Is not the surjective map $\pi$ associated with a vector bundle infact a bijection?

Question

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.

Answer 1

The map $\pi$ is not a bijection.

The elements of $TM$ can be viewed as pairs $(p, v)$ where $p \in M$, $v \in T_pM$, then $\pi((p, v)) = p$, so for $p \in M$, $\pi^{-1}(p) = \{(p, v) \mid v \in T_pM\} \cong T_pM$.

A section is a map $s : M \to TM$ such that $\pi\circ s = \operatorname{id}_M$, so for any $p \in M$, we have $(\pi\circ s)(p) = p$. As $s(p) \in TM$, we can write $s(p) = (q, v)$ for some $q \in M$ and $v \in T_qM$. Note that $$(\pi\circ s)(p) = \pi(s(p)) = \pi((q, v)) = q,$$ but by assumption $(\pi\circ s)(p) = p$, so $q = p$. That is, $s(p) = (p, v)$ where $v \in T_pM$. So you can think of a section $s$ of $TM$ as a (smoothly varying) choice of vector in each tangent space $T_pM$.

Answer 2

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!