# Showing the projection map is a submersion.

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 $$\mathbb{R}^{2n}$$?

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 $$U_x\times \mathbb{R}^p\rightarrow U_x$$.
• If you mean that the question is to show at at each point, that $p$ is a sumbersion then your answer is very similar to (math.stackexchange.com/questions/1890209/…) which I understand. Could you explain errors/miss understandings in my attempt instead of giving an entirely different solution please? Thank you for your answer though, I appreciate your help and do not mean to come across as rude. – T. Stark Jun 30 at 14:03