[proofplan] Let $X$ be the rose with petals indexed by $S$, and let $p: \tilde{X} \to X$ be the universal cover of $X$, which we describe as a Cayley-type tree: a connected graph whose vertices correspond to reduced words in $F(S)$ and whose edges encode multiplication by a generator. Using the [deck transformation theorem](/theorems/???), $\pi_1(X, x_0)$ acts simply transitively on the fibre $p^{-1}(x_0)$, giving a bijection between $\pi_1(X, x_0)$ and $p^{-1}(x_0)$. On the other hand, the universal-property construction of [universal covers of graphs](/theorems/???) identifies $p^{-1}(x_0)$ with the set of reduced words $F(S)$. Composing gives a bijection $F(S) \leftrightarrow \pi_1(X, x_0)$, and we verify directly that the bijection is a group homomorphism by showing concatenation of words corresponds to concatenation of loops. [/proofplan] [step:Construct the universal cover as a Cayley tree] The space $X$ is the rose with one $1$-cell $e_s$ for each $s \in S$ attached at the single vertex $x_0$; equivalently, $X$ is the CW complex with one $0$-cell $x_0$ and $|S|$ oriented $1$-cells $e_s$ with both endpoints attached to $x_0$. As a CW complex, $X$ is path connected, locally path connected, and semi-locally simply connected (every point has a neighbourhood deformation retracting to the vertex), so [the universal cover exists](/theorems/???). Define the graph $T_S$ as follows: \begin{itemize} \item \emph{Vertices:} the set $F(S)$ of reduced words (including the empty word $e$). \item \emph{Edges:} for each reduced word $w$ and each $s \in S$, an oriented edge labelled $s$ from $w$ to $ws$ provided $ws$ is reduced (equivalently, the last letter of $w$ is not $s^{-1}$); we require the edge from $w$ to $ws$ and the edge from $ws$ to $w$ (labelled $s^{-1}$) to be the same geometric $1$-cell traversed in opposite directions. \end{itemize} Realise $T_S$ as a $1$-dimensional CW complex whose $0$-cells are the vertices and whose $1$-cells are the edges above. Define \begin{align*} p: T_S &\to X \end{align*} by sending every vertex to $x_0$ and sending the edge from $w$ to $ws$ to the $1$-cell $e_s$ (traversed in the direction of its orientation). This is a well-defined continuous cellular map. [claim:$T_S$ is a simply connected CW complex] [proof] \emph{Connectedness.} Any vertex $w = x_1 \cdots x_n$ can be joined to the vertex $e$ by the path following edges labelled $x_n^{-1}, x_{n-1}^{-1}, \ldots, x_1^{-1}$ in succession: this traces $w \to x_1 \cdots x_{n-1} \to \cdots \to e$. The edge path is valid because at each stage we are removing the last letter of a reduced word, which reverses a reduced edge. So every vertex connects to $e$; connectedness follows. \emph{Absence of cycles.} Suppose $v_0, v_1, \ldots, v_k = v_0$ is a non-trivial loop in $T_S$ (no consecutive repeated edges). Each step $v_{j} \to v_{j+1}$ multiplies $v_j$ on the right by some letter $\ell_j \in S \cup S^{-1}$ (where $\ell_j = s$ if the edge is labelled $s$ in the forward direction, $\ell_j = s^{-1}$ if traversed in reverse). Since the loop has no immediate backtracking, the sequence $(\ell_0, \ell_1, \ldots, \ell_{k-1})$ has no consecutive pair $(s, s^{-1})$ or $(s^{-1}, s)$, i.e., it is a reduced word. But then $v_k = v_0 \ell_0 \ell_1 \cdots \ell_{k-1}$ and the product $\ell_0 \ell_1 \cdots \ell_{k-1}$ is a non-trivial reduced word, so $v_k \ne v_0$ — contradiction. Hence $T_S$ has no non-trivial loops. A connected graph with no non-trivial loops is a [tree](/theorems/???), and [trees are simply connected](/theorems/???). Hence $\pi_1(T_S) = \{e\}$. [/proof] [/claim] [claim:$p: T_S \to X$ is a covering map] [proof] For each $s \in S$ let $U_s \subseteq X$ be the open neighbourhood consisting of the interior of the $1$-cell $e_s$ together with a small open arc at each endpoint (both endpoints equal $x_0$), and let $V \subseteq X$ be a small open star neighbourhood of $x_0$ meeting each $e_s$ in a small open arc. Over the interior of $e_s$: the preimage $p^{-1}(\text{int } e_s)$ is a disjoint union of open intervals, one for each edge of $T_S$ labelled $s$ in the forward direction; each is mapped homeomorphically to $\text{int } e_s$ by $p$. Over a star neighbourhood $V$ of $x_0$: the preimage $p^{-1}(V)$ is a disjoint union of open star neighbourhoods of the vertices of $T_S$, one for each vertex $w \in F(S)$, each mapped homeomorphically to $V$. Combining these: every point of $X$ has an evenly covered open neighbourhood, so $p$ is a covering map. [/proof] [/claim] Since $p: T_S \to X$ is a covering map with $T_S$ simply connected (and path connected), $p$ is a universal covering. Picking $\tilde{x}_0 := e \in T_S$ (the vertex corresponding to the empty word) as basepoint with $p(\tilde{x}_0) = x_0$, we have \begin{align*} p^{-1}(x_0) = \{w : w \in F(S)\} \end{align*} as a set, with the explicit identification $w \leftrightarrow w$ at the level of vertices. [guided] The universal cover of the rose is the \emph{Cayley graph} of the free group $F(S)$ with respect to its generating set $S$. Vertices of this Cayley graph are reduced words (= elements of $F(S)$), and edges record multiplication by generators. The Cayley graph of a free group is famously a tree, and our construction makes this explicit. Why does $T_S$ cover $X$? Locally, $X$ looks like the union of $2|S|$ half-open arcs meeting at the vertex $x_0$ (two per petal). At each vertex $w \in T_S$ there are the same $2|S|$ half-open arcs — one for each $s \in S$ going forward to $ws$, one going backward to $ws^{-1}$, and similarly for every generator — matching up exactly. Away from the vertex, each petal is a line segment, and each corresponding edge of $T_S$ is also a line segment mapping homeomorphically. Why is $T_S$ simply connected? Because it has no loops — and we saw no loops because any loop's edge labels multiply to a non-trivial reduced word, moving the "current vertex" away from its starting point. [/guided] [/step] [step:Set up the bijection $F(S) \to \pi_1(X, x_0)$ via the fibre $p^{-1}(x_0)$] By the previous step, the fibre identifies as \begin{align*} \Theta: F(S) &\xrightarrow{\sim} p^{-1}(x_0), & w &\mapsto w \text{ (vertex of } T_S\text{)}. \end{align*} By the [deck-transformation characterisation of the fundamental group](/theorems/???) for the universal cover, there is a bijection \begin{align*} \Lambda: \pi_1(X, x_0) &\xrightarrow{\sim} p^{-1}(x_0) \end{align*} obtained as follows: given a loop $\gamma$ at $x_0$, lift it uniquely to a path $\tilde{\gamma}$ in $T_S$ with $\tilde{\gamma}(0) = \tilde{x}_0 = e$; then $\Lambda([\gamma]) := \tilde{\gamma}(1)$. This is well-defined on homotopy classes by the [homotopy lifting property](/theorems/???) and bijective because $T_S$ is simply connected. Define \begin{align*} \Psi: F(S) &\to \pi_1(X, x_0), & \Psi := \Lambda^{-1} \circ \Theta. \end{align*} Since both $\Theta$ and $\Lambda$ are bijections, $\Psi$ is a bijection. [/step] [step:Describe $\Psi$ explicitly on generators and reduced words] For $s \in S$, let $\ell_s: [0, 1] \to X$ be the loop that traverses the petal $e_s$ once in its orientation. Its unique lift to $T_S$ starting at $e$ runs along the edge from $e$ to $s$ (the vertex corresponding to the reduced word $s$), so $\Lambda([\ell_s]) = s$ and therefore $\Psi(s) = [\ell_s]$. For a reduced word $w = x_1 x_2 \cdots x_n \in F(S)$ with each $x_j \in S \cup S^{-1}$, consider the loop \begin{align*} \gamma_w := \ell_{x_1} \cdot \ell_{x_2} \cdot \ldots \cdot \ell_{x_n} \end{align*} in $X$ (concatenation of petal loops, where $\ell_{s^{-1}}$ is the reverse of $\ell_s$). The lift $\tilde{\gamma}_w$ to $T_S$ starting at $e$ traces the edges corresponding to $x_1, x_2, \ldots, x_n$ in order: at stage $j$, the lift has reached the vertex $x_1 \cdots x_{j-1}$ (a reduced word because $w$ is reduced) and the edge labelled $x_j$ from $x_1 \cdots x_{j-1}$ to $x_1 \cdots x_j$ exists in $T_S$ precisely because $x_1 \cdots x_j$ is reduced. Hence $\tilde{\gamma}_w(1) = x_1 \cdots x_n = w$, giving $\Lambda([\gamma_w]) = w$, and therefore \begin{align*} \Psi(w) = [\gamma_w] = [\ell_{x_1} \cdot \ell_{x_2} \cdot \ldots \cdot \ell_{x_n}]. \end{align*} [guided] Concretely, $\Psi$ sends a reduced word $w = x_1 \cdots x_n$ to the homotopy class of the loop that traces the $n$ petals indicated by $x_1, \ldots, x_n$ in order (with the convention that $\ell_{s^{-1}}$ is $\ell_s$ traversed backwards). Why does the lift $\tilde{\gamma}_w$ to $T_S$ end at the vertex $w$? Because we designed $T_S$ so that the edge labelled $s$ at vertex $u$ points to $us$ exactly when $us$ is reduced. The reducedness of $w$ ensures that at each intermediate stage $x_1 \cdots x_j$, we are at a reduced word, and the next letter $x_{j+1}$ gives an edge of $T_S$ to follow. So the lift traces a path from $e$ to $w$ using the letters of $w$ as "driving instructions". This is the crucial geometric content: the reduced-word structure of $F(S)$ is literally the vertex-structure of $p^{-1}(x_0)$. [/guided] [/step] [step:Verify $\Psi$ is a group homomorphism] Let $u, v \in F(S)$ be reduced words. We show $\Psi(uv) = \Psi(u) \Psi(v)$ in $\pi_1(X, x_0)$. Following the argument of the [universal property of the free group](/theorems/1900), Step 3, write \begin{align*} u &= x_1 \cdots x_{p-k} \, a_k \cdots a_1, \\ v &= a_1^{-1} \cdots a_k^{-1} \, y_{k+1} \cdots y_q, \end{align*} where $k \ge 0$ is the maximal seam cancellation, so that \begin{align*} uv = x_1 \cdots x_{p-k} \, y_{k+1} \cdots y_q \end{align*} is the reduced form. By the explicit formula in the previous step: \begin{align*} \Psi(u) \Psi(v) = [\ell_{x_1} \cdot \ldots \cdot \ell_{x_{p-k}} \cdot \ell_{a_k} \cdot \ldots \cdot \ell_{a_1} \cdot \ell_{a_1^{-1}} \cdot \ldots \cdot \ell_{a_k^{-1}} \cdot \ell_{y_{k+1}} \cdot \ldots \cdot \ell_{y_q}]. \end{align*} For each $j = 1, \ldots, k$, the concatenation $\ell_{a_j} \cdot \ell_{a_j^{-1}}$ is the loop $\ell_{a_j}$ immediately followed by its reverse; this is null-homotopic in $X$ rel endpoints by the [standard homotopy from $\alpha \cdot \bar{\alpha}$ to the constant loop](/theorems/???). Cancelling these $k$ pairs from the inside out (each cancellation is a homotopy rel endpoints), \begin{align*} \ell_{x_1} \cdot \ldots \cdot \ell_{x_{p-k}} \cdot \ell_{a_k} \cdot \ldots \cdot \ell_{a_1} \cdot \ell_{a_1^{-1}} \cdot \ldots \cdot \ell_{a_k^{-1}} \cdot \ell_{y_{k+1}} \cdot \ldots \cdot \ell_{y_q} \simeq \ell_{x_1} \cdot \ldots \cdot \ell_{x_{p-k}} \cdot \ell_{y_{k+1}} \cdot \ldots \cdot \ell_{y_q}. \end{align*} The right-hand side is exactly the loop representing $\Psi(uv)$ from the previous step. Hence $\Psi(u) \Psi(v) = \Psi(uv)$ in $\pi_1(X, x_0)$. [guided] The verification mirrors the algebraic cancellation in $F(S)$ exactly. In $F(S)$, the product $uv$ reduces by cancelling $k$ pairs of mutually inverse letters at the seam. In $\pi_1(X, x_0)$, the concatenated loop $\gamma_u \cdot \gamma_v$ homotopes to $\gamma_{uv}$ by the $k$ "loop-and-reverse" cancellations: each pair $\ell_{a_j} \cdot \ell_{a_j^{-1}}$ is null-homotopic, so it can be excised from the concatenation without changing the homotopy class. The bookkeeping is identical to Step 3 of the free-group universal property proof: the seam structure of reduced-word multiplication is exactly mirrored by the homotopy structure of concatenated loops. One way to see the whole argument more abstractly: by the universal property of $F(S)$, the set map $s \mapsto [\ell_s] \in \pi_1(X, x_0)$ extends uniquely to a homomorphism $F(S) \to \pi_1(X, x_0)$; this homomorphism is $\Psi$. So $\Psi$ is a homomorphism by construction. [/guided] [/step] [step:Conclude that $\Psi$ is an isomorphism] The map $\Psi: F(S) \to \pi_1(X, x_0)$ is a bijection (Step 2) and a group homomorphism (Step 4), hence a group isomorphism. This establishes $F(S) \cong \pi_1(X, x_0)$. [/step]