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The boundary operator of a chain complex is its degree-minus-one differential $\partial :C_i\to C_{i-1}$. The identity ${\partial }^2=0$ says that every chain boundary is a chain cycle.

An element of a chain group is a boundary when it is the image of a higher-degree element under the boundary operator. Since ${\partial }^2=0$, every boundary is a chain cycle. Thus $H_n(C)=Z_n(C)/B_n(C)$, and a chain cycle represents zero in homology exactly when it is a boundary.

A filled triangle is a disk with three vertices, three edges and one two-dimensional topological cell, so $\chi =3-3+1=1$. Its simplicial homology has $(b_0,b_1,b_2)=(1,0,0)$: the filled triangle is contractible, and the three-edge chain cycle is the chain boundary of its face. Keeping only the cycle graph on those three vertices gives $\chi =3-3=0$ and $(b_0,b_1)=(1,1)$, since that chain cycle no longer bounds a face. The surrounding plane's exterior is not part of either simplicial complex; do not add it to $F$.

A homology class is an equivalence class of chain cycles modulo chain boundaries in a chain complex.

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.

The chain map identity $d_{D}f=f{d}_C$ gives, for $(x,y)\in M_i$,Thus $M$ is a chain complex; it is a sign convention for the algebraic mapping cone.

Put $E_i=C_{i-1}$ with differential $d_E=d_C$. Inclusion into the second summand and projection onto the first give a short exact sequence of chain complexesIts connecting homomorphism is particularly explicit. If $x\in C_{i-1}$ is a chain cycle, lift it to $(x,0)\in M_i$ and take its boundary. The second component is $(-1{)}^{i}f(x)$, so the long exact sequence in homology containsIf every $H_i(M)$ vanishes, the exact sequence makes every $f_{\ast }$ both injective and surjective. Conversely, if every $f_{\ast }$ is an isomorphism, the map $H_i(D)\to H_i(M)$ has zero image, while $H_i(M)\to H_{i-1}(C)$ has zero image and zero kernel; hence $H_i(M)=0$.

One can also see the converse directly on chain cycles. If $(x,y)\in M_i$ is a chain cycle, then $d_{C}x=0$ and $d_{D}y=-(-1{)}^{i}f(x)$. Injectivity of the induced map on homology $f_{\ast }$ gives $x=d_{C}z$. Subtracting the chain boundary $d_f(z,0)$ leaves $(0,t)$ with $d_{D}t=0$. Surjectivity of the induced map on homology $f_{\ast }$ lets us choose a chain cycle $w\in C_i$ and $v\in D_{i+1}$ such thatThen $(0,t)=d_f(w,v)$. Thus every chain cycle in $M$ is a chain boundary, proving the mapping cone acyclicity criterion:

For chain maps $f,g:C_{\ast }\to D_{\ast }$, a chain homotopy from $g$ to $f$ is a family of group homomorphisms $s_q:C_q\to D_{q+1}$ satisfyingIf $z$ is a chain cycle, then $(f-g)(z)=d^{D}s(z)$ is a chain boundary. Consequently $[f(z)]=[g(z)]$, so $\boxed{f_{\ast }=g_{\ast }}$ on every homology group.

Order the vertices of the simplex as $v_0,\dots ,v_n$. Its simplicial chain complex hasand zero groups otherwise. An arbitrary ordering represents the sign of the permutation times the increasing ordering. Its boundary operator isIn $d^2$ each face obtained by deleting the vertices in positions $r<t$ occurs twice. Deleting $r$ first gives sign $(-1{)}^r(-1{)}^{t-1}$, whereas deleting $t$ first gives $(-1{)}^t(-1{)}^r$. These signs are opposite, so all terms cancel. The case $d_0{d}_1=0$ follows directly from $d_0=0$. Thus $\boxed{d^2=0}$.

For the homology calculation, add the augmentation $\epsilon :S_0\to \mathbb{Z}$ sending every vertex to $1$. Regard $\mathbb{Z}$ as the group in degree $-1$ of an augmented chain complex. Define the simplicial cone chain contraction byIf the simplex does not contain $v_0$, expansion of its coned boundary gives $ds(\sigma )=\sigma -sd(\sigma )$. If it contains $v_0$, the only face on which $s$ is nonzero is the face omitting $v_0$, and $sd(\sigma )=\sigma$. In degree zero this uses the augmentation, and in degree $-1$ it says $\epsilon s_{-1}=1$. HenceEvery positive-degree chain cycle is therefore a chain boundary. In degree zero, the same identity says $z-\epsilon (z)[v_0]=d{s}_0(z)$, so the augmentation induces an isomorphism $H_0\to \mathbb{Z}$, with inverse $1\mapsto [v_0]$. Thus, including $n=0$,This is a direct chain homotopy calculation and uses no result about cellular homology.

Represent a class in $H_k(K)$ by a chain cycle $z\in Z_k(K)$. Since every simplex of $K=M\cup N$ belongs to at least one of the two simplicial subcomplexes, split the chain in the simplicial chain complex asBecause $\partial z=0$, we haveThe left side lies in $C_{k-1}(M)$ and the right side in $C_{k-1}(N)$, so this common chain is supported in $L=M\cap N$. It is a cycle because ${\partial }^2=0$. The connecting homomorphism is thereforeChanging the decomposition or the representative changes $\partial m$ only by a boundary in $L$, which is why the construction descends to homology classes.

Inclusion and passage to the quotient give a short exact sequence of chain complexesDefine the connecting homomorphism as follows. If $[c]\in H_k(K,L)$, then $\bar{\partial }[c]=0$, so $\partial c\in C_{k-1}(L)$. Since ${\partial }^{2}c=0$, it is a chain cycle in $L$, and we setChanging $c$ by a relative boundary or a chain in $L$ changes $\partial c$ by a boundary in $L$, so $\delta$ is well defined.

The resulting sequence isWe verify exactness directly. If a cycle of $K$ maps to zero in the relative homology, it differs from a boundary by a cycle in $L$, so it lies in $\mathrm{im}{i}_{\ast }$. If a relative cycle $[c]$ satisfies $\delta [c]=0$, then $\partial c=\partial \ell$ for some $\ell \in C_k(L)$; hence $c-\ell$ is a cycle in $K$ mapping to $[c]$. Finally, if a cycle $z\in C_{k-1}(L)$ becomes a boundary in $K$, say $z=\partial c$, then $[c]$ is a relative cycle with $\delta [c]=[z]$. The reverse containments follow immediately from consecutive composites being zero. This proves the long exact sequence in relative homology.

The Snake lemma associates to a commutative diagram with exact rows an exact sequenceFor $K=L_1\cup L_2$, apply it degree by degree to the short exact sequence of chain complexesThe resulting connecting maps splice the kernels modulo boundaries into the Mayer--Vietoris sequence

Now let $K$ satisfy the stated simplicial pseudomanifold conditions. In an $n$-cycle, cancellation at a common $(n-1)$-face determines the coefficient of either incident $n$-simplex from the other, up to the orientation sign. Connectivity through such faces therefore determines every top-simplex coefficient from one integer. If the signs are globally compatible, their oriented sum is a cycle and generates $H_n(K)\cong \mathbb{Z}$; if they are inconsistent, that integer must be zero and $H_n(K)=0$.

Let $x$ be the resulting fundamental class, represented by $z$. Since the intersection has dimension below $n$, split uniquely
$z=z_1+z_2$ into top chains in $L_1$ and $L_2$. Thenand the Mayer--Vietoris boundary isIt is nonzero exactly when both $L_1$ and $L_2$ contain top-dimensional simplices; if one side contains none, the fundamental cycle already lies in the other side, while if both do, connectedness forces a nonempty interface and its oriented boundary represents a nonzero class.

Finally take $K\cong S^3$, $L_1\cong S^1\times D^2$, and
$L_1\cap L_2\cong T^2$. The preceding boundaryis a nonzero map between copies of $\mathbb{Z}$ and sends the fundamental class to the oriented boundary torus, hence is an isomorphism. Exactness then givesThe next part of the sequence isSince $H_1(L_1)\cong \mathbb{Z}$, it follows that
$H_1(L_2)\cong \mathbb{Z}$. The degree-zero sequence makes $L_2$ connected. Therefore

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.

In an augmented chain complex of a simplex with vertices $v_0,\dots ,v_n$, put $s(1)=[v_0]$ and $s[v_{i_0},\dots ,v_{i_q}]=[v_0,v_{i_0},\dots ,v_{i_q}]$ when $v_0$ is absent, and zero when it is present. Direct expansion of the boundary operator gives $ds+sd=1$. This proves that every positive-degree chain cycle is a chain boundary, without invoking cellular homology.