TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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The fundamental group of a finite wedge of circles is free of that rank

Statement

Let $Q = R / Z$ be pointed at $[0]$ , and for $r \in N$ put

$W_{r} : = ⋁_{j < r} (Q, [0]) .$

Then $π_{1} (W_{r}, w)$ is the free group on the $r$ standard loops, one traversing each circle summand once. In particular it has rank $r$ (The rank of a free group admitting a finite basis). For $r = 0$ , $W_{0}$ is a point and the basis is empty.

Facts & Assumptions

Given: The finite quotient-circle wedges $W_{r}$ and their standard based loops.

[L1]

The successor wedge $W_{r + 1} = W_{r} \lor Q$ has a two-set van Kampen cover whose members deformation retract to $W_{r}$ and $Q$ and whose overlap is simply connected (Finite wedges of quotient circles have van Kampen covers at the wedge point).

[L2]

A two-set van Kampen cover with simply connected overlap has fundamental group the free product of the two factor fundamental groups (A simply connected overlap turns the van Kampen pushout into a free product).

[F1]

The degree map is an isomorphism $π_{1} (Q, [0]) ≅ (Z, +)$ and sends the standard once-around loop to $1$ ( $Deg : π_{1} (R / Z, [0]) \to (Z, +)$ is an isomorphism).

[F2]

The free product of free groups on disjoint bases is the free group on the disjoint union of those bases (Free groups on disjoint bases freely multiply to the free group on their union).

[F3]

If a property holds at $0$ and passes from every natural $r$ to $r + 1$ , then it holds for every natural number (The principle of mathematical induction).

Proof

technique · induction

1.1

The empty wedge $W_{0}$ is a point by definition. Every based loop in a point is constant, so $π_{1} (W_{0}, w)$ is the one-element group, which is the free group on the empty basis and has rank $0$ .

base

1.2

The group of one circle is infinite cyclic by [F1], so its standard loop is a one-element free basis. This is the first successor case and fixes the basis convention used below.

1.3

Assume $π_{1} (W_{r}, w)$ is free on the $r$ standard circle loops. By [L1] and [L2], the successor wedge satisfies $π_{1} (W_{r + 1}, w) ≅ π_{1} (W_{r}, w) * π_{1} (Q, [0]) .$

L1L2ih

2.1

The induction hypothesis and [F1] identify the two factors as free groups on disjoint bases consisting of the old $r$ standard loops and the new standard loop. By [F2], their free product is free on the union, exactly the $r + 1$ standard loops of $W_{r + 1}$ .

step 1.3F1F2

3.1

Step 1.1 is the base case and steps 1.3 and 2.1 prove the successor implication, so [F3] gives the result for every $r \in N$ . The basis has $r$ elements, hence the rank is $r$ by definition.

step 1.1step 1.3step 2.1F3discharge-induction∎

Depends on

Used by

π₁(S¹∨ S¹) is the free group on two generators Corollary

Dependency tree · two levels

34 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen Hatcher, Algebraic Topology, Example 1.21 (standard reference, not scraped)
J. Peter May, A Concise Course in Algebraic Topology, Chapter 2, Section 8 (standard reference, not scraped)