Loops in the Base and Paths Upstairs
A loop in the base space lifts to a path in the covering space, and the endpoint of that lift may be a different point of the original fiber. This gives the fundamental group a way to move points within the fiber. The central fact is that this action records exactly which points of the fiber can be joined by paths in the covering space.
Let \(p:E\to B\) be a covering map and fix \(b_0\in B\). A covering map means that every point of \(B\) has an open neighborhood \(U\) for which \(p^{-1}(U)\) is a disjoint union of open sets, each mapped homeomorphically onto \(U\) by \(p\). Give \(E\) and \(B\) their stated topologies, and let \(F=p^{-1}(\{b_0\})\) be the fiber over \(b_0\). A path is a continuous map from \(I=[0,1]\). A loop based at \(b_0\) is a path \(\gamma:I\to B\) with \(\gamma(0)=\gamma(1)=b_0\). The fundamental group \(\pi_1(B,b_0)\) consists of based loops modulo homotopy that keeps both endpoints at \(b_0\).
We use the path-product convention that \(\alpha*\beta\) traverses \(\alpha\) first and then \(\beta\). For \(e\in F\) and a based loop \(\gamma\), the path lifting theorem gives a unique lift \(\widetilde{\gamma}_e:I\to E\) with \(\widetilde{\gamma}_e(0)=e\). Since \(p(\widetilde{\gamma}_e(1))=b_0\), its endpoint is in \(F\). Define \(e\cdot[\gamma]=\widetilde{\gamma}_e(1)\). The homotopy lifting property implies that based-homotopic loops give the same endpoint, so this is well defined on \(\pi_1(B,b_0)\). With the stated path-product convention it is a right action: lifting \(\alpha\) and then \(\beta\) gives the lift of \(\alpha*\beta\), so \((e\cdot[\alpha])\cdot[\beta]=e\cdot([\alpha][\beta])\).
The Orbit–Path Component Theorem
Proof. Suppose first that \(e\) and \(e'\) are in the same orbit. By the definition of orbit, there is a based loop \(\gamma\) at \(b_0\) such that \(e'=e\cdot[\gamma]\). Its lift \(\widetilde{\gamma}_e\) starts at \(e\) and ends at \(e'\). Since a lift is a continuous path in \(E\), this path joins \(e\) to \(e'\); hence they lie in the same path component.
Conversely, suppose a path \(\lambda:I\to E\) joins \(e\) to \(e'\), so \(\lambda(0)=e\) and \(\lambda(1)=e'\). The composite \(\gamma=p\circ\lambda:I\to B\) is continuous, and \(\gamma(0)=p(e)=b_0=p(e')=\gamma(1)\), so it is a based loop. The path \(\lambda\) is a lift of \(\gamma\) starting at \(e\). By uniqueness of path lifts, \(\lambda=\widetilde{\gamma}_e\). Therefore \(e\cdot[\gamma]=\widetilde{\gamma}_e(1)=\lambda(1)=e'\), which puts \(e'\) in the orbit of \(e\). \(\square\)
The proof uses only two covering-space facts: a based loop can be lifted from any chosen point of its initial fiber, and a lift with a specified starting point is unique. In the reverse direction, a path upstairs supplies its own projected loop downstairs. Thus the two descriptions are not merely compatible; each directly constructs the object required by the other.
Components That Meet a Fiber
Proof. Each orbit is contained in a single path component by the theorem. Conversely, if a path component meets \(F\), choose \(e\) in its intersection with \(F\). Every other point \(e'\) in that intersection is joined to \(e\) by a path in the component, so the theorem puts \(e'\) in the orbit of \(e\). Hence the component's intersection with \(F\) is exactly one orbit. If two such components gave the same orbit, that orbit would lie in both components; since path components are equivalence classes under the relation of being joined by a path, the components would be equal. \(\square\)
This statement concerns only components that meet the chosen fiber. If \(B\) is path-connected, every path component of \(E\) does meet \(F\): given \(x\in E\), choose a path in \(B\) from \(p(x)\) to \(b_0\), and lift it starting at \(x\). Its endpoint lies in \(F\) and is joined to \(x\) by that lifted path. Therefore, when \(B\) is path-connected, the orbits on \(F\) correspond to all path components of \(E\). Without that hypothesis, some components may not meet the fiber at all.
Worked Examples
Worked Example: The Two-Sheeted Product Cover of an Interval
Let \(B=I=[0,1]\) with its standard topology, let \(\{0,1\}\) have the discrete topology, and give \(E=I\times\{0,1\}\) the product topology. Define \(p(x,i)=x\). Verify the covering and determine the orbits in the fiber over \(0\).
For any open neighborhood \(U\) in \(I\), its preimage is \(U\times\{0\}\,\cup\,U\times\{1\}\). These are disjoint open sets, and the restriction of \(p\) to either one is a homeomorphism onto \(U\). Thus \(p\) is a covering map. The fiber over \(0\) is \(F=\{(0,0),(0,1)\}\).
A lift of any path \(\gamma:I\to I\) starting at \((0,i)\) is \(t\mapsto(\gamma(t),i)\): it projects to \(\gamma\) and starts at \((0,i)\), so uniqueness identifies it as the lift. If \(\gamma\) is a loop based at \(0\), then \(\gamma(1)=0\), and this lift ends at \((0,i)\). Thus every based loop fixes both fiber points. The two orbits are the singleton sets \(\{(0,0)\}\) and \(\{(0,1)\}\).
The path components are also the two sheets \(I\times\{0\}\) and \(I\times\{1\}\). Each sheet is path-connected because \(I\) is path-connected. No path can join points with distinct second coordinates: the second-coordinate projection of such a path would be a continuous map from the connected interval \(I\) into the discrete space \(\{0,1\}\), and every such map is constant. In particular, the components meet \(F\) in exactly the two orbits found above.
Worked Example: A Cover Over an Indiscrete Two-Point Space
Let \(B=\{u,v\}\) have the indiscrete topology \(\{\varnothing,B\}\), let \(D=\{0,1\}\) have the discrete topology, and set \(E=B\times D\) with the product topology. For the projection \(p(x,i)=x\), find the path components and the orbits over \(u\).
The only nonempty open set in \(B\) is \(B\), and \(p^{-1}(B)=B\times\{0\}\,\cup\,B\times\{1\}\). Each sheet is open and its projection to \(B\) is a homeomorphism: both a sheet and \(B\) have the indiscrete topology, and the projection is a bijection. Hence \(p\) is a covering map. The fiber over \(u\) is \(\{(u,0),(u,1)\}\).
Every map into an indiscrete space is continuous, so every path in \(B\) is continuous. But a path in \(E\) cannot change its second coordinate: its composite with the projection \(E\to D\) is continuous, and a continuous image of the connected interval in the discrete space \(D\) is a singleton. Within each sheet, any two points can be joined by a path, since a map into the indiscrete space \(B\) is continuous. Thus the path components are exactly \(B\times\{0\}\) and \(B\times\{1\}\).
A loop based at \(u\) lifts from \((u,i)\) to a path whose second coordinate is constantly \(i\). Since its projected endpoint is \(u\), the lift ends at \((u,i)\). Both orbits are therefore singletons, matching the intersections of the two path components with the fiber.
Worked Example: A Component That Misses the Chosen Fiber
Let \(B=E=\{a,b\}\) have the discrete topology and let \(p:E\to B\) be the identity map. Choose \(b_0=a\). Identify the fiber orbits and explain why they do not describe every component of \(E\).
The identity map is a covering: each singleton \(\{x\}\) is an open neighborhood whose preimage is the same singleton, mapped homeomorphically onto it. The fiber over \(a\) is \(F=\{a\}\). Every based loop in \(B\) at \(a\) is constant, because any continuous map from the connected interval into a discrete space is constant. Consequently, the only orbit in \(F\) is \(\{a\}\).
In a discrete space every singleton is open, and no path can move from \(a\) to \(b\), again because a continuous image of the connected interval in a discrete space is a singleton. The path components of \(E\) are \(\{a\}\) and \(\{b\}\). Only \(\{a\}\) meets the chosen fiber. This verifies why the corollary refers to components that meet \(F\); the base here is not path-connected.
A Common Pitfall
Common pitfall. A frequent mistake is to infer that two points are path-connected merely because they have the same image under \(p\). For the two-sheeted product cover above, both \((0,0)\) and \((0,1)\) map to \(0\), but they lie in different path components. In terms of the action, the calculation for any loop \(\gamma\) is \((0,i)\cdot[\gamma]=(\gamma(1),i)=(0,i)\), for \(i=0,1\). Neither orbit contains the other point. The correct test is to ask whether a path joins the points in \(E\), or equivalently, by the theorem, whether a based loop has a lift from one point to the other.
Check Your Understanding
Use the orbit–path component theorem to organize your reasoning.
- In the two-sheeted product cover of \(I\), what is the endpoint of the lift of a based loop starting at \((0,1)\), and why?
- Suppose \(e,e'\) lie over \(b_0\) and a path in \(E\) joins them. Describe the based loop that proves they lie in the same orbit.
- Why does the corollary identify only components meeting the chosen fiber when \(B\) is not assumed path-connected?
- For the identity covering of a discrete space \(B\), describe the path components and the orbits in the fiber over a chosen point \(b_0\).