TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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For a path-connected locally path-connected semilocally simply connected base, the deck group of a universal cover is isomorphic to the fundamental group

Statement

For a path-connected, locally path-connected, semilocally simply connected base $B$ , the deck group of a universal cover is isomorphic to $π_{1} (B, b_{0})$ . With the library's traversal-order path product the monodromy action is a right action, and the assignment carrying a loop class to the deck transformation that moves the chosen point of the fibre to the corresponding lifted endpoint is itself an isomorphism; no path reversal is inserted.

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

For a covering $p : E \to B$ , a deck transformation is an isomorphism $h : E \to E$ over $B$ , so $p \circ h = p$ (def-map-and-isomorphism-of-covering-spaces). Deck transformations form the deck group $Deck (p)$ under composition, and this group acts on $E$ by evaluation (def-group, def-group-action). (Deck transformations and the deck-transformation group of a covering).

[F2]

Fix a covering $p : E \to B$ , a basepoint $b_{0} \in B$ , and $e \in p^{- 1} (b_{0})$ . For $[α] \in π_{1} (B, b_{0})$ , define $e \cdot [α]$ as the endpoint of the unique lift of $α$ beginning at $e$ (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 $[α] \cdot e : = e \cdot [α]^{- 1}$ (def-group-action). (The monodromy right action on a covering fibre and its equivalent left-action convention).

[F3]

Let $B$ be path-connected and locally path-connected. After basepoints over the same point are fixed, a universal cover of $B$ admits a unique based continuous map over $B$ to every connected covering of $B$ ; in particular any two universal covers of $B$ are uniquely isomorphic over $B$ . (For a path-connected locally path-connected base, a universal cover maps uniquely over the base to every connected covering, and any two universal covers are uniquely isomorphic).

[F4]

For a covering with connected total space, two deck transformations agreeing at one point are equal. Consequently the deck group acts freely on the total space. (On a connected covering space, a deck transformation is determined by one point and the deck action is free).

[F5]

For every pointed topological space $(X, x_{0})$ , the product $[α] [β] = [α * β]$ is well defined and makes $π_{1} (X, x_{0})$ a group. Its identity is the class of the constant loop $c_{x_{0}}$ , and $[α]^{- 1} = [\overset{ˉ}{α}]$ . (Loop classes form the group $π_{1} (X, x_{0})$ under concatenation).

Proof

technique · direct

1.1

For a path-connected locally path-connected semilocally simply connected base and a chosen point upstairs, each loop class determines the endpoint of its lift.

givenF3F2F4F5

2.1

The universal-cover lifting criterion gives the unique deck transformation taking the chosen point to that endpoint.

step 1.1F1F3F4

3.1

Because the library multiplies loops in traversal order, [F2] makes the monodromy a right action, and no path reversal is needed. A deck transformation $h$ satisfies $h (x \cdot g) = h (x) \cdot g$ : since $p \circ h = p$ , composing $h$ with the lift of a loop starting at $x$ gives a lift of that loop starting at $h (x)$ , and lifts from a given point are unique, so the endpoints correspond. Writing $h_{a}$ for the deck transformation of step 2.1 with $h_{a} (e) = e \cdot a$ , this gives $(h_{a} \circ h_{b}) (e) = h_{a} (e \cdot b) = h_{a} (e) \cdot b = (e \cdot a) \cdot b = e \cdot (a b) = h_{a b} (e)$ , so $h_{a} \circ h_{b}$ and $h_{a b}$ agree at $e$ and [F4] makes them equal. Hence $a \mapsto h_{a}$ is a homomorphism as it stands; assigning inverse classes instead would reverse products and give an antihomomorphism.

step 2.1F2F3F1F4

4.1

For injectivity, suppose two loop classes give the same lifted endpoint. Their quotient then fixes the chosen point of the fibre, so it lies in the stabiliser of that point under the monodromy action [F2]. That stabiliser is the image of the upstairs fundamental group, which is trivial because the total space of a universal cover is simply connected; so the two classes are equal. Freeness of the deck action then makes the induced assignment injective as a map of groups, and path-connectedness of the total space gives surjectivity.

step 3.1F2F3

5.1

The preceding construction and implications establish the assertion.

step 4.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 71 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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)