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A contraction is a homotopy from the identity map of $X$ to a constant map. Its existence says that $X$ is a contractible space. For a tree, choosing a root and moving each point along its unique finite path to that root gives a contraction, with continuity understood in the weak topology of a CW complex.

A topological graph is a CW complex with only zero-dimensional and one-dimensional cells: its vertices form the zero-skeleton, and each edge is a copy of an open interval attached by its two endpoints to vertices, possibly the same vertex. It has the weak topology of a CW complex. A subgraph is a CW subcomplex, obtained by choosing vertices and edges and retaining the endpoints of every chosen edge. Loops and multiple edges are allowed.

A spanning tree is a connected spanning subgraph containing no cycle in a graph. To construct one in any connected topological graph $\Gamma$, consider all acyclic spanning subgraphs, ordered by inclusion; the subgraph with every vertex and no edges is one. The union of a chain is still acyclic, since any cycle uses only finitely many edges and would already occur in a member of the chain. The Zorn lemma therefore gives a maximal such subgraph $\Delta$.

If $\Delta$ were disconnected, a finite edge path in $\Gamma$ between vertices of two of its components would contain an edge joining different components. Adding that edge creates no cycle, contradicting maximality. Thus $\Delta$ is a spanning tree. Fix a root vertex. Each point of $\Delta$ has a unique finite path to the root; give every edge length one and move the point a fraction $s$ of the way along that path at time $s\in [0,1]$. This defines a contraction of a topological space of $\Delta$. On each closed edge times $[0,1]$, the construction takes place in a finite tree and is continuous. Since the interval is a locally finite CW complex, the product with $\Delta$ has the corresponding weak topology of a CW complex, so these restrictions establish continuity globally. ConsequentlyThis argument includes infinite graphs and does not assume local finiteness of $\Gamma$.

If $p:\tilde{\Gamma }\to \Gamma$ is a covering map, take $p^{-1}({\Gamma }^0)$ as the vertices upstairs. Each component of the inverse image of an open edge is an open interval mapping homeomorphically to that edge, because an interval is simply connected. Its parametrization lifts along the closed edge to attach its endpoints to lifted vertices, by path lifting theorem. Small star neighborhoods at vertices are lifted homeomorphically by the covering map, and give exactly the neighborhoods of the resulting topological graph. Thus the cell structure has the topology of the original covering space. The total space may be disconnected, but every component is a topological graph.

The Nielsen–Schreier formula isFor a proof, realize $F_n$ as the fundamental group of the wedge sum of $n$ circles. Construct its connected covering graph with vertices the right cosets $Hg$, and an oriented edge labeled $x_i$ from $Hg$ to $Hg{x}_i$ for each free generator $x_i$. There is exactly one incoming and one outgoing edge of each label at every vertex, which establishes the local covering map property. Reading a word from $H$ ends at its coset, so the closed paths at $H$ give exactly the words in $H$. The induced map of fundamental groups is injective by the homotopy lifting property of a covering map, and its image is $H$.

For index $d$, this covering graph has $d$ vertices and $dn$ edges. A spanning tree in it has $d-1$ edges. The permitted rank formula for a finite connected graph now givesIn particular this proves both freeness and the index formula.

If $n\ge 2$ and $d>1$, then $1+d(n-1)>n$, so a proper finite-index subgroup cannot be isomorphic to $F_n$. For $n=1$, $F_1\cong \mathbb{Z}$, and $d\mathbb{Z}$ is a proper index-$d$ subgroup isomorphic to $\mathbb{Z}$ for every $d\ge 2$. Hence