# Relation of fiber and image of a section (Fiber bundles)

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?

• 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 $M$.
– J126
Commented Oct 18, 2016 at 13:54
• Then why is in most tests the fiber defined as $\pi^{-1}(x)$? Is it because it is defined locally?
– JDoe
Commented Oct 18, 2016 at 13:56
• 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.
– J126
Commented Oct 18, 2016 at 13:57
• Ah this makes things clearer. So I can't call the image of a section a (part of a) fiber?
– JDoe
Commented Oct 18, 2016 at 13:59
• 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.
– J126
Commented Oct 18, 2016 at 14:04

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.