# Natural projection of tangent bundle is submersion

I am wondering how to see that the natural projection from $\pi:TX \to X$ is a submersion. To see that $\pi$ is surjective I can identify $TX$ as product space $X \times F$ where $F$ is a vector space with the same dimension as $X$ and so for all $(x,v) \in TX \simeq X \times F$, $\pi(x,v) = x$ is surjective. Correct? But I don't now how to proof that $d\pi$ is surjective. Do I need some concept of the double tangent space Or can I avoid the map $d\pi: TTX \to TX$?

• You can't really identify $T(X)$ as $X\times F$. Locally, yes. Globally, not always. But luckily, locally is good enough in this case. EDIT: I'm hesitant to give out the entire solution straight away because something tells me that this is a homework problem – funktor Aug 15 '16 at 17:02
• I am actually try to understand this for my bachelor thesis. I am slightly passing this topic and don't know how far I have to go into the theory. That's why I asked if I need the concept of the double tangent space. So your help is really appreciated. – JDoe Aug 16 '16 at 8:17
• @funktor I have another question regarding your first comment. If $X$ is a $C^k$ manifold ($k \in \mathbb{N}$) isn't it correct that the Tangent fiber bundle $(T(X), \pi_{TX}, X)$ is globally trivial and therefore I can identify it with $X \times F$, $F$ is a vector space or is it just possible for $X = F = \mathbb{R}^n$? – JDoe Aug 19 '16 at 15:30
• Your tangent bundle may not be always trivial. As an example, take $S^2$, the sphere. In this case the tangent bundle $T(S^2)$ is not diffeomorphic to $S^2\times \mathbb R^2$ globally. The fact that its tangent bundle is not trivial is what the the Hairy Ball theorem is I believe. en.wikipedia.org/wiki/Hairy_ball_theorem – funktor Aug 31 '16 at 7:01

Let $\pi:T(X)\to X$, be the projection map in question.
Let $U\subset T(X)$ be a neighborhood of some $(x',v')\in T(X)$ and $V\subset X$ a neighborhood of $\pi(x',v')=x'\in X$. We know that there exists a chart $\psi:V\to \psi(V)\subset \mathbb R^k$.
• Note that you can directly show that $\nu$ is a submersion by computing its Jacobian matrix. – Sou Dec 28 '17 at 17:35