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For a nonempty path-connected total space, a covering fibre is in bijection with the right cosets of the induced fundamental-group subgroup
Statement
For a covering with nonempty path-connected total space, put . With traversal-order multiplication, the fibre is in bijection with the set of right cosets . Thus its finite number of sheets equals the subgroup index, and one is infinite exactly when the other is recorded as .
Facts & Assumptions
Given: The objects, hypotheses, and choice principles stated above.
Fix a covering , a basepoint , and . For , define as the endpoint of the unique lift of beginning at (thm-path-lifting-for-covering-maps). Endpoint homotopy invariance makes this well defined (cor-lifted-path-endpoints-depend-only-on-path-homotopy). With the library's traversal-order product this is a right action; the corresponding left action is (def-group-action). (The monodromy right action on a covering fibre and its equivalent left-action convention).
For a covering , the induced homomorphism is injective. (A covering map induces an injective homomorphism on fundamental groups).
Monodromy acts on each covering fibre by bijections. Its orbit through is exactly the intersection of the path component of with that fibre. (Monodromy acts by fibre bijections, and its orbits are the intersections of path components with the fibre).
Let be a group and let be a subgroup (def-group, def-subgroup). For , the left coset and right coset of represented by are The element is a representative of these cosets. The notation denotes subsets of ; it does not assert that either subset is a subgroup. (Left and right cosets and of a subgroup).
Let . The left coset set is By lem-coset-partition, its elements are exactly the blocks of the coset partition of . The index of in is when is finite, with finite cardinality as in def-finite-cardinality. If is not finite, write . Here is a symbol, not a natural number, and no arithmetic with it is defined. The right coset set has the same finite or infinite size, because lem-left-and-right-cosets-equinumerous gives an explicit bijection between the two coset sets; thus the index does not depend on choosing left rather than right cosets. (The coset set and the index of a subgroup).
Proof
Fix a point in the fibre.
Send a loop class to the endpoint of its lifted path.
Endpoint homotopy invariance makes this well defined; with traversal-order multiplication, two classes have the same endpoint exactly when their quotient lies in the image of the upstairs fundamental group, and holds exactly when . By the convention of [F4] the sets are right cosets, so the fibres of the endpoint map are precisely the members of .
Path-connectedness of the total space gives surjectivity onto the fibre.
Translate the bijection into the published index convention. That convention defines from the left coset set , so use the clause of [F5] giving an explicit bijection between the left and right coset sets: and have the same finite or infinite size. Composing with step 4.1, the finite cardinalities agree with the number of sheets, and one side is infinite exactly when the other is recorded as .
The preceding construction and implications establish the assertion.
Depends on
- The monodromy right action on a covering fibre and its equivalent left-action convention
- A covering map induces an injective homomorphism on fundamental groups
- Monodromy acts by fibre bijections, and its orbits are the intersections of path components with the fibre
- Left and right cosets $gH$ and $Hg$ of a subgroup
- The coset set $G/H$ and the index $[G:H]$ of a subgroup
- Inversion induces a bijection $gH\mapsto Hg^{-1}$ from left cosets to right cosets
Used by
Nothing in the library uses this result yet.
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Sources
- Allen Hatcher, Algebraic Topology, §1.3 (standard reference, not scraped)
- J. Peter May, A Concise Course in Algebraic Topology, Ch. 3 (standard reference, not scraped)
- Marco Gualtieri, MAT1300 Week 4 Term 2, §1.6 (standard reference, not scraped)