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I am a bit confused about the definition of a section and its relation to a fiber. To wrap it up: Let $(E, \pi, M)$ be a fiber bundle. A (local) section is a map $\phi: W \to E$ such that $\pi \circ \phi = id_M$ and $W \subset M$. If a section is defined on all $M$ it is called a global section.

Can't I always take (at least locally) take the inverse of the projection $\pi$, i.e $\pi^{-1}$ since the composition equals the identity?

Furthermore we call the set $\pi^{-1}(x)$ the fiber over $x \in M$. Is the image of a section also a fiber? Or is it maybe a fiber if a section is evaluated at a single point?

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    $\begingroup$ A function only has an inverse if it is a bijection. Most fiber bundles aren't injective. So, locally, you do have an inverse. But, this inverse does not always extend to all of $M$. $\endgroup$
    – J126
    Commented Oct 18, 2016 at 13:54
  • $\begingroup$ Then why is in most tests the fiber defined as $\pi^{-1}(x)$? Is it because it is defined locally? $\endgroup$
    – JDoe
    Commented Oct 18, 2016 at 13:56
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    $\begingroup$ The fiber is the set of points in $E$ that map to $x$. So, each point in $M$ has a fiber that maps to it. When they write $\pi^{-1}(x)$, they mean the pre-image of $x$. They are not claiming the existence of an inverse function. $\endgroup$
    – J126
    Commented Oct 18, 2016 at 13:57
  • $\begingroup$ Ah this makes things clearer. So I can't call the image of a section a (part of a) fiber? $\endgroup$
    – JDoe
    Commented Oct 18, 2016 at 13:59
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    $\begingroup$ A section is a choice of exactly one element in $\pi^{-1}(x)$ for each $x \in M$. Note, since fiber bundles are (usually) assumed to be surjective, you can always do this. The issue comes when you demand your sections be continuous, differentiable, etc. $\endgroup$
    – J126
    Commented Oct 18, 2016 at 14:04

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Joe Johnson has made some very good points in his comments. I'm adding this answer just for the sake of a nice picture.

A fiber bundle

In this figure, the green circle is our manifold $M$ and the orange spiral (which is assumed to carry on forever) is our manifold $E$. (Notice that $E\cong\mathbb{R}$.) The projection $\pi\colon E\to M$ is simply projection along radial line segments emanating from the origin. If $x\in M$ is the blue point shown on the green circle, then $\pi^{-1}(x)$, the fiber over $x$, is the disjoint collection of red points you see in $E$. It's the set of points that get projected to $x$.

A section over the circle is then a choice of a single fiber element for each point in the circle. One way to obtain such a section (for the fiber bundle pictured) is to say, for instance, that $\phi(x)$ is the element of $\pi^{-1}(x)$ which is nearest the origin among those points outside the circle, for each $x\in M$. This means $\phi$ will trace out the first loop of the spiral that sits outside the circle.

As Joe points out, we still have issues such as continuity, differentiability, etc. that need working out.

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