\cdots + x_n^2},\cdots, \dfrac{2x_{n}}{x_1^2+\cdots + x_n^2}, \dfrac{x_1^2+\cdots + x_n^2 -1}{x_1^2+\cdots + x_n^2 +1}\right)$ is a homeomorphism, which is clear to me. Let's denote the inverse of $f$ by $g$.

To show that $S^n$ is homeomorphic to $\mathbb{R}^n\cup\lbrace\infty\rbrace$, we can define the function $\bar{f}\colon\mathbb{R}^n\cup\lbrace\infty\rbrace\longrightarrow S^n$ given by $\bar{f}(x)=f(x)$ if $x\in\mathbb{R}^n$ and $f(\infty)=N$, which is clearly a bijection with inverse $\bar{g}\colon S^n\longrightarrow\mathbb{R}^n\cup\lbrace\infty\rbrace$ given by $\bar{g}(x)=g(x)$ if $x\in S^n\setminus\lbrace N\rbrace$ and $g(N) = \infty$.

So, my question is: how can I prove that $\bar{f}$ and $\bar{g}$ are continuous at $\infty$ and $N$ respectively?

Initially, to show the continuity of $\bar{f}$, my idea was to take an open neighborhood $V$ of $N$. By the definition of relative topology, this implies the existence of an open set $A$ of $\mathbb{R}^{n+1}$ such that $V = A\cap S^n$. But upon analyzing the inverse image $\bar{f}^{-1}(V) = \bar{f}^{1}(A)\cap\bar{f}^{-1}(S^n)$ I couldn't conclude anything. How can I proceed?

Answer 1

Let us define a one-point compactfication of a locally compact Hausdorff space $X$ to be a pair $(X', \iota)$ consisting of a compact Hausdorff space $X'$ and an embedding $\iota: X \to X'$ such that $X' \setminus X$ is one-point space. The point in $X' \setminus X$ will be called the "point at infinity" denoted by $\infty_{X'}$.

Lemma. Let $(X'_i,\iota_i)$ be one-point compactifications of $X_i$, $i = 1,2$, and let $h : X_1 \to X_2$ be a homeomorphism. Then there exists a unique homeomorphism $h' : X'_1 \to X'_2$ such that $h' \circ \iota_1 = \iota_2 \circ h$. This maps $\infty_{X'_1}$ to $\infty_{X'_2}$.

Proof. Define $$h'(x)= \begin{cases} \iota_2(h(\iota_1^{-1}(x))) & x \in \iota_1(X_1) \\ \infty_{X'_2} & x = \infty_{X'_1} \end{cases}$$ This is of course the only bijective function with the property $h' \circ \iota_1 = \iota_2 \circ h$. Clearly $h'$ is continuous at all points of the open subset $\iota_1(X_1)$ of $X'_1$. It remains to show that $h'$ is continuous at $\infty_{X'_1} $; then we know that $h'$ is a continuous bijection between compact Hausdorff spaces and hence a homeomorphism.

Let $U_2$ be an open neigborhood of $\infty_{X'_2} $ in $X'_2$. Then $C_2 = X'_2 \setminus U_2$ is a closed subset of $X_2$, hence compact. It is contained in $\iota_2(X_2)$. Thus $C_1 = \iota_1(h^{-1}(\iota_2^{-1}(C_2)))$ is a compact subset of $\iota_1(X_1)$ and $U_1 = X'_1 \setminus C_1$ is an open neigborhood of $\infty_{X'_1} $ in $X'_1$. We get $h'(U_1) = U_2$. To see this, observe that $h'$ is a bijection so that it suffices to show $h'(X'_1 \setminus U_1) = X'_2 \setminus U_2$. But $X'_i \setminus U_i = C_i$ and $h'(C_1) = C_2$ by construction.

You can apply the Lemma to the inclusions $\mathbb R^n \to \mathbb R^n \cup \{\infty\}$ and $S^n \setminus \{ N \} \to S^n$. For $h$ you can take the stereographic projection $\mathbb{R}^{n}\longrightarrow S^n\setminus \lbrace N\rbrace$.

For a related and more general result see Is one point compactification functorial? and Functoriality of the one-point compactification.

Answer 2

It is conceptually nicer to ask; is $S^n$ the one-point compactification of its puncture $S^n\setminus\{N\}$? And the answer to this is automatically yes, since $S^n$ is compact Hausdorff. The only thing to check is that open neighbourhoods of $N$ are characterised as complements of compact subsets of $S^n\setminus\{N\}$, but this is surely true; an open neighbourhood of $N$ in $S^n$ is the complement of some closed set avoiding $N$ - by definition - and closed subsets of a compact space are compact.

Directly, if $U$ is an open neighbourhood of $N$ and $K:=S^n\setminus U$, $f^{-1}(K)$ is easily bounded and thus compact; $f$ is a proper mapping, all notions being equivalent in locally compact Hausdorff space. Now it follows $f^{-1}(U)=\{\infty\}\cup\Bbb R^n\setminus\text{something compact}$ i.e. $f$ is continuous at infinity. And continuous bijections of compact Hausdorff spaces are homeomorphisms.