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The singular chain complex of a product is naturally chain-homotopy equivalent to the tensor product of the factors' chain complexes. The shuffle map and the front-face/back-face map implement the equivalence. It is the chain-level input to the Künneth theorem.

A chain complex consists of abelian groups $C_q$ and homomorphisms ${\partial }_q:C_q\to C_{q-1}$ with ${\partial }_{q-1}{\partial }_q=0$. Its groups of chain cycles and chain boundaries are $Z_q=\ker {\partial }_q$ and $B_q=\mathrm{im}{\partial }_{q+1}$; the identity ensures $B_q\subseteq Z_q$. Its homology groups areFor a nonnegative complex, take $C_q=0$ for $q<0$.

The singular chain group $C_q(X)$ is the free abelian group on continuous maps $\sigma :{\Delta }^q\to X$, called singular simplices. With ${\delta }_i$ the inclusion of the face opposite vertex $i$, defineDeleting any two vertices in the two possible orders gives the same face with opposite signs, proving ${\partial }^2=0$. This defines the singular chain complex and its singular homology. A continuous map $f$ induces the chain map $f_{\#}(\sigma )=f\circ \sigma$; it commutes with the boundary because composition commutes with restriction to faces.

To prove homotopy invariance of homology, let $F:X\times I\to Y$ have endpoints $f,g$. Triangulate ${\Delta }^q\times I$ by the $q+1$ simplices with ordered verticesLet $p_i:{\Delta }^{q+1}\to {\Delta }^q\times I$ be their affine parameterizations. The singular prism operator isOn taking boundaries, shared internal faces of consecutive prism simplices cancel. The top and bottom faces give $g_{\#}\sigma -f_{\#}\sigma$, and the remaining faces are $-P_{q-1}(\partial \sigma )$ with their induced signs. ThusFor a chain cycle $z$, $g_{\#}z-f_{\#}z=\partial Pz$ is a chain boundary, so $f_{\ast }=g_{\ast }$ on homology. If $f$ is a homotopy equivalence, its homotopy inverse therefore induces an inverse to $f_{\ast }$, proving that singular homology depends only on homotopy type.

Use the standard CW complex for an orientable genus-three surface: one vertex, six edges $a_1,b_1,a_2,b_2,a_3,b_3$, and a two-cell whose attaching word is ${\prod }_{j=1}^3[a_j,b_j]$. The cellular homology theorem identifies this cellular calculation with singular homology. The boundary of every edge is zero, and each edge's total exponent in the attaching word is zero, so the two-cell boundary is zero. HenceChoose $C_1$ to be the nonseparating curve represented by $a_1$, and choose $C_2$ as a small circle bounding a disc in a surface patch. Then $[C_1]$ is a nonzero primitive homology class, whereas the inclusion of $C_2$ induces zero on $H_1$. If a homotopy equivalence carried $C_1$ onto $C_2$, its restriction would factor through $C_2$ and kill $[C_1]$, contradicting injectivity on first homology. This is the homology obstruction to carrying surface curves onto one another.

There is nevertheless such a continuous map. Map the edge $a_1$ homeomorphically onto $S^1$ and all other edges to its base point. The two-cell attaching word then has winding number zero in $S^1$, hence is null-homotopic and the map extends to $r:{\Sigma }_3\to S^1$. Compose $r$ with a homeomorphism $S^1\to C_2$ and the inclusion of $C_2$ in the surface. This gives $g(C_1)=C_2$. Thus an arbitrary map can carry the first curve onto the second, but a homotopy equivalence cannot.

For an unambiguous drawing, realize ${\Sigma }_3$ as the double of a manifold formed from a disc with three holes. Its two copies are glued along all four boundary circles. The red curve below follows an arc from the outer boundary to the first hole on the front copy and returns along the matching arc on the back copy; it is a nonseparating curve. The blue circle is entirely in a disc patch on the front copy and is contractible.

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.

Use the cellular chain complex or simplicial chain complex over $\mathbb{Q}$, rather than the infinite singular chain complex. Put $r_i=\mathrm{rank}{\partial }_i$, with $r_0=r_{d+1}=0$. The rank-nullity theorem gives $\dim \ker {\partial }_i=\dim C_i-r_i$; quotienting the chain cycles by the chain boundaries then gives $b_i=\dim C_i-r_i-r_{i+1}$. Each boundary operator's rank occurs in adjacent degrees with opposite signs, soFiniteness makes this cancellation an ordinary finite sum. This is the algebra behind the cell-count formula for Euler characteristic.