TITLE: Is not the surjective map $\pi$ associated with a vector bundle infact a bijection?
QUESTION [0 upvotes]: I am reading John M Lee's Riemannian Manifolds : An Introduction to Curvature, which is very well written.
On page 16 : "Vector bundles are defined", quoting 

A (smooth) $k$-dimensional vector bundle is a pair of smooth manifolds $E$ (total space) and $M$ (the base), together with the surjective map $\pi : E \to M $ (the projection) satisfying the following conditions : [...] 

Because later defining smooth sections on page 19 : (quoting again)

If $\pi : E \rightarrow M $ is a vector bundle over M, a section of E is a map $ F : M \rightarrow E $ such that $ \pi \circ F = Id_M. $ [...]

So the point is : when one says $E$ is $TM$, i.e. the tangent bundle, then a point in E is $(p, T_pM)$ (in which case it will be a bijection) or $(p, V)$ where $V \in T_pM$ ?
I am confused between these two concepts : especially "section".
Some reference or explanation with a clear example will be really helpful.

REPLY [1 votes]: In down to earth terms a vector bundle over say a manifold $M$ is a structured datum of a vector space $V_P$ for each point $P\in M$. To choose a section means to choose for each point $P\in M$ a vector $v_P\in V_P$. The simplest non trivial case is that of
$$
\pi:\Bbb R^2\longrightarrow\Bbb R,
\qquad\pi(x,y)=x.
$$
Here, choosing a (continuous) section $\sigma:\Bbb R\rightarrow\Bbb R^2$ amounts to choosing a (continuous) function $y=f(x)$ defined for all $x\in\Bbb R$.
It is clear that $\pi$ is surjective but not a bijection!
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