Let $\pi$ be the quotient map $S^n\to\mathbb R\mathrm{P}^n$. Is $\pi_{\star p}:T_pS^n\to T_{\pi(p)}\mathbb R\mathrm P^n$ an isomorphism?

Question

I am trying to prove the following statement:

Let $\pi$ be the quotient map $S^n\to\mathbb R\mathrm{P}^n$. Then $\pi_{\star p}: T_pS^n\to T_{\pi(p)}\mathbb R\mathrm P^n$ an isomorphism

Here is my attempt:

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Proof. Since $\dim S^n=\dim \mathbb R\mathrm P^n= n$ we only have to check that $\pi_{\star p}$ is surjective. Indeed, let $Y_{\pi(p)}$ be a vector in $T_{\pi(p)}\mathbb R\mathrm P^n$ and $\alpha:(-\xi,\xi)\to\mathbb R\mathrm P^n$ be a smooth curve such that $\alpha(0)=\pi(p)$ and $\alpha'(0)=Y_{\pi(p)}$. Then there exists another curve $\beta:(-\delta,\delta)\to S^n$ such that $\pi\circ\beta=\alpha$. This implies that: $$\pi_\star \beta'(0)=(\pi\circ\beta)'(0)=\alpha'(0)=Y_{\pi(p)}$$ Thus $\pi_{\star p}$ must be surjective.

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How can I ensure the existence of that curve $\mathbf{\beta}$?

Answer 1

It is true because $\pi$ is a local diffeomorphism. See Why does the projection $\pi:(a,b,c) \in S^2 \to[a,b,c] \in P^2$ have rank 2 everywhere? for $n =2$. You can easily generalize this to arbitrary $n$.