Regularity and the Deck Group
A covering can have several points above each point of its base. The deck transformations are the symmetries of the covering that leave the base unchanged. A covering is called regular when these symmetries can move any point of a fiber to any other point in that fiber. The criterion proved here translates this geometric condition into an algebraic one: normality of a subgroup of the fundamental group.
Let \(p:E\to B\) be a covering map, let \(b_0\in B\), and choose \(e_0\in p^{-1}(b_0)\). Assume throughout the main theorem that \(B\) is path-connected and locally path-connected and that \(E\) is path-connected. A covering over a locally path-connected base is locally path-connected, so the lifting criterion for maps applies to \(E\). Write \(G=\pi_1(B,b_0)\), with loop products concatenated in the order traversed, and define the associated subgroup \[ H=p_*\bigl(\pi_1(E,e_0)\bigr)\leq G. \]
We use two results established in the earlier material on covering-space lifting. First, path lifting gives a unique lift of a path after its starting point is prescribed. Second, the lifting criterion says that if \(X\) is path-connected and locally path-connected, \(f:X\to B\) is continuous, \(x_0\in X\), and \(y_0\in p^{-1}(f(x_0))\), then a lift \(\widetilde f:X\to E\) with \(\widetilde f(x_0)=y_0\) exists if and only if \(f_*(\pi_1(X,x_0))\subseteq p_*(\pi_1(E,y_0))\). We also use uniqueness of lifts on a connected domain when their values agree at one point.
Comparing the Subgroups at Two Points of a Fiber
The subgroup based at \(e_0\) is not literally the subgroup based at every other point of the same fiber: the basepoint in \(E\) has changed. The precise relation is conjugation by a loop in the base.
Proof. If \(\beta\) is a loop in \(E\) based at \(e_1\), then \(\alpha*\beta*\alpha^{-1}\) is a loop based at \(e_0\). Its projection represents \(g[p\circ\beta]g^{-1}\), which belongs to \(H\). Thus \(gH_1g^{-1}\subseteq H\). Conversely, if \(\delta\) is a loop in \(E\) based at \(e_0\), then \(\alpha^{-1}*\delta*\alpha\) is based at \(e_1\). Its projection represents \(g^{-1}[p\circ\delta]g\), so \(g^{-1}Hg\subseteq H_1\). The first inclusion gives \(H_1\subseteq g^{-1}Hg\), and the second gives the reverse inclusion. Hence \(H_1=g^{-1}Hg\). \(\square\)
The Normal-Subgroup Criterion
- \(H=p_*(\pi_1(E,e_0))\) is normal in \(\pi_1(B,b_0)\);
- \(\operatorname{Deck}(p)\) acts transitively on \(p^{-1}(b_0)\);
- \(p\) is a regular covering, meaning the deck group acts transitively on every fiber.
Proof. First suppose \(H\) is normal. Choose any \(e_1\in p^{-1}(b_0)\), and choose a path \(\alpha\) from \(e_0\) to \(e_1\); such a path exists because \(E\) is path-connected. With \(g=[p\circ\alpha]\), the proposition gives \(H_1=g^{-1}Hg=H\). Apply the lifting criterion to the map \(f=p:E\to B\), with domain basepoint \(e_0\) and prescribed lift value \(e_1\). Its required subgroup inclusion is \(p_*(\pi_1(E,e_0))\subseteq p_*(\pi_1(E,e_1))\), which here is \(H\subseteq H\). Therefore there is a lift \(T:E\to E\) of \(p\) such that \(T(e_0)=e_1\); in other words, \(p\circ T=p\).
This lift is a deck transformation, not merely a continuous self-map. Apply the same criterion with domain basepoint \(e_1\) and prescribed value \(e_0\). Since \(H_1=H\), it gives a lift \(S:E\to E\) with \(S(e_1)=e_0\). The maps \(S\circ T\) and \(\operatorname{id}_E\) are lifts of \(p\) and agree at \(e_0\), so uniqueness of lifts gives \(S\circ T=\operatorname{id}_E\). Likewise, \(T\circ S\) and \(\operatorname{id}_E\) are lifts of \(p\) agreeing at \(e_1\), so \(T\circ S=\operatorname{id}_E\). Thus \(T\) is a homeomorphism with inverse \(S\), and it is a deck transformation. As \(e_1\) was arbitrary, the deck group is transitive on the fiber over \(b_0\).
Now suppose the deck group is transitive on \(p^{-1}(b_0)\). Let \(g\in G\) be arbitrary, represented by a loop at \(b_0\). Lift this loop starting at \(e_0\), and call its endpoint \(e_1\). The path-lifting property ensures that \(e_1\in p^{-1}(b_0)\). By transitivity, some deck transformation \(T\) sends \(e_0\) to \(e_1\). Because \(p\circ T=p\), applying \(T\) to loops based at \(e_0\) shows \(H\subseteq H_1\), where \(H_1=p_*(\pi_1(E,e_1))\). Applying \(T^{-1}\) gives the reverse inclusion, so \(H=H_1\). The proposition now gives \(H=g^{-1}Hg\). This holds for every \(g\in G\), so \(H\) is normal.
It remains to relate transitivity on the fiber over \(b_0\) to transitivity on every fiber. Suppose the deck group is transitive over \(b_0\), choose \(b\in B\), and choose a path \(\gamma\) from \(b_0\) to \(b\). Lifting \(\gamma\) gives a bijection from the fiber over \(b_0\) to the fiber over \(b\): its inverse is given by lifting the reverse path. Given \(y_1,y_2\in p^{-1}(b)\), lift the reverse of \(\gamma\) from each to obtain \(x_1,x_2\in p^{-1}(b_0)\). Choose a deck transformation \(T\) with \(T(x_1)=x_2\). A deck transformation carries a lift of \(\gamma\) to a lift of \(\gamma\), since \(p\circ T=p\). Uniqueness of path lifts therefore implies \(T(y_1)=y_2\). Thus the action is transitive over \(b\), and \(b\) was arbitrary. The converse follows immediately from the definition of regularity. \(\square\)
Worked Examples
Worked Example: The Identity Covering
Let \(B\) be any path-connected, locally path-connected space and give \(E=B\) the same topology. Take \(p=\operatorname{id}_B\), with \(e_0=b_0\). The fiber over \(b_0\) is \(\{b_0\}\), and the associated subgroup is \(H=(\operatorname{id}_B)_*(\pi_1(B,b_0))=\pi_1(B,b_0)=G\). The whole group \(G\) is normal in itself. The only deck transformation is the identity: if \(T:B\to B\) satisfies \(p\circ T=p\), then \(\operatorname{id}_B\circ T=\operatorname{id}_B\), so \(T=\operatorname{id}_B\). Its action on the one-point fiber is transitive, since the only point must be sent to itself. This verifies both sides of the criterion directly.
Worked Example: A Connected Two-Sheeted Graph Covering
Let \(B\) be the graph consisting of two circles joined at one vertex \(v\), and call the two loop edges \(a\) and \(b\). Give the graph its usual topology. Form \(E\) from two vertices \(u_0,u_1\), two edges \(a_0,a_1\) with \(a_0\) running from \(u_0\) to \(u_1\) and \(a_1\) from \(u_1\) to \(u_0\), and one \(b\)-loop at each vertex. Give \(E\) its graph topology. Map both vertices to \(v\), each \(a_i\) to \(a\), and each \(b\)-loop to \(b\), preserving the edge parameter. This defines \(p:E\to B\).
At each vertex of \(E\), the four incident edge directions map one-to-one onto the four directions at \(v\) in \(B\). Small open neighborhoods of the vertices therefore map homeomorphically onto a neighborhood of \(v\); open subintervals inside edges have the same property. These neighborhoods verify the covering condition. The space \(E\) is path-connected because the edge \(a_0\) joins its two vertices, and each edge point can be joined to a vertex. The fiber over \(v\) is exactly \(\{u_0,u_1\}\).
Swapping \(u_0\) and \(u_1\), swapping \(a_0\) and \(a_1\) at the same edge parameter, and swapping the two \(b\)-loops at the same parameter defines a homeomorphism \(T:E\to E\). It satisfies \(p\circ T=p\) and sends \(u_0\) to \(u_1\), so the deck group is transitive on the fiber over \(v\). The fiber-stabilizer result from the earlier material on fundamental-group actions on covering fibers says that for a connected covering the index \([G:H]\) equals the size of the fiber. Here \([G:H]=2\). Every subgroup of index two is normal: if \(g\notin H\), then \(gH\) and \(Hg\) are each the complement of \(H\), because there is only one coset other than \(H\); if \(g\in H\), both cosets equal \(H\). Thus \(gH=Hg\) for all \(g\in G\), proving \(H\trianglelefteq G\), in agreement with the theorem.
Worked Example: Why Connectedness of the Covering Space Matters
Use the connected two-sheeted covering \(q:E\to B\) just described, and form the disjoint union \(X=B\bigsqcup E\) with the disjoint-union topology. Define \(P:X\to B\) to equal the identity on the \(B\) component and \(q\) on the \(E\) component. This is a covering: each base neighborhood evenly covered by either map is evenly covered in the disjoint union. The base \(B\) remains path-connected and locally path-connected, but \(X\) is not path-connected.
Choose \(x_0\) in the identity component above \(v\). Every loop based at \(x_0\) maps under \(P\) to itself, so the associated subgroup is \(H=G\), which is normal. But the fiber over \(v\) has three points: one in the identity component and two in the \(E\) component. No deck transformation can carry the first point to either of the other two. Indeed, a homeomorphism maps connected components to connected components; if a deck transformation mapped the identity component to the \(E\) component, its restriction would give a bijection between their fibers over \(v\), which have respectively one and two points. Such a bijection cannot exist. Therefore the deck group is not transitive on this fiber. This example shows why the connectedness hypothesis on the covering space cannot be dropped.
Cosets, Conventions, and a Common Pitfall
For a connected covering, lifting loops from a point in the fiber gives a right action of \(G\) on the fiber: first lift \(g\), then lift \(h\), which corresponds to the product \(gh\) under the stated traversal convention. Starting at \(e_0\), the endpoints of \(g\) and \(g'\) agree exactly when \(gg'^{-1}\in H\). Consequently the fiber is identified with the left cosets \(Hg\). This coset orientation matters when \(H\) is not normal.
Common pitfall. A tempting but incorrect identification is to use right cosets \(gH\) for this right action. For a concrete distinction, in the symmetric group \(S_3\) let \(H=\{e,(12)\}\) and \(g=(13)\), with permutation products composed right to left. Direct calculation gives \((12)(13)=(132)\) and \((13)(12)=(123)\). Therefore \(Hg=\{(13),(132)\}\), whereas \(gH=\{(13),(123)\}\); the cosets differ. The correction is to use \(Hg\) for the loop-lifting right action in this convention. If \(H\) is normal, left and right cosets agree, but one must not assume normality before proving the regularity criterion.
The equivalence is useful because it lets us test a geometric symmetry condition using the subgroup \(H\), or interpret normality as the condition that deck symmetries reach every point of a fiber. The two viewpoints depend on connectedness and on keeping basepoints and coset conventions straight.
Check Your Understanding
Use the hypotheses and conventions stated in the tutorial when answering.
- State the associated subgroup for a covering \(p:E\to B\), including the basepoints that define it.
- Why does normality of \(H\) imply \(p_*(\pi_1(E,e_0))\subseteq p_*(\pi_1(E,e_1))\) for \(e_1\) in the fiber over \(b_0\)?
- In the subgroup-comparison proposition, explain why the loop in \(B\) conjugating \(H_1\) to \(H\) is \(g=[p\circ\alpha]\).
- For the right action defined by lifting loops in traversal order, what is the correct coset description of the fiber, and what endpoint condition gives it?
- Identify the connectedness hypothesis that fails in the disjoint-union example, and explain why its associated subgroup can still be normal.