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A CW subcomplex is a union of cells of a CW complex containing the attaching boundary of every included cell. In a combinatorial 2-complex, including a two-cell therefore includes every edge in its boundary. A connected CW subcomplex has a connected 1-skeleton, whose intrinsic path metric may differ from the ambient graph metric.

A topological graph is a CW complex with only zero-cells and one-cells, called vertices and edges. Edge endpoints attach to vertices; loops, multiple edges and infinitely many edges are allowed. A subgraph in this setting is a CW subcomplex. A connected acyclic graph is a tree and is a contractible space.

Let $C_{\ast }(X)$ be the integral singular chain complex. Augment it by the homomorphism $C_0(X)\to \mathbb{Z}$ sending every point simplex to one. The homology of this augmented complex is reduced homology: for a nonempty space, it agrees with ordinary homology in positive degrees and ${\tilde{H}}_0(X)$ is the kernel of $H_0(X)\to \mathbb{Z}$. The relative homology $H_{\ast }(X,A)$ is the homology of $C_{\ast }(X)/C_{\ast }(A)$; a relative cycle has boundary in $A$, and chains differing by a boundary or a chain in $A$ represent the same class.

Excision theorem: if $U\subset A\subset X$ and $\overline{U}\subset {\mathrm{int}}_{X}A$, the inclusion $(X\setminus U,A\setminus U)\hookrightarrow (X,A)$ induces isomorphisms on all relative homology groups. In particular, for a nonempty CW subcomplex $A$, choose a neighborhood $N$ deforming onto $A$, with the deformation fixing $A$. Such neighborhoods exist for CW pairs. Then $N/A$ is contractible. The long exact sequence of a pair gives $H_{\ast }(X,A)\cong H_{\ast }(X,N)$ and $H_{\ast }(X/A,\{\ast \})\cong H_{\ast }(X/A,N/A)$. Excision removes $A$ and the collapsed point respectively; the resulting pairs $(X\setminus A,N\setminus A)$ and $((X/A)\setminus \{\ast \},(N/A)\setminus \{\ast \})$ are homeomorphic. ConsequentlyHere the nonempty-subcomplex condition is the usual quotient convention in this statement. If $A=\varnothing$, use the pointed cofiber $X_{+}$ with a disjoint basepoint; the literal unpointed quotient $X/\varnothing =X$ would not give the asserted degree-zero identity.

For one embedded circle, its long exact sequence of a pair reduces toA separating circle bounds a subsurface, so its homology class is zero. A nonseparating circle has a transverse closed curve meeting it once, showing that its class is nonzero and primitive. Thus $i_{\ast }$ is either zero or an injection with torsion-free cokernel. The resulting possibilities arewith all other relative groups zero. These ranks determine whether the circle is null-homologous.

For $g\ge 2$, a disk boundary and a separating circle bounding a once-punctured torus have the same relative groups. The first is null-homotopic; the second is not. For a direct verification, use the surface-group presentation $⟨a_1,b_1,\dots ,a_g,b_g\mid {\prod }_i[a_i,b_i]=1⟩$. Send $a_1,b_1$ to free generators $u,v$, send $a_2,b_2$ to $v,u$, and all remaining generators to one. The relation maps to $[u,v][v,u]=1$, so this defines a homomorphism to the free group. The separating circle represents $[a_1,b_1]$ and maps to the nonempty reduced word $uv{u}^{-1}{v}^{-1}$; it is therefore not null-homotopic. This proves the requested failure to detect nullhomotopy. There is a necessary low-genus qualification: on ${\Sigma }_0$ or ${\Sigma }_1$, an embedded circle is nullhomologous exactly when it is nullhomotopic. On the sphere every circle bounds a disk; on the torus every essential embedded circle is nonseparating. The printed nondetection assertion therefore needs $g\ge 2$.

For two disjoint embedded circles let $r$ be the rank of their image in $H_1({\Sigma }_g)$, and let $c$ be the number of connected components of the cut surface. Removing open annular neighborhoods and applying excision gives the compact cut surface $W$ relative to its boundary. Poincare-Lefschetz duality gives $H_2(W,\partial W)\cong H^0(W)={\mathbb{Z}}^c$ and $H_1(W,\partial W)\cong H^1(W)$, which is free. Thus there is no hidden torsion. The long exact sequence gives $c=3-r$, and the Euler characteristic of the pair is $2-2g$, givingThe possibilities are $r=0,1,2$: two separating circles give zero; a nonseparating circle with a disk boundary, or two parallel essential circles, give one; circles around two different handles give two. For genus zero only $r=0$ occurs; for genus one $r=0,1$ occur; for every genus at least two all three occur. This is relative homology of a surface modulo disjoint circles.

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

Let $L$ be the maximum perimeter of a two-cell of $X$, taking $L=0$ if there are none. Compactness of the combinatorial complex makes $L$ finite. We show that the vertex-endpoint metric geodesic $\gamma$ stays within $L/2$ of ${\tilde{Y}}^{(1)}$.

Fix a shortest path $\sigma$ in ${\tilde{Y}}^{(1)}$ and choose the reduced diagram of minimum area, then minimum edge count, as in the preceding part. There is no spur in the interior of either boundary side, since both sides are reduced metric geodesic paths in their respective graphs.

There can be no shell with its entire exterior arc $Q$ on $\gamma$: the Greendlinger lemma gives a complementary path $P$ shorter than $Q$, contradicting the ambient metric geodesic property. There can be no such shell with $Q$ entirely on $\sigma$ either. If its image cell $R$ belongs to $\tilde{Y}$, the shorter complementary path belongs to ${\tilde{Y}}^{(1)}$ and contradicts the intrinsic metric geodesic property of $\sigma$. If $R$ is outside $\tilde{Y}$, the given perimeter decomposition has an arc $I$ of length at least half the perimeter all of whose edges are outside $\tilde{Y}$. Any contiguous boundary arc whose edges lie in $\tilde{Y}$ must then be contained in the complementary arc $O$ and has length at most half the perimeter. This contradicts $|Q|>|\partial R|/2$.

Thus every shell or spur of the diagram must straddle one of the two marked boundary corners $p,q$. Distinct shell exterior arcs have disjoint interiors, so there can be at most two such exposed features. The Greendlinger ladder theorem now says that the diagram is a single cell or a ladder, apart from common edge paths and the zero-area case. Its two ends contain the marked corners; each intervening cell has one boundary arc on $\gamma$ and one on $\sigma$. Common edge paths already map into $\tilde{Y}$.

For a point $z$ of the $\gamma$ arc of a cell, choose a vertex on its $\sigma$ arc. One of the two paths around that cell boundary to the chosen vertex has length at most half the perimeter. Its image is an ambient path to ${\tilde{Y}}^{(1)}$, soThe same argument treats a single cell. In the zero-area case the reduced boundary sides coincide and already lie in ${\tilde{Y}}^{(1)}$.

Finally allow endpoints in the interiors of edges of ${\tilde{Y}}^{(1)}$. The initial and final partial edges of a graph metric geodesic lie in ${\tilde{Y}}^{(1)}$; what remains is a metric geodesic between vertices of that CW subcomplex. If both endpoints are joined without reaching a vertex, the whole segment is already in its edge. Thus the same bound applies to every graph metric geodesic, andThis is quasiconvexity from no missing shells. The intrinsic path $\sigma$ was never assumed to be an ambient metric geodesic, and the bound is independent of the endpoints and of its length.

Let $X$ be a finite combinatorial 2-complex satisfying $C^{\prime }(1/6)$, and let $Y$ be a connected CW subcomplex of its universal cover with no missing shells. If the largest cell perimeter is $L$, then $Y^{(1)}$ is $(L/2)$-quasiconvex in the universal cover's unit-edge 1-skeleton. Bound an ambient metric geodesic and an intrinsic metric geodesic of $Y^{(1)}$ by a reduced disc diagram. A shell on the ambient side would shorten a metric geodesic. A shell on the intrinsic side belongs to $Y$ and would shorten its intrinsic metric geodesic. Spurs are likewise excluded from side interiors. The Greendlinger ladder theorem forces a ladder between the two marked corners. Each cell gives a path of length at most half its perimeter from either side to the other.

A CW subcomplex has no missing shells if a cell with a shell-length boundary arc contained in the CW subcomplex is itself contained in the CW subcomplex. For $C^{\prime }(1/6)$ it suffices that every excluded cell have a complementary boundary arc of length at least half its perimeter with no edge in the CW subcomplex: an arc lying in the CW subcomplex then has length at most half and cannot be a shell's exterior arc.