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For a pair $A\subseteq X$, the short exact sequence of chain complexes $0\to C_{\ast }(A)\to C_{\ast }(X)\to C_{\ast }(X,A)\to 0$ induces

For a chain map $f:C\to D$, use $M_i=C_{i-1}\oplus D_i$ and $d(x,y)=(d_{C}x,(-1{)}^{i}f(x)+d_{D}y)$. The chain map identity makes $d^2=0$. The short exact sequence of chain complexes $0\to D\to M\to C_{\ast -1}\to 0$ has connecting homomorphism $(-1{)}^i{f}_{\ast }$ in degree $i$. Its long exact sequence in homology shows that the homology of $M$ vanishes precisely when every $f_{\ast }$ is an isomorphism.

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.

The Excision theorem for singular homology states that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq {\mathrm{int}}_{X}A$, inclusion induces isomorphisms $H_q(X\setminus Z,A\setminus Z;\mathbb{Z})\cong H_q(X,A;\mathbb{Z})$ for all $q$. For an open cover $X=U\cup V$, its small-chain form says that the subcomplex of singular chains whose individual simplices lie in $U$ or $V$ computes $H_{\ast }(X)$. Concretely, repeated barycentric subdivision puts each finite chain in that subcomplex, and subdivision is chain-homotopic to the identity. There is then a short exact sequence of chain complexesIts long exact sequence in homology, with excision identifying the last complex's homology with $H_{\ast }(X)$, is the Mayer–Vietoris sequence:

The Real projective space ${\mathbb{RP}}^1$ is a circle. The projective plane has one cell in each dimension $0,1,2$, with cellular homology boundary maps $d_2=2$ and $d_1=0$. Thus, writing $A_q=H_q({\mathbb{RP}}^2;\mathbb{Z})$, we have $A_0=\mathbb{Z}$, $A_1=\mathbb{Z}/2$ and $A_q=0$ for $q\ge 2$. Cover the circle by two open arcs with a two-component intersection and multiply this cover by the projective plane. In the Mayer–Vietoris sequence the overlap-to-cover map isIts kernel and cokernel are each $A_q$. Consequently $0\to A_q\to H_q(X)\to A_{q-1}\to 0$ for $q\ge 1$. In degree one the quotient is free, so this sequence splits; in degree two it identifies the group directly with $A_1$. The integral homology of a circle times a real projective plane is therefore

Without a compactness restriction, the answer to the final existence question is yes: there is a noncompact oriented four-manifold. Explicitly, takeThe involution is free and reverses orientation on both $S^2$ and $\mathbb{R}$, hence preserves their product orientation. Its quotient is consequently an oriented three-manifold, the orientable total space of the orientation line bundle of the projective plane. The contraction $[v,t]\mapsto [v,(1-s)t]$ is a deformation retraction onto ${\mathbb{RP}}^2$, so $Y\simeq X$.

If a closed oriented manifold is intended, the answer instead is no. A closed connected oriented $d$-manifold has $H_d\cong \mathbb{Z}$ and no homology above $d$. Here the highest nonzero positive-dimensional homology is the torsion group $H_2=\mathbb{Z}/2$, so no dimension can supply that required top fundamental class. The distinction is necessary because the printed wording does not require the oriented manifold to be closed or three-dimensional.

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 each prime number $p$, reduction of the mapping cone is the mapping cone of the reduced chain map. Part (a), whose proof works over any coefficient ring, therefore givesEvery $M_i$ is a finitely generated free abelian group. Multiplication by $p$ consequently gives a short exact sequence of chain complexesThe neighboring terms $H_{i+1}(M\otimes {\mathbb{F}}_p)$ and $H_i(M\otimes {\mathbb{F}}_p)$ in its long exact sequence in homology both vanish. Thus multiplication by $p$ on $H_i(M)$ is an isomorphism for every prime number $p$.

The group $H_i(M)$ is a finitely generated abelian group, since it is a quotient of a subgroup of $M_i$. Write it, using the Fundamental theorem of finitely generated abelian groups, aswith $T_i$ finite. If $b_i>0$, multiplication by any prime number is not surjective. If $T_i\ne 0$, choose a prime number dividing its order; multiplication by that prime is not injective. Both contradict the preceding isomorphism. Hence $H_i(M)=0$ for every $i$.

Applying part (a) integrally now proves the stronger natural conclusionThis is prime coefficient detection of quasi-isomorphisms. The finite-generation hypothesis is used precisely when excluding nonzero groups on which every prime acts invertibly.

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.

First check the signs directly. For $(x,y)\in M_i$, applying the proposed differential twice givesbecause $f$ is a chain map. Thus $M$ is a chain complex.

Let $C^{\prime }$ be the shifted chain complex with $C_i^{\prime }=C_{i-1}$ and differential $d_C$; here there is no minus sign in that differential. Inclusion in the second summand and projection onto the first give a short exact sequence of chain complexesTo compute the connecting map in its long exact sequence in homology, lift a cycle $x\in C_{i-1}$ to $(x,0)\in M_i$. Its boundary is $(0,(-1{)}^{i}f(x))$. Hence the connecting map $H_i(C^{\prime })\to H_{i-1}(D)$ is $(-1{)}^i{f}_{\ast }$, and the relevant exact portion isIf every $f_{\ast }$ is an isomorphism, exactness makes $H_i(M)=0$. Conversely, if every $H_i(M)$ is zero, the adjacent exact portions make every $f_{\ast }$ both injective and surjective. ThereforeThis is the acyclicity criterion for a mapping cone, with the degree-dependent signs adjusted to the given convention.

For the assertion about spaces, use the cellular approximation theorem to replace $f$ by a cellular map, and take the induced map of the finite free cellular chain complexes. Tensoring these complexes and their cone with ${\mathbb{F}}_p$ gives the corresponding mod-$p$ complexes and cone. The same exact-sequence argument works over ${\mathbb{F}}_p$, so the assumed homology isomorphisms implyThe universal coefficient theorem for homology givesIn particular $H_i(M)\otimes {\mathbb{F}}_p=0$ for every prime. Each $H_i(M)$ is a finitely generated abelian group. A nonzero free summand would survive tensoring with every ${\mathbb{F}}_p$, while a nonzero finite cyclic summand would survive for a prime dividing its order. Thus $H_i(M)=0$ in every degree, by detection of integral acyclicity modulo primes. Applying the cone criterion once more proves the integral homology map is an isomorphism in every degree. Finite generation is what makes detection by all prime fields sufficient.

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.

Let $f:(C,d)\to (C^{\prime },d^{\prime })$ be a chain map, soIf $x\in C_i$ is a cycle, then $d^{\prime }{f}_i(x)=f_{i-1}d(x)=0$, so $f_i(x)$ is a cycle. If $x=d_{i+1}y$ is a boundary, thenis a boundary. Henceis a well-defined induced map on homology.

The maps $f$ and $g$ are chain homotopic if there are homomorphismssuch thatFor a cycle $x$, this giveswhich is a boundary. Thus $f_{\ast }[x]=g_{\ast }[x]$ and $f_{\ast }=g_{\ast }$ on homology.

Now consider the proposed mapping cone $M(f)$. Applying its differential twice givesThe diagonal entries vanish because $C$ and $C^{\prime }$ are chain complexes, and the lower-left entry isby the chain-map identity. Therefore $d_f^2=0$ and $M(f)$ is a chain complex.

There is a short exact sequence of chain complexeswhereIts long exact sequence in homology isTo identify the connecting map, represent a class in $H_{i-1}(C)$ by a cycle $x$ and lift it to $(x,0)\in M(f{)}_i$. ThensoAt the preceding occurrence the degree is $i+1$, giving $(-1{)}^{i+1}{f}_{\ast }$. Hence the sequence is exactly

Finally suppose $f-g=d^{\prime }h+hd$. DefineA direct calculation givesThese are equal precisely because $f-g=d^{\prime }h+hd$. Thus $\Phi$ is a chain map. Replacing the plus sign in its definition by a minus sign gives its inverse, so $M(f)$ and $M(g)$ are isomorphic as chain complexes.

Let $f$ be a chain map between degreewise finitely generated free abelian groups. If its reduction modulo every prime number is a quasi-isomorphism, the mapping cone acyclicity criterion makes $M(f)\otimes {\mathbb{F}}_p$ acyclic. The short exact sequence of chain complexes $0\to M\stackrel{p}{\to }M\to M\otimes {\mathbb{F}}_p\to 0$ implies that multiplication by $p$ is an isomorphism on every $H_i(M)$. These are finitely generated abelian groups. Their free summands would make multiplication by $p$ fail to be surjective, and any nonzero finite summand would make it fail to be injective for a prime dividing its order. Thus $H_i(M)=0$ and $f$ is a quasi-isomorphism. Degreewise finite generation is essential to this argument.