Coverings Under Pullback and Restriction
A covering map describes a space as several locally separate copies of its base. Pullbacks let us transfer this local structure along a continuous map: over each point of a new space, the pullback records the points in the original covering that lie above its image. This construction is useful when a map into a base space is given and we want a covering over its domain. Restriction requires care: taking the full preimage of a subspace always works, but deleting selected points from the covering space can destroy the covering property.
We use the following definition. A map \(p:E\to B\) is a covering map if it is continuous and surjective and every \(b\in B\) has an open neighborhood \(U\) such that \(p^{-1}(U)\) is a disjoint union of open subsets \(V_\alpha\) of \(E\), with each restriction \(p|_{V_\alpha}:V_\alpha\to U\) a homeomorphism. Such a neighborhood \(U\) is evenly covered. All subspaces below have the subspace topology, and products have the product topology.
The defining equation ensures that the pair \((x,e)\) records a point \(e\) lying over \(f(x)\). The central result is that this construction produces a covering over \(X\), without requiring \(f\) to be injective, surjective, or a covering map.
Proof. First, \(q\) is continuous because it is the restriction to the subspace \(E_f\) of the product projection \(X\times E\to X\), which is continuous. It is surjective: for any \(x\in X\), surjectivity of \(p\) gives \(e\in E\) with \(p(e)=f(x)\), so \((x,e)\in E_f\) and \(q(x,e)=x\).
To prove the local covering condition, fix \(x_0\in X\). Choose an evenly covered open neighborhood \(U\) of \(f(x_0)\) in \(B\). Write \(p^{-1}(U)=\bigsqcup_{\alpha\in J}V_\alpha\), where each \(p|_{V_\alpha}:V_\alpha\to U\) is a homeomorphism. The set \(W=f^{-1}(U)\) is open in \(X\) by continuity of \(f\), and it contains \(x_0\). For each \(\alpha\in J\), define \(O_\alpha=E_f\cap(W\times V_\alpha)\). These sets are open in \(E_f\), are pairwise disjoint, and their union is \(q^{-1}(W)\).
For each \(\alpha\), the restriction \(q|_{O_\alpha}:O_\alpha\to W\) is a homeomorphism. Indeed, for \(x\in W\), the unique point in \(V_\alpha\) lying above \(f(x)\) is \((p|_{V_\alpha})^{-1}(f(x))\). Thus the inverse of this restriction is \(x\mapsto (x,(p|_{V_\alpha})^{-1}(f(x)))\), a continuous map because \(f|_W\), \((p|_{V_\alpha})^{-1}\), and the product pairing are continuous. The restriction and this map are inverse by their formulas. Hence \(W\) is evenly covered by \(q\). This holds for every \(x_0\), proving that \(q\) is a covering map. \(\square\)
Computing Pullbacks
Worked Example: Pulling Back the Exponential Cover to an Interval
Let \(p:\mathbb{R}\to S^1\) be \(p(t)=e^{it}\), the standard covering of the circle, and let \(X=I=[0,1]\) with its usual topology. Define \(f:I\to S^1\) by \(f(s)=e^{2\pi i s}\). The pullback is \(E_f=\{(s,t)\in I\times\mathbb{R}:e^{it}=e^{2\pi i s}\}\).
For real angles, \(e^{it}=e^{2\pi i s}\) holds exactly when \(t-2\pi s=2\pi k\) for some \(k\in\mathbb{Z}\). Therefore $$ E_f=\bigcup_{k\in\mathbb{Z}}\{(s,2\pi s+2\pi k):s\in I\}. $$ For each integer \(k\), the set \(G_k=\{(s,2\pi s+2\pi k):s\in I\}\) is the graph of a continuous function on \(I\); it is homeomorphic to \(I\) by projection onto the first coordinate. The pullback projection \(q(s,t)=s\) restricts to a homeomorphism \(G_k\to I\), with inverse \(s\mapsto(s,2\pi s+2\pi k)\). The graphs are disjoint: if \(2\pi s+2\pi k=2\pi s+2\pi \ell\), then \(k=\ell\). Thus the pullback consists of countably many copies of \(I\), each mapping homeomorphically onto \(I\).
Check: At \(s=0\), the points of the fiber have \(t=2\pi k\), and \(e^{i(2\pi k)}=1=f(0)\). At \(s=1\), they have \(t=2\pi+2\pi k=2\pi(k+1)\), and \(e^{i(2\pi(k+1))}=1=f(1)\). The description satisfies the pullback equation at both endpoints as well as for every \(s\in I\).
Worked Example: Pulling Back the Squaring Covering
Let \(p:S^1\to S^1\) be \(p(w)=w^2\), and use \(f:S^1\to S^1\), \(f(z)=z^2\). Here \(S^1\) has the subspace topology from \(\mathbb{C}\). The standard squaring map is a covering; locally, an open arc of angles of length less than \(2\pi\) has two disjoint inverse arcs, each mapped homeomorphically onto it. The pullback consists of pairs \((z,w)\) satisfying \(w^2=z^2\).
Since \(z\in S^1\), it is nonzero, and the equation becomes \((w/z)^2=1\). Factoring gives \((w/z-1)(w/z+1)=0\), so \(w=z\) or \(w=-z\). Consequently the pullback is exactly the union of the two sets \(G_+=\{(z,z):z\in S^1\}\) and \(G_-=\{(z,-z):z\in S^1\}\). They are disjoint: if \((z,z)=(z,-z)\), then \(z=-z\), which would imply \(z=0\), contrary to \(z\in S^1\). Each graph maps homeomorphically onto \(S^1\) under \(q(z,w)=z\); the inverse maps are \(z\mapsto(z,z)\) and \(z\mapsto(z,-z)\).
Check: For every \(z\in S^1\), both \((z,z)\) and \((z,-z)\) satisfy the equation, since \(z^2=z^2\) and \((-z)^2=z^2\). They are distinct because \(z\ne -z\) for nonzero \(z\). The fiber over each \(z\) therefore has exactly two points, one in each graph.
Restricting a Covering to a Subspace
Proof. The restricted map is continuous as a restriction of \(p\), and it is surjective because \(p\) is surjective: for each \(a\in A\), some \(e\in E\) satisfies \(p(e)=a\), so \(e\in p^{-1}(A)\). Fix \(a\in A\) and choose an evenly covered open neighborhood \(U\subseteq B\) of \(a\), with \(p^{-1}(U)=\bigsqcup_{\alpha\in J}V_\alpha\) and each \(p|_{V_\alpha}\) a homeomorphism onto \(U\). The set \(U\cap A\) is open in \(A\), contains \(a\), and $$ p^{-1}(U\cap A)=\bigsqcup_{\alpha\in J}\bigl(V_\alpha\cap p^{-1}(A)\bigr). $$ The pieces on the right are disjoint and open in \(p^{-1}(A)\). The restriction of \(p\) maps each piece homeomorphically onto \(U\cap A\): it is the restriction of the homeomorphism \(p|_{V_\alpha}\) to the preimage of \(U\cap A\). Thus every point of \(A\) has an evenly covered neighborhood in the subspace topology, and the restriction is a covering map. \(\square\)
Worked Example: Restricting the Squaring Cover to an Arc
Again let \(p:S^1\to S^1\) be \(p(w)=w^2\), and take the subspace \(A=\{e^{it}:-\pi/3<t<\pi/3\}\) of the target circle. The two inverse arcs are \(V_+=\{e^{it/2}:-\pi/3<t<\pi/3\}\) and \(V_-=\{-e^{it/2}:-\pi/3<t<\pi/3\}\). Squaring either expression gives \(e^{it}\), so each arc maps onto \(A\).
Each restriction is a homeomorphism: the parameter \(t\) uniquely specifies the point in its arc, and the inverse on \(A\) sends \(e^{it}\) to \(e^{it/2}\) for \(V_+\), or to \(-e^{it/2}\) for \(V_-\). These inverse formulas are continuous on the arc \(A\). The arcs are disjoint. In fact, equality \(e^{it/2}=-e^{is/2}\) would require \((t-s)/2\) to be an odd multiple of \(\pi\), whereas \(|(t-s)/2|<\pi/3\). Since the squaring map has exactly two roots over each point of \(A\), the two arcs exhaust \(p^{-1}(A)\).
Check: At the point \(1=e^{i0}\in A\), the two preimages are \(e^{i0/2}=1\) and \(-e^{i0/2}=-1\); both square to \(1\). Thus the restricted map has two points over this point, one in each disjoint sheet, as the formulas predict.
The full-preimage condition matters. A restriction to an arbitrary subset of the covering space need not remain a covering, even if its image is the whole base. One quick test is available for finite fibers: if a covering has a finite fiber over one point, the number of points in the fiber is constant in some neighborhood of that point. Indeed, an evenly covered neighborhood is split into sheets, and every point in that neighborhood has exactly one preimage in each sheet. The number of sheets is therefore the same at every point there.
Worked Example: Deleting One Point Can Break a Covering
Let \(p:S^1\to S^1\) be the squaring map, and restrict its domain to \(E'=S^1\setminus\{1\}\), with target still \(S^1\). This restricted map is surjective. The point \(1\) in the target has just one preimage in \(E'\), namely \(-1\), because its two preimages under \(p\) are \(1\) and \(-1\), and \(1\) was deleted. Every target point \(z\ne1\) has two preimages, both in \(E'\): neither preimage can equal \(1\), since \(p(1)=1\).
Thus the fiber has one point over \(1\) and two points over every other point. Every open neighborhood of \(1\) in \(S^1\) contains a point different from \(1\), so no neighborhood of \(1\) has constant finite fiber size. By the local constancy test above, the restricted map cannot be a covering map. This failure occurs even though the original map is a covering and the restriction remains onto.
Check: The deleted point is indeed a preimage only of \(1\), because \(p(1)=1^2=1\). For any \(z\ne1\), both roots remain: if either root were \(1\), then squaring it would give \(z=1\), a contradiction.
A Local Test for More General Restrictions
There is also a useful way to recognize when a restriction to a selected part of the covering space works. Let \(A\subseteq B\), and suppose \(E'\subseteq p^{-1}(A)\) is given the subspace topology. To prove that \(p|_{E'}:E'\to A\) is a covering, one must check that it is onto and that near each \(a\in A\) the selected points split into open pieces, each mapping homeomorphically onto a neighborhood of \(a\) in \(A\). Equivalently, for each \(a\) there must be an open neighborhood \(U\) of \(a\) in \(B\) such that \(p^{-1}(U\cap A)\cap E'\) is a disjoint union of open subsets of \(E'\), each carried homeomorphically onto \(U\cap A\).
For \(E'=p^{-1}(A)\), the restriction theorem supplies exactly these pieces by intersecting the original sheets with the full preimage of \(A\). For a smaller \(E'\), some pieces may be truncated or removed, and then a piece may no longer map onto a neighborhood. The example of deleting \(1\) from the domain of the squaring map shows that this is a genuine obstruction, not merely a limitation of the proof.
Common pitfall. It is incorrect to assume that every restriction of a covering map is a covering just because the original map is one. For the squaring map, deleting the domain point \(1\) changes the fiber size from two to one over the target point \(1\), while nearby target points still have two preimages. The correction is to use the full preimage \(p^{-1}(A)\) when restricting to a subspace \(A\) of the base, or to verify the local sheet condition directly for any selected subset of the domain.
Check Your Understanding
For each question, specify the topology being used and justify the local covering condition when it is relevant.
- For a covering \(p:E\to B\) and a continuous \(f:X\to B\), write the pullback space and its projection. Which hypotheses on \(f\) are needed for the pullback theorem?
- In the pullback of \(p(t)=e^{it}\) along \(f(s)=e^{2\pi i s}\) on \(I\), describe the point over \(s=1/2\) in the sheet indexed by \(k\in\mathbb{Z}\).
- Prove that restricting a covering to the full preimage of a subspace \(A\subseteq B\) gives a covering over \(A\).
- For the squaring map \(S^1\to S^1\), determine the number of preimages of \(z=1\) after deleting the domain point \(-1\), and compare it with the number over \(z\ne1\). Does this restriction define a covering?
- State the local condition that a selected subset \(E'\subseteq p^{-1}(A)\) must satisfy for its restriction to be a covering over \(A\).