Showing the projection map is a submersion.

Question

I am working through a question in Guillemin and Pollacks Differential topology. Similar questions to this have been asked here before ( Natural projection of tangent bundle is submersion), (Natural projection of tangent bundle is submersion). I understand how these suggested solutions work. However when I attempted the question I had a different approach and correct my attempt to help improve my understanding.

Question (1.8.5): Prove that the projection map $p:T(X)\rightarrow X$, $p(x,v)=x$ is a submersion.

Attempt: Let $X$ be a manifold in $\mathbb{R}^n$. Then its tangent bundle $T(X)$ is a manifold in $\mathbb{R}^{2n}$. Consider the projection map $\pi:\mathbb{R}^{2n}\rightarrow \mathbb{R}^n$, $(x_1,\ldots,x_n,\ldots,x_{2n})\mapsto (x_1,\ldots,x_n)$. We claim that this is a submersion. Note that for any $p\in \mathbb{R}^{2n}$, $T_p(\mathbb{R}^{2n})=\mathbb{R}^{2n}$ and $T_{\pi(p)}(\mathbb{R}^{n})=\mathbb{R}^n$. Hence $d\pi_p:\mathbb{R}^{2n}\rightarrow \mathbb{R}^n$. This is clearly surjective since for any $h\in \mathbb{R}^{2n}$, $d\pi_p(h)=\pi(h)$, hence $\pi$ is a sumbersion.

Part that I am unsure about: Since $p$ is the restriction of $\pi$ to $T(x)$ it must also be that $p$ is a submersion. I think this is true for submanifolds? But is it enought to say that since $T(X)$ is a manifold and a subset of $\mathbb{R}^{2n}$, then it is a submanifold of $\mathbb{R}^{2n}$?

Answer 1

The question id local, if $X$ is a $p$-dimensional manifold, each element $x$ of $X$ has a neighborhood $U_x$ diffeomorphic to an open subset of $\mathbb{R}^p$, you deduce that $TU_{x}=U_{x}\times \mathbb{R}^p$ and the restriction of $p$ to $U$ is the projection $U_x\times \mathbb{R}^p\rightarrow U_x$.