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I have to prove that the compactification of $\Bbb R^n$ (i.e. $\Bbb R^n\cup\{{\infty}\}) $ is diffeomorphic to $S^n=\{x \in \Bbb R^{n+1}: ||x||=1\} $.

Firstly, I proved the stereographic projection function on these sets:$ \pi: S^n \backslash\{(0,...,0,1)\} \to \Bbb R^n,\,\pi(x_1,...,x_{n+1}) = \frac{1}{1-x_{n+1}}(x_1,...,x_n)$ is a diffeomorphism, with inverse

$\pi^{-1}:\Bbb R^n \to S^n \backslash\{(0,...,0,1)\},\,\pi^{-1}(y_1,...,y_n)=\frac{2}{1+||y||^2}(y_1,...,y_n,\frac{||y||^2-1}{2}) $.

But I don't know how to prove $\pi$ is differentiable in $(0,...,0,1)$ and $\pi^{-1}$ in $\infty$.

I'm supossing that this $\pi$ function is the suitable one to prove the diffeomorpfism.

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    If you want to prove that $\mathbb{R}^n \cup \{ \infty \}$ is diffeomorphic to $S^n$, you first need to decide how to put a smooth structure on $\mathbb{R}^n \cup \{ \infty \}$ (otherwise, the statement doesn't make sense). The answer to your question will depend on how you do that. Namely, you will extend $\pi$ to the whole of $S^n$ by defining $\pi((0,\dots,0,1)) = \infty$ and then you will have to show that $\pi$ is smooth as a map from $S^n$ to $\mathbb{R}^n \cup \{ \infty \}$ by checking that it is smooth with respect to the appropriate charts. – 2017-01-05

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