Use covering spaces to show that any free group is a subgroup of \(F_2\), the free group on \(2\) generators.
Solution
1. Let\[ R=S^1_a\vee S^1_b \]be the rose with two oriented loops \(a,b\) and one vertex \(v\).Then\[ \pi_1(R,v)\cong F_2=\langle a,b\rangle. \]
The rose is a connected graph with one vertex and two edges. Collapsing a maximal tree, which here is just the vertex, gives the standard computation \[ \pi_1(S^1\vee S^1,v)\cong\mathbb Z*\mathbb Z=F_2. \]
2. For every integer \(n\ge2\), construct a connected \((n-1)\)-sheeted covering\[ p_n:Y_n\to R \]as follows.
Put \[ d=n-1. \] Let \(Y_n\) have vertices \[ v_0,\dots,v_{d-1}. \] For each \(i\) modulo \(d\), include
one oriented \(a\)-edge from \(v_i\) to \(v_{i+1}\), and
one \(b\)-loop at \(v_i\).
Map every \(a\)-edge homeomorphically to the \(a\)-loop of \(R\), every \(b\)-loop homeomorphically to the \(b\)-loop, and every vertex to \(v\). At each vertex of \(Y_n\) there is exactly one incoming and one outgoing lift of each oriented edge germ at \(v\). Hence a sufficiently small star neighborhood of each vertex maps homeomorphically to the corresponding star neighborhood of \(v\). Thus \(p_n\) is a covering map.
The \(a\)-edges form a \(d\)-cycle through all the vertices, so \(Y_n\) is connected. Each point of the interior of either loop of \(R\) has exactly \(d\) preimages, so the covering has \(d=n-1\) sheets.
3. The group \(\pi_1(Y_n)\) is free of rank \(n\).
The graph \(Y_n\) has \[ V=d \] vertices and \[ E=2d \] edges. For a connected finite graph, a maximal tree has \(V-1\) edges, and collapsing it leaves a wedge of \[ E-(V-1)=E-V+1 \] circles. Therefore \[ \operatorname{rank}\pi_1(Y_n) = 2d-d+1 = d+1 = n. \] Hence \[ \pi_1(Y_n)\cong F_n. \]
4. Consequently \(F_n\) embeds in \(F_2\) for every finite \(n\ge2\).
A covering map induces an injective homomorphism on fundamental groups. Indeed, if a loop in \(Y_n\) projects to a null-homotopic loop in \(R\), lift a null-homotopy of its projection starting with the given loop; uniqueness of lifts forces the original loop to be null-homotopic. Thus \[ (p_n)_*:\pi_1(Y_n,v_0)\hookrightarrow\pi_1(R,v) \] is injective. Using 1 and 3 gives \[ F_n\cong\pi_1(Y_n,v_0) \hookrightarrow \pi_1(R,v) \cong F_2. \]
5. The remaining finite ranks also occur: \(F_1\) and the trivial group embed in \(F_2\).
The cyclic subgroup \[ \langle a\rangle\le F_2 \] is isomorphic to \(\mathbb Z=F_1\), and the trivial subgroup is the free group of rank \(0\).
6. There is also a covering whose fundamental group is free of countably infinite rank.
Let \(Y_\infty\) have vertices \[ \{v_k:k\in\mathbb Z\}, \] an oriented \(a\)-edge from \(v_k\) to \(v_{k+1}\) for every \(k\in\mathbb Z\), and a \(b\)-loop at every \(v_k\). Map all \(a\)-edges to \(a\), all \(b\)-loops to \(b\), and all vertices to \(v\). The same local-star check as in 2 shows that \[ p_\infty:Y_\infty\to R \] is a connected covering.
The union of all \(a\)-edges is a bi-infinite line and hence a maximal tree in \(Y_\infty\). Collapsing this tree leaves one circle for each \(b\)-loop, indexed by \(k\in\mathbb Z\). Therefore \[ \pi_1(Y_\infty,v_0) \cong F_{\aleph_0}. \] Since \((p_\infty)_*\) is injective, this gives \[ F_{\aleph_0}\hookrightarrow F_2. \] Concretely, its free generators may be represented in \(F_2\) by \[ a^kba^{-k}, \qquad k\in\mathbb Z. \]
7. The phrase ``any free group’’ cannot literally include free groups of uncountable rank.
The group \(F_2\) is countable: every element is represented by a finite reduced word in the four-letter alphabet \[ \{a,a^{-1},b,b^{-1}\}. \] Hence every subgroup of \(F_2\) is countable.
If \(I\) is uncountable, the free group \(F_I\) on the set \(I\) contains \(I\) itself as its set of free generators and is therefore uncountable. Thus no \(F_I\) with uncountable \(I\) can embed in \(F_2\).
So the statement holds exactly for free groups of at most countable rank: \[ \boxed{F_2\text{ contains a subgroup isomorphic to every free group of finite or countable rank.}} \] In particular, it contains every finitely generated free group.