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Covering Spaces and Lifting — Examples
1 · Prerequisites
- Binary Operations, Monoids, Groups and Subgroups
- Compactness
- Compactness in Metric Spaces
- Completeness, Completion, and Uniform Continuity
- Connectedness
- Construction of the Natural Numbers
- Construction of the Real Numbers via Cauchy Sequences
- Construction of the Real Numbers via Dedekind Cuts
- Continuity, IVT, EVT, and Uniform Continuity
- Countability and Uncountability
- Covering Spaces and Lifting
- Filters and Ultrafilters
- Foundations of the Real Numbers for Analysis
- Group Actions, Orbits, Stabilisers and Cayley's Theorem
- Homotopy and Homotopy Equivalence
- Limits of Real Functions
- Linear Independence, Bases and Dimension
- Linear Transformations, Rank-Nullity and Quotient Spaces
- Metric Spaces
- Monotone Sequences, Bolzano-Weierstrass, and Cauchy Completeness
- Order, Zorn's Lemma, and the Axiom of Choice
- Relations, Functions, and Quotients
- Rings, Subrings, Integral Domains and Fields
- Rⁿ as a Normed Space; Vector-Valued Functions
- Roots, Rational Powers, and Classical Inequalities
- Sequences and Limits
- Subspaces, Products, and Quotients
- Suprema and Infima
- The Fundamental Group
- The ZFC Axioms and the Basic Set Constructions
- Topological Spaces and Continuity
- Topology of ℝ
- Vector Spaces, Linear Subspaces, Span and Direct Sums
2 · Summary
3 · Logical flowchart
4 · Definitions, theorems and proofs
None yet.
5 · Examples, counterexamples and false statements
Trivial coverings are products with a discrete fibre
Example
If is any space and is a nonempty discrete space, the projection is a trivial covering with fibre . If , the same holds for ; for nonempty , an empty fibre would violate surjectivity.
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
A covering map is a continuous surjection such that every has an open neighbourhood for which is a disjoint union of open sets , called sheets, and each restriction is a homeomorphism (def-continuous-map-top, def-homeomorphism-and-open-maps, def-disjoint-union-topology). Such a is evenly covered, and is the fibre over . A covering is trivial when it is isomorphic over to a product projection with discrete. (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).
The product set. Let be a set and let be a set for each . The product is and we write , the -th coordinate of . Two elements of the product are equal exactly when they agree at every index, functions being equal when they have the same domain and the same values. For the -th projection is . The product topology on is the initial topology of the projections: the topology generated by the subbasis . Finite intersections of subbasic sets form a basis for it, and they are exactly the boxes with every open in and for all but finitely many . (The product set of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).
Throughout, a topology is as in def-topological-space, and finite, at most countable and uncountable are as in def-countable, so that "countable" always means "at most countable" and every finite set is countable. Let be a set. The six families below are topologies on ; that each really satisfies (T1), (T2) and (T3) is discharged in full after the list. Among those six is the discrete topology , in which every subset is open and hence every subset is also closed. (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).
Verification
For any space and discrete set , verify that the projection is a covering, with every open subset of the base evenly covered.
Identify its sheets and fibre, including : the projection then fails surjectivity unless , so state the nonempty-fibre convention explicitly.
The preceding construction and implications establish the assertion.
A surjective local homeomorphism need not be a covering map
Statement refuted
Let and map both summands by inclusion to . This map is a surjective local homeomorphism but is not a covering map.
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
Every covering map is a surjective local homeomorphism, and each of its fibres is discrete in the subspace topology. (Covering maps are surjective local homeomorphisms with discrete fibres).
For a covering , the cardinality of is locally constant as a function of . If is connected, all fibres are equinumerous. (The cardinality of a covering fibre is locally constant and is constant on a connected base).
The underlying set. Let be a set and let be a set for each . The disjoint union is whose elements are the pairs with and . For the -th canonical injection is The construction is what makes the word "disjoint" honest. Each is injective (def-injection-surjection-bijection), since forces ; the images are pairwise disjoint, since the second coordinate determines ; and their union is the whole set. So no assumption that the are disjoint as sets is needed, and none is made: the tag separates the copies even when for . (The disjoint union (coproduct) with the final topology of the canonical injections: a set is open exactly when each of its traces is).
Throughout, is the complete ordered field (def-complete-ordered-field, def-ordered-field) with its order (def-real-order). A subset is order-convex when and imply , and the intervals of are the nine listed forms, among them and . (Intervals of : the nine order-convex forms, nondegeneracy, and length).
Counterexample
Let the domain be the disjoint union of and and map both components by inclusion onto the base .
The first component makes the map surjective and each inclusion is a local homeomorphism.
The fibre has one point at and two immediately to its right, so local constancy of sheet number rules out a covering; equivalently, the second component supplies only a one-sided partial sheet above every neighbourhood of .
The preceding construction and implications establish the assertion.
A covering over a disconnected base can have different sheet numbers on different components
Statement refuted
There is a covering of a two-point discrete space whose fibre over one point has one element and whose fibre over the other has two elements. Thus connectedness is necessary for global constancy of sheet number.
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
For a covering , the cardinality of is locally constant as a function of . If is connected, all fibres are equinumerous. (The cardinality of a covering fibre is locally constant and is constant on a connected base).
Throughout, a topology is as in def-topological-space, and finite, at most countable and uncountable are as in def-countable, so that "countable" always means "at most countable" and every finite set is countable. Let be a set. The six families below are topologies on ; that each really satisfies (T1), (T2) and (T3) is discharged in full after the list. Among those six is the discrete topology , in which every subset is open and hence every subset is also closed. (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).
Counterexample
Map a three-point discrete space onto a two-point discrete space with one point over the first basepoint and two over the second.
Each singleton base neighbourhood is evenly covered, while the fibre cardinal is not globally constant.
This isolates exactly why connectedness appears in the sheet-number theorem.
The preceding construction and implications establish the assertion.
The quotient is a covering with integer translations as deck transformations
Example
For the quotient by integer translation, is a covering map, and every deck transformation is a unique translation with .
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
For a covering-space action of on , the orbit map is a covering. If is path-connected, the deck group of this covering consists exactly of the transformations supplied by . (The orbit map of a covering-space action is a covering, with the acting group equal to the deck group when the total space is path-connected).
For a covering , a deck transformation is an isomorphism over , so (def-map-and-isomorphism-of-covering-spaces). Deck transformations form the deck group under composition, and this group acts on by evaluation (def-group, def-group-action). (Deck transformations and the deck-transformation group of a covering).
The quotient topology. Let be a topological space (def-topological-space), let be a set and let be a surjection (def-injection-surjection-bijection). The quotient topology on induced by is the final topology of the one-element family (def-initial-and-final-topology): That this is a topology is discharged in def-initial-and-final-topology, where every final topology is verified to satisfy (T1), (T2) and (T3). Dually, is closed in exactly when is closed in , because . (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).
On the set of pairs of natural numbers, define This is an equivalence relation (lem-int-equivalence). The integers are the quotient and we write for the equivalence class of . (The integers as equivalence classes of pairs of naturals).
Identify with its canonical copy inside along the embeddings ; then for every real there is exactly one integer with , written (Integer part: for every real there is exactly one integer with ).
Verification
By [F5] the integers sit inside as a subgroup under the ordered-field operations, so iff is an equivalence relation on ; [F4] supplies only the abstract construction of and not its copy in , so the embedding of [F5] is what makes the relation and the translations below meaningful. Take the quotient topology of [F3] on .
For an open interval of length below the translates , , are pairwise disjoint, since two points of differ by less than while distinct integer translates differ by at least in the order of [F5]; the preimage is then a union of open sets, so is open and evenly covered. Openness of is required and not cosmetic: for the preimage is not open, so is not even a neighbourhood.
Verify directly that translations are deck transformations and that every deck transformation is the unique integer translation determined by the image of zero, without computing the fundamental group of the quotient.
The preceding construction and implications establish the assertion.
The projected unit interval is not nullhomotopic in
Example
The loop in is not nullhomotopic.
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
For the quotient by integer translation, is a covering map, and every deck transformation is a unique translation with . (The quotient is a covering with integer translations as deck transformations).
Let be a covering, a homotopy, and a lift of . There is a unique lift of extending . (Existence and uniqueness of homotopy lifts through a covering map).
Endpoint-fixed homotopic paths in the base have lifts with the same endpoint whenever their lifts begin at the same point. (The endpoint of a lifted path depends only on its endpoint-fixed homotopy class).
Verification
The loop lifts from zero to the path , whose endpoint is one.
If it had an endpoint-fixed nullhomotopy, homotopy lifting would keep the terminal lift in the discrete fibre while deforming the initial lift to the constant path at zero, forcing endpoints one and zero to agree.
This proves essentiality without asserting .
The preceding construction and implications establish the assertion.
The maps on are -sheeted coverings for
Example
For every integer , the map given by is a well-defined -sheeted covering.
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
For the quotient by integer translation, is a covering map, and every deck transformation is a unique translation with . (The quotient is a covering with integer translations as deck transformations).
A covering map is a continuous surjection such that every has an open neighbourhood for which is a disjoint union of open sets , called sheets, and each restriction is a homeomorphism (def-continuous-map-top, def-homeomorphism-and-open-maps, def-disjoint-union-topology). Such a is evenly covered, and is the fibre over . A covering is trivial when it is isomorphic over to a product projection with discrete. (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).
For a covering , the cardinality of is locally constant as a function of . If is connected, all fibres are equinumerous. (The cardinality of a covering fibre is locally constant and is constant on a connected base).
On the set of pairs of natural numbers, define This is an equivalence relation (lem-int-equivalence). The integers are the quotient and we write for the equivalence class of . (The integers as equivalence classes of pairs of naturals).
Let with . Then there exist integers and with and , and this pair is unique (Division with remainder in : for and there are unique with and ).
Verification
Check well-definedness modulo integer translation.
Around a class take an interval short enough that its inverse branches are disjoint; these branches give the evenly covered neighbourhood of [F2]. The fibre has exactly points because the branches are indexed by the residues of the integers modulo , and by the division algorithm [F5] every integer has exactly one residue with : existence gives distinct branch labels and uniqueness stops two labels from coinciding. [F4] constructs but supplies no division algorithm.
At the map is the identity, so no zero-sheet or division-by-zero case is hidden.
The preceding construction and implications establish the assertion.
Pulling a covering back to an evenly covered open set gives a trivial covering
Example
If is evenly covered by , then the pullback of along the inclusion is a trivial covering of .
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
For a covering and a continuous map , define with the subspace topology, and let be (def-product-topology, def-subspace-topology-top). This is the pullback covering space; its covering property is proved in prop-covering-spaces-are-stable-under-restriction-finite-products-and-pullback. (The pullback of a covering space along a continuous map).
A covering map is a continuous surjection such that every has an open neighbourhood for which is a disjoint union of open sets , called sheets, and each restriction is a homeomorphism (def-continuous-map-top, def-homeomorphism-and-open-maps, def-disjoint-union-topology). Such a is evenly covered, and is the fibre over . A covering is trivial when it is isomorphic over to a product projection with discrete. (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).
If is any space and is a nonempty discrete space, the projection is a trivial covering with fibre . If , the same holds for ; for nonempty , an empty fibre would violate surjectivity. (Trivial coverings are products with a discrete fibre).
Verification
For the inclusion of an evenly covered open set of the base, identify the pullback with the disjoint union of the sheets over .
Write the explicit mutually inverse maps over and verify their continuity from the pullback subspace topology.
The preceding construction and implications establish the assertion.
The Hawaiian earring is locally path-connected but has no universal cover
Example
The Hawaiian earring is locally path-connected but is not semilocally simply connected at its wedge point. Consequently it has no universal cover.
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
The loop in is not nullhomotopic. (The projected unit interval is not nullhomotopic in ).
If a space admits a universal covering, then it is semilocally simply connected. No local path-connectedness hypothesis is required. (A space admitting a universal covering is semilocally simply connected).
A space is semilocally simply connected at when there is a neighbourhood of and a basepoint-preserving inclusion whose induced map on fundamental groups is trivial (def-neighbourhood-top, def-induced-homomorphism-on-fundamental-groups, def-based-loops-and-fundamental-group). It is semilocally simply connected when this holds at every point. The neighbourhood need not itself be simply connected. (Semilocally simply connected spaces with explicit basepoint convention).
The quotient topology. Let be a topological space (def-topological-space), let be a set and let be a surjection (def-injection-surjection-bijection). The quotient topology on induced by is the final topology of the one-element family (def-initial-and-final-topology): That this is a topology is discharged in def-initial-and-final-topology, where every final topology is verified to satisfy (T1), (T2) and (T3). Dually, is closed in exactly when is closed in , because . (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).
Verification
Model the earring as countably many copies of with diameters tending to zero and all zero classes identified, using the standard shrinking-wedge metric.
Away from the wedge point, sufficiently short open arcs are path-connected neighbourhoods.
At the wedge point, every open neighbourhood contains a smaller metric ball whose intersection with each circle is an arc through the wedge point and which contains every sufficiently small circle, so that ball is path-connected.
Retraction to one such small circle and the essential unit loop show its inclusion carries a nontrivial loop, so semilocal simple connectedness fails at the wedge point.
The necessity theorem then rules out a universal cover.
The preceding construction and implications establish the assertion.
Sources
Standard references
Recommended treatments; not extraction sources.