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Due to some reason I do not know what diffeomorphisms are (and just now learned the definition from Wikipedia), so I am very unsure about whether my current argument is correct.

Obviously $F$ is a bijection (with inverse as itself). Therefore I think I only need to prove that $F$ is differentiable on $S^2$. My proof is as follows:

Fix a point $p \in S^2$ with a corresponding chart $x$ (whose image contains $p$). I want to show that $F \circ x$ is differentiable, which is true trivially by $F$ differentiable in $\mathbb{R}^n$ and the regularity of $x$ (implying $x$ differentiable). Then $F \circ x$ differentiable. I can now conclude that $F$ is a diffeomorphism on $S^2$.

Is my argument correct?

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If I understand correctly, your argument is this: because the map $F:\Bbb R^3\to \Bbb R^3$ given by $F(p)=-p$ is a smooth map and $S^2$ is an embedded submanifold of $\Bbb R^3$ such that $F({S^2})=S^2$, therefore $F|_{S^2}:S^2\to S^2$ is smooth. This is actually true in general. It is a combination (3) of two results (1) and (2) below.

(1) Let $f:M\to N$ be a smooth map between smooth manifolds. If $X$ is a submanifold (immersed or embeded) of $M$, then $f|_X:X\to N$ is also a smooth map.

To show this, we note that the embedding $\iota:X\to M$ is a smooth map. Since the composition of two smooth maps is smooth and $f\circ \iota=f|_X$, $f|_X$ is therefore smooth.

(2) Let $f:M\to N$ be a smooth map between smooth manifolds. If $Y$ is an embedded submanifold of $N$ such that $f(M)\subseteq Y$, then $f:M\to Y$ is smooth. If $Y$ is an immersed submanifold of $N$ such that $f(M)\subseteq Y$ and $f:M\to Y$ is a continuous map, then $f:M\to Y$ is smooth.

The proof is a bit long, so I refer you to Theorem 5.29 of John Lee's Introduction to Smooth Manifolds.

(3) Let $f:M\to N$ be a smooth map between smooth manifolds and $X$ is a submanifold (immersed or embedded) of $M$. If $Y$ is an embedded submanifold of $N$ such that $f(X)\subseteq Y$, then $f|_X:X\to Y$ is smooth. If $Y$ is an immersed submanifold of $N$ such that $f(X)\subseteq Y$ and $f|_X:X\to Y$ is a continuous map, then $f|_X:X\to Y$ is smooth.

This is simply a corollary of (1) and (2).


Here is a more hands-on proof. Let $S^2=\big\{(X,Y,Z)\in \Bbb R^3\ :\ X^2+Y^2+Z^2=1\big\}$. To show that $F$ is differentiable (smooth in this case) is to show that for an atlas $\big\{(\phi_\alpha:U_\alpha\to V_\alpha)\ :\ {\alpha\in J}\big\}$ of $S^2$, the maps $$\psi_{\alpha,\beta}:\phi_\alpha^{-1}\big(U_\alpha\cap U_\beta\cap F^{-1}(U_\alpha\cap U_\beta)\big) \to \phi^{-1}_\beta\big(U_\alpha\cap U_\beta\cap F(U_\alpha\cap U_\beta)\big)$$ are smooth, where $\psi_{\alpha,\beta}(x)=\phi_\beta^{-1}\circ F\circ \phi_\alpha(x)$. Recall that $S^2$ has an atlas $$\big\{(\phi_s:U_s\to V_s),(\phi_n:U_n\to V_n)\big\}$$ given by stereographic projections from the south pole $s=(0,0,-1)$ and the north pole $n=(0,0,1)$. To be precise, $U_s=S^2\setminus \{s\}$, $U_n=S^2\setminus\{n\}$, $V_s=V_n=\Bbb R^2$, $$\phi_s(X,Y,Z)=\left(\frac{X}{1+Z},\frac{Y}{1+Z}\right)$$ and $$\phi_n(X,Y,Z)=\left(\frac{X}{1-Z},\frac{Y}{1-Z}\right).$$

From the given definition of $F$, we can see that $$\psi_{s,n}(x,y)=(-x,-y)=\psi_{n,s}(x,y)$$ for all $(x,y)\in \Bbb R^2$. Likewise $$\psi_{s,s}(x,y)=\left(-\frac{x}{x^2+y^2},-\frac{y}{x^2+y^2}\right)=\psi_{n,n}(x,y)$$ for all $(x,y,z)\in\Bbb R^2\setminus\big\{(0,0)\big\}$. These are smooth maps.

Since $F^{-1}=F$, $F^{-1}$ is also smooth. That is, $F$ is a diffeomorphism on $S^2$.

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