I've been struggling with this problem for several hours now and I would appreciate some assistance. This is the only part of the larger problem that I am stuck on.
The Tangent Bundle $TM$ of a differentiable manifold $M$ is defined by the following:
$TM := \sqcup_{x \in M} T_xM = \{(x,y)|x \in M \text{ and } y \in T_xM\}$.
$TM$ inherits a manifold structure from $M$ and has dimension $2m$.
I wish to prove the following claim:
Claim: Given a differentiable manifold $M$ of dimension $m$, the projection operator $\pi: TM \mapsto M$ defined by $\pi(x,v) = x$ is continuous.
How do I proceed with the proof? Clearly the projection operator is continuous, correct? My first instinct is to take an open set $\chi \subseteq M$ in the sense that the coordinate charts/patches for $M$ form a countable basis for the (Hausdorff) topology on $M$. Then I would show that $\pi^{-1}(\chi) \subseteq TM$ is open. However, I am not finding properties of $\pi$ in a suitable way to show continuity.
Any hints would be appreciated.
The projection is not only continuous, it is smooth.
Recall the charts for $TM.$ Given a chart $(U,\varphi)=(U,x^i)$ for $M$ we have a chart $(\pi^{-1}(U),\tilde{\varphi})$ for $TM,$ where $\tilde{\varphi}:\pi^{-1}(U)\to \varphi(U)\times \mathbb{R}^n$ is given by $\tilde{\varphi}(p,v^i\partial/\partial x_i|_p)=(\varphi(p),v^1,\ldots,v^n).$
Then $\pi$ in the coordinates $(U,\varphi)$ for $M,$ $(\pi^{-1}(U),\tilde{\varphi})$ for $TM$ is given by $$ \varphi\circ\pi\circ \tilde{\varphi}^{-1}(q,v^1,\ldots,v^n)=\varphi\circ\pi(\varphi^{-1}(q),v^i\partial/\partial x^i|_p)=\varphi(\varphi^{-1}(q))=q,$$ so it is smooth. Therefore $\pi$ is smooth as a map $TM\to M.$