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For a locally compact space, replace finite singular chains by chains for which every compact subset meets the images of only finitely many simplices with nonzero coefficients. The boundary operator preserves this local finiteness, giving the Borel-Moore chain complex and its homology. On a compact space it reduces to ordinary singular homology. A noncompact manifold has a locally finite fundamental class, with its orientation local system when necessary. The chain groups and homology are zero in negative degrees.

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.

An element of a chain group is a cycle when its boundary operator kills it. A cycle need not itself be a chain boundary. The homology group $H_n(C)$ identifies cycles whose difference is a chain boundary.

A chain subcomplex $D_{\bullet }\subseteq C_{\bullet }$ consists of subgroups $D_i\le C_i$ preserved by the boundary operator: $\partial (D_i)\subseteq D_{i-1}$.

An element of a cycle module is null-homologous when it represents zero in homology, equivalently when it lies in the image of the boundary operator. A closed submanifold bounding a submanifold one dimension higher is null-homologous.

Use integral singular homology. The singular chain group $C_n(B)$ is the free abelian group on continuous maps ${\Delta }^n\to B$, and its boundary operator isFor $i<j$, deleting vertex $j$ and then vertex $i$ has sign $(-1{)}^{i+j}$, while deleting $i$ and then the shifted vertex $j-1$ has sign $(-1{)}^{i+j-1}$. Both yield the same codimension-two face. Pairing these terms proves ${\partial }^2=0$. The inclusion $A\subset B$ identifies $C_n(A)$ with the subgroup spanned by those simplices having image in $A$, so it is injective and preserved by $\partial$.

The relative chain complex is $C_n(B,A)=C_n(B)/C_n(A)$ with induced boundary $\partial \bar{b}=\overline{\partial b}$. Thus there is a degreewise short exact sequence of chain complexesInclusion and quotient take homological cycles to homological cycles and boundaries to boundaries, hence define the group homomorphisms $i_{\ast }$ and $q_{\ast }$ on homology. We construct and prove exactness of the resulting sequence directly, rather than invoke an unproved algebraic lemma:ending in $H_0(A)\to H_0(B)\to H_0(B,A)\to 0$.

For the relative homology connecting homomorphism, lift a relative cycle $\bar{b}\in C_n(B,A)$ to $b\in C_n(B)$. Since $\partial \bar{b}=0$, the chain $\partial b$ belongs to $C_{n-1}(A)$; it is a homological cycle because ${\partial }^2=0$. DefineChanging the lift by $a\in C_n(A)$ changes the boundary by $\partial a$, hence does not change its homology class. More generally, if a different relative cycle represents the same relative homology class, lifts satisfy $b^{\prime }-b=\partial c+a$ for some $c\in C_{n+1}(B)$ and $a\in C_n(A)$. Then $\partial b^{\prime }-\partial b=\partial a$. This proves independence both of the lift and of the homology representative. Lifting a sum by the sum of lifts proves additivity, so $\delta$ is a well-defined group homomorphism.

At $H_n(A)$, an element in the image of $\delta$ is represented by $\partial b$ for some $b\in C_{n+1}(B)$, so its image in $H_n(B)$ is zero. Conversely, if a homological cycle in $A$ $a$ has $i_{\ast }[a]=0$, then $a=\partial b$ for a $B$-chain $b\in C_{n+1}(B)$. Its quotient $\bar{b}$ is a relative cycle and $\delta [\bar{b}]=[a]$. ThusFor completeness, the other terms are exact by the same elementary chain argument. At $H_n(B)$, clearly $q_{\ast }{i}_{\ast }=0$. If a homological cycle in $B$ $b$ has zero relative class, then $\bar{b}=\partial \bar{c}$ for a quotient chain $\bar{c}$. Lifting $c$ gives $a=b-\partial c\in C_n(A)$, and $\partial a=0$. Hence $[b]=i_{\ast }[a]$, proving $\ker q_{\ast }=\mathrm{im}{i}_{\ast }$. At $H_n(B,A)$, absolute homological cycles have zero connecting image. Conversely, if $\delta [\bar{b}]=0$, then $\partial b=\partial a$ for some $a\in C_n(A)$; the chain $b-a$ is an absolute homological cycle and maps to $[\bar{b}]$. Therefore $\ker \delta =\mathrm{im}{q}_{\ast }$. In degree zero every quotient chain lifts to an absolute homological cycle, because the differential to degree $-1$ is zero, so $q_{\ast }:H_0(B)\to H_0(B,A)$ is surjective. These proofs establish the whole long exact sequence in relative homology and all the algebraic facts used in its construction.

Use integral homology throughout. The singular chain group $C_k(X)$ is free on the continuous maps from the standard $k$-simplex to $X$. Since the boundary operator preserves chains lying in $A$, these form a chain subcomplex. The relative chain complex and its relative homology areThe short exact sequence of chain complexes gives the long exact sequence in relative homology.

A sufficient hypothesis for the quotient comparison is a nonempty good pair: $A$ is closed and has an open neighbourhood $U$ which admits a deformation retraction onto $A$ while fixing $A$ throughout. A CW pair is another standard sufficient setting. These hypotheses hold for a simple closed curve in the surface here, using its annular collar neighbourhood. The induced quotient map $q:X\to Y=X/A$ collapses $A$ to a point $a$.

Here is the proof of the collapsing a pair theorem. Put $V=U/A\subset Y$. The deformation retraction makes $H_{\ast }(U,A)=0$, so the long exact sequence of the triple gives $H_{\ast }(X,A)\cong H_{\ast }(X,U)$. It also contracts $V$ onto $a$, giving $H_{\ast }(Y,\{a\})\cong H_{\ast }(Y,V)$. Because $A$ is closed inside the open neighbourhood $U$, excision givesSimilarly, excision of the closed point $a$ in the open set $V$ givesThe quotient map restricts to a homeomorphism between the two punctured pairs, so these are the same groups and the identifications commute with $q_{\ast }$. Finally $H_{\ast }(Y,\{a\})\cong {\tilde{H}}_{\ast }(Y)$, including degree zero. ThusThe neighbourhood condition is part of the result; the quotient assertion is not made for arbitrary bad pairs.

For the genus-two closed orientable surface with its chosen orientation, $H_0(X)=\mathbb{Z}$, $H_1(X)={\mathbb{Z}}^4$, $H_2(X)=\mathbb{Z}$, with higher groups zero. The curve has $H_0(A)=H_1(A)=\mathbb{Z}$. Its relevant long exact sequence in relative homology isThe final map is an isomorphism because both spaces are connected. Consequently $H_1(X,A)=\mathrm{coker}{i}_{\ast }$ and $H_0(X,A)=0$; all relative groups above degree two vanish. The quotient topological space is connected, so its unreduced $H_0$ is $\mathbb{Z}$.

In the separating case, the curve is the oriented boundary of one of the two subsurfaces. Its homology class in $X$ is therefore zero, so $i_{\ast }=0$. The exact sequence becomesThe first sequence splits since its quotient is a free abelian group. Hence the homology after collapsing a separating surface curve isGeometrically, collapsing the boundary of each once-punctured torus fills its puncture with a cone on the circle, which is a disk. The quotient is a wedge sum of two tori at the collapsed point. Their two independent fundamental classes explain the extra second-homology generator.

In the nonseparating case, join the two new boundary components of the cut surface by an arc. Upon regluing, this supplies a closed curve meeting $A$ once transversely. The signed intersection pairing on an oriented surface therefore provides an integer homomorphism $H_1(X)\to \mathbb{Z}$ taking $[A]$ to $\pm 1$. Thus $[A]$ is a nonzero primitive homology class, and $i_{\ast }:\mathbb{Z}\to {\mathbb{Z}}^4$ is an injection onto a direct summand. Its kernel is zero and its cokernel is ${\mathbb{Z}}^3$. The same exact sequence gives $H_2(X,A)\cong H_2(X)\cong \mathbb{Z}$. Therefore the homology after collapsing a nonseparating surface curve isAnother description starts from the genus-one surface with two boundary circles. Collapse those two circles separately to obtain a closed torus with two marked points, then identify the two points. Identifying two distinct points in a connected CW complex adds a loop up to homotopy, so this quotient has the homotopy type of $T^2\vee S^1$. The preceding relative homology calculation proves the groups without requiring that homotopy description.

For the sphere homology calculation use singular homology, homotopy invariance of homology, and the reduced Mayer–Vietoris sequence, not cellular homology. For a point, its chain group for singular homology is $\mathbb{Z}$ in every nonnegative degree, and its boundary operator is multiplication by ${\sum }_{i=0}^q(-1{)}^i$, hence alternately zero and one in positive degrees. Its homology is therefore $\mathbb{Z}$ in degree zero and zero above. Singular chains of a disjoint union split into the summand chain complexes. Start with $S^0$, the disjoint union of two points: $H_0(S^0;\mathbb{Z})={\mathbb{Z}}^2$, all positive groups vanish, and ${\tilde{H}}_0(S^0;\mathbb{Z})=\mathbb{Z}$. For $n\ge 1$, remove the north and south poles to obtain an open cover $U,V$ of $S^n$. Each member is contractible, while $U\cap V$ has a deformation retraction to $S^{n-1}$, including the disconnected intersection when $n=1$. The reduced exact sequence therefore suppliesAlso $S^n$ is connected, so its reduced degree-zero group vanishes. Induction givesThe separate $S^0$ calculation avoids conflating its two degree-zero generators.

Choose an orientation of $S^n$, with $n\ge 1$, and let $[S^n]$ be its fundamental class. The mapping degree is the integer determined by $f_{\ast }[S^n]=\mathrm{deg}(f)[S^n]$. To define the local degree at $x_0$, set $y_0=f(x_0)$ and require $x_0$ to be isolated in $f^{-1}(y_0)$. Choose an open neighborhood $B$ containing no other point of that fiber. Then $f$ defines a map of pairsThe degree-$n$ relative homology groups on both sides are $\mathbb{Z}$, by a manifold chart and excision. The local orientation of a manifold chooses their generators. The induced map multiplies these generators by an integer, denoted ${\mathrm{deg}}_{x_0}f$. Excision and naturality show independence of the choice of $B$. Differentiability is unnecessary; if $f$ is smooth and $D{f}_{x_0}$ is invertible, the inverse function theorem gives ${\mathrm{deg}}_{x_0}f=\mathrm{sign}\det D{f}_{x_0}$ relative to the orientations. Without isolation in the fiber this pointwise definition need not apply.

For the degree as a sum of local degrees, assume $y\in S^n$ has finite fiber $F=\{x_1,\dots ,x_s\}$. Disjoint coordinate neighborhoods and excision identifyThe absolute fundamental class maps to the tuple of local orientation generators. Under the relative map induced by $f$, the $j$th generator goes to ${\mathrm{deg}}_{x_j}f$ times the generator at $y$, and the resulting map from the direct sum adds these contributions. On the other hand, naturality says that first applying the absolute map $f_{\ast }$ and then passing to the local group at $y$ gives $\mathrm{deg}(f)$ times that generator. ThusIf the fiber is empty the sum is zero: the map factors through the contractible punctured sphere and has degree zero. If every point of a fiber is isolated, compactness makes that closed fiber finite, so the theorem applies. In particular, a regular value of a smooth map gives the signed count of its inverse images. For $n=0$, define a mapping degree on the reduced generator $[p_{+}]-[p_{-}]$: the identity has mapping degree one, the swap of the two points minus one, and either constant map zero. With local orientation signs positive at $p_{+}$ and negative at $p_{-}$, the same sum formula holds. The connected-sphere convention above is the standard positive-dimensional one.

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.

Because $L$ is a simplicial subcomplex of $K$, every face of a simplex of $L$ also belongs to $L$. The simplicial boundary operator therefore satisfiesso $C_{\bullet }(L)$ is a chain subcomplex of the simplicial chain complex $C_{\bullet }(K)$.

Define a map on the quotient groups byIf $c^{\prime }=c+\ell$ with $\ell \in C_k(L)$, then $\partial c^{\prime }-\partial c=\partial \ell \in C_{k-1}(L)$, so this definition is independent of the representative. Moreover,Thus $C_{\bullet }(K,L)$ is the relative simplicial chain complex.

Write ${\mathbb{F}}_2=\mathbb{Z}/2$. A double covering has a deck transformation $\tau$ that exchanges the two points in every fibre. Every singular simplex $\sigma :{\Delta }^q\to Y$ has exactly two lifts $\tilde{\sigma }$ and $\tau \tilde{\sigma }$. Define the mod-two transfer chain map of a double covering byand let $f_{\#}$ send a simplex of $X$ to its composite with $f$. Uniqueness of lifted faces shows that $T$ commutes with the boundary operator, so both maps are chain maps.

For each base simplex, its two lifts span a copy of ${\mathbb{F}}_2^2$. On this summand, $T$ is $a\mapsto (a,a)$ and $f_{\#}$ is $(a,b)\mapsto a+b$. Hence the sequence is exact in every degree:

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.

For a subspace $A\subseteq X$, the relative chain complex is the quotient group in each degree,with boundary operator $[c]\mapsto [\partial c]$. This is well defined because $C_{\ast }(A)$ is a chain subcomplex of $C_{\ast }(X)$.

A relative homology class is represented by a singular chain whose boundary lies in $A$. Two representatives give the same class when their difference is a boundary operator image plus a chain in $A$. Thus a path with endpoints in $A$ can represent a class even when its endpoints are different.

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.

A singular $n$-simplex in a space $X$ is a continuous map $\sigma :{\Delta }^n\to X$ from the standard $n$-simplex. Its alternating restrictions to the faces form its boundary.