# Fiber Bundle (Trivial Fibration)

Let $S^{1}$ be the unit circle with basepoint $1 \in S^{1}$. Consider the map $f_{n} : S^{1} \rightarrow S^{1}$ given by $f_{n}(z)=z^{n}$. Then $f_{n} : S^{1} \rightarrow S^{1}$ is locally trivial fibration with fiber a set of n distinct points (the nth root of unity).

What I don't understand here is that why is the fiber the n distinct points (nth root of unity)? Also, for this example if we fix an x in the base space then locally it is like $U_{x} \times F$ where $U_{x}$ is an open set and F is the fiber. How do we visualize this? $U_{x}$ in this case is an arc am I right? Then why is it "locally" the cartesian product of $U_{x}$ and the set of roots?

The fiber is by definition $f_n^{-1}(1)$, which is the set of all solutions to $z^n=1$. These are by definition the $n$th roots of unity.

You are right that the "correct" $U_x$ is a little teeny arc $U_x$. Since $F$ is finite (and discrete), $U_x \times F$ is the same as a disjoint union of copies of those teeny arcs. Since $f_n$ wraps the circle around itself $n$ times, $U_x \times F$ is a bunch of equally spread out little arcs around the (domain) circle.

• Thanks Randall! So if n=2, then we only have two roots i.e. 1 and -1, then how is it possible to cover the whole circle? Or I cannot fix the n here? – LanaDR Sep 7 '17 at 13:48
• @user478267 The roots of unity are only the fibers of $1$. Every element of the base space has a fiber. In this case, the fibers will be cosets of the roots of unity. That is, the solutions to $z^n=w$ are a coset of the group of units satisfying $z^n=1$. (Can you prove this?) Thus, while the fiber of $1$ is an $n$-gon with $1$ as a vertex, the other fibers are rotations of that $n$-gon. Clearly they collectively include all points of the total space. – anon Sep 7 '17 at 13:51
• Ahh...I think I get it! Thank you all! – LanaDR Sep 7 '17 at 13:55
• ^Correct. "The" fiber is the preimage of 1, but the other fibers are necessarily homemorphic to this special fiber. This is a by-product of choosing a basepoint for your cover/fibration. – Randall Sep 7 '17 at 13:55

(1) The solutions to $z^n=1$ are the $n$th root of unity. More generally, for any $w$ in the base space, if we have one solution $z_0$ to $z^n=w$, then all of the solutions are $\{z_0\xi:\xi^n=1\}$, i.e. a coset of the group of $n$th roots of unity. How do you prove this? Well, let $z_1$ be any other solution to $z^n=w$, then notice $(z_1/z_0)^n=z_1^n/z_0^n=w/w=1$, so $z_1/z_0=\xi$ is a root of unity, hence $z_1=z_0\xi$, and conversely if $\xi$ is any root of unity then $(z_0\xi)^n=z_0^n\xi^n=w\cdot1=w$, so $z_0\xi$ is also a root.

(2) The preimage of a (sufficiently small) arc in the base space is indeed $n$ disjoint arcs in the total space. A disjoint union of $n$ arcs is, topologically, the same as the product of the arc and $\{1,\cdots,n\}$ interpreted as a discrete space. In other words, $U\times F$, where $U$ is the arc and $F$ is a set of size $n$.

This only works locally because if you make the arc in the base space too big, it's no longer true. Consider what happens as we make the arc in the base space bigger. The biggest we can go before it's the whole circle is the whole circle minus a point, and the preimage will be all of the circle minus $n$ points. Once we fill in the missing point in the base space, that fills in the missing $n$ points in the total space, which connects the $n$ disjoint arcs into one connected circle.

Clearly the whole circle is not the same as $n$ disjoint circles (which would be the trivial bundle, an actual direct product $S^1\times\mathbb{Z}_n$). If it helps the visualization, consider a rubber band that is tight enough that when it sits on the counter, it wraps around a given imaginary circle $n$ times. That illustrates how $S^1$ covers itself in an $n$-fold fashion.

A similar example occurs with infinitely many wraps. The space $\mathbb{R}$ (the number line) can be interpreted as a spiral winding around a cylinder; by projecting to a circle around the cylinder we get a map $\mathbb{R}\to S^1$ (which, if we use the angles of elements of $S^1$ to interpret $S^1$ as $\mathbb{R}/2\pi\mathbb{Z}$, this is just the quotient map $\mathbb{R}\to\mathbb{R}/2\pi\mathbb{Z}$). Clearly $\mathbb{R}$ is not an infinite collection of disjoint circles, it's one big connected space.