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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.

Write $v$ for the cone vertex. The quotient is the topological mapping cone $C_f=Y{\cup }_{f}CX$. First assume $X\ne \varnothing$, as is implicit in having this vertex; a map then also forces $Y\ne \varnothing$.

Take the open cover consisting of the image of $X\times [0,2/3)$ and the image of $Y\bigsqcup (X\times (1/3,1])$. Denote these sets by $U,V$. The first contracts to $v$, the second has a deformation retraction onto $Y$, and $U\cap V\cong X\times (1/3,2/3)$ retracts onto $X$. Under these identifications the two inclusion maps induce the zero map into positive-degree homology of $U$ and $f_{\ast }$ into the homology of $V$. The Mayer–Vietoris theorem consequently gives, in positive degrees,The same assertion holds in degree zero, but needs care: the ordinary Mayer-Vietoris map isSince ${\epsilon }_Y{f}_{\ast }a={\epsilon }_{X}a$, its kernel equals $\ker f_{\ast }$. On quotienting $H_0(Z;G)$ by the direct summand $G[v]$, its cokernel becomes $\mathrm{coker}{f}_{\ast }$. That quotient is canonically identified with reduced homology by subtracting the augmentation times $[v]$. Thus the last nonzero part isIn particular the map from $H_0(Y;G)$ sends a component class $[y]$ to $[y]-[v]$. This proves the mapping cone exact sequence; finite generation of $G$ is not needed for this construction.

There is a literal empty-space exception in the printed hypotheses. If $X=\varnothing$ and $Y$ is a point, the given quotient is just $Y$, and the asserted degree-zero exact sequence would be $0\to G\to 0$. Thus the displayed reduced sequence presupposes nonempty $X$. It holds also for empty $X$ if one defines its cone to include a new isolated vertex. The later homology-isomorphism claim itself remains true for empty finite complexes, with the cases involving one empty complex dealt with directly in degree zero.

Now take nonempty finite CW complexes $X,Y$. Put $A_q={\tilde{H}}_q(Z;\mathbb{Z})$. The exact sequence giveswhere $H_{-1}=0$. All $A_q$ are finitely generated abelian groups, since the groups for $X,Y$ are finitely generated. Moreover, by exactness,The reduced universal coefficient theorem for homology givesIf all $A_q$ vanish, so does the middle term for every prime, and the exact sequence with coefficients proves that $f_{\ast }$ is an isomorphism modulo every prime. Conversely, those isomorphisms imply that the middle terms vanish, hence $A_q\otimes {\mathbb{F}}_p=0$ for every prime $p$. By the Fundamental theorem of finitely generated abelian groups, a nonzero free summand would survive modulo every prime, and a cyclic torsion summand would survive modulo a prime dividing its order. Thus $A_q=0$ for every $q$, provingThis argument uses the exact sequence to obtain finite generation; it does not assume in advance that the cone of an arbitrary continuous map is a finite CW complex.

Finally suppose $f$ is a cellular map. Keep the cells of $Y$, add the vertex $v$, and add a $(q+1)$-cell for the cone on every $q$-cell of $X$. Its top face attaches to $Y$ through $f$, and its other faces attach to cones on lower-dimensional cells. The cellular condition puts these attachments in the appropriate skeleton. This constructs a finite CW complex structure on $Z$.

Let $A_{\ast }=C_{\ast }^{\mathrm{cell}}(X)$ and $B_{\ast }=C_{\ast }^{\mathrm{cell}}(Y)$ be the ordinary cellular chain complexes, with $A_{-1}=0$. Orient each coned cell so that its boundary has the form $f_{\#}(e)-c(d_{A}e)$. Using $[y]-[v]$ as the reduced generator for each vertex of $Y$ givesIn particular ${\tilde{C}}_0^{\mathrm{cell}}(Z)=B_0$, and the boundary of a cone edge is its image vertex minus $v$. The chain map identity $d_B{f}_{\#}=f_{\#}{d}_A$ givesThis is the cellular chain complex of a mapping cone, equivalently the algebraic mapping cone with the displayed summand and sign conventions.

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.

For a positively oriented local parametrization $X(u,v)$ with first fundamental formthe area element of a surface isUnder an orientation-preserving coordinate change, the Jacobian from $du\wedge dv$ cancels the inverse Jacobian in the square root of the metric determinant. Hence these local expressions agree and define a global two-form.

The Euler characteristic may be defined from any finite triangulation byor equivalently by the alternating sum of the dimensions of the rational homology groups. Subdivision leaves $V-E+F$ unchanged, and the homological formula shows that it is a topological invariant, so the definition does not depend on the triangulation.

Give the boundary its induced orientation and parametrize it by arc length. If $T$ is its unit tangent and $N$ is the chosen unit normal to the surface, its signed geodesic curvature iswhere $D_s$ is the surface covariant derivative. The Gauss-Bonnet theorem for a compact oriented surface with smooth boundary and no corners iswhere $K$ is the Gaussian curvature.

The Schur multiplier of a group $G$ isthe second group homology group with trivial integral coefficients. If $G\cong F/R$ is a free presentation, Hopf's formula states that

Write $I_F=\ker (\mathbb{Z}F\to \mathbb{Z})$ for the augmentation ideal. The presentation relation sequence iswhere $r[R,R]\mapsto 1\otimes (r-1)$. If $F$ is free on a set $S$, then $I_F$ is free as a left $\mathbb{Z}F$-module on the elements $s-1$, so the two modules immediately preceding $\mathbb{Z}$ are free $\mathbb{Z}G$-modules. Resolving the relation module $R/[R,R]$ by free modules and splicing produces a free resolution of $\mathbb{Z}$.

Apply the right-exact functor $\mathbb{Z}{\otimes }_{\mathbb{Z}G}-$ to this partial resolution. Its degree-two homology is the kernel ofThe coinvariant module on the left is $R/[F,R]$. On the right, the map $f-1\mapsto f[F,F]$ identifies the coinvariants with the abelianization $F/[F,F]$. The displayed map is induced by the inclusion $R\hookrightarrow F$, so its kernel isThis proves Hopf's formula.

For an abelian group $A$, the Schur multiplier of an abelian group is $M(A)\cong {\bigwedge }^{2}A$. One way to see the direct-sum rule is the degree-two Künneth theorem:A cyclic group has zero second integral group homology, whileConsequently

Choose an orientation for each simplex. The chain group $C_k$ is the free abelian group on the oriented $k$-simplices, with orientation reversal changing sign. On an ordered simplex setIts face maps respect the identifications in a Delta-complex. In ${\partial }_{k-1}{\partial }_k$, deleting any pair of vertices occurs in the two possible orders with signs $(-1{)}^{i+j-1}$ and $(-1{)}^{i+j}$, which cancel. Hence ${\partial }^2=0$. The groups of homological cycles and boundaries are $Z_k=\ker {\partial }_k$ and $B_k=\mathrm{im}{\partial }_{k+1}$; the identity just proved gives $B_k\subseteq Z_k$, so $H_k=Z_k/B_k$ is defined. Mod-two homology uses the same chain complex with coefficients reduced modulo two.

Here the edge identifications put all six vertices into one vertex class. Let the three boundary edge classes be $a,b,c$ in cyclic order, and orient the internal diagonals as $d=[v_1,v_3]$, $e=[v_3,v_5]$, $f=[v_5,v_1]$. With the four face orientations listed in the source, the boundaries areThus $C_2={\mathbb{Z}}^4$, $C_1={\mathbb{Z}}^6$, $C_0=\mathbb{Z}$, and ${\partial }_1=0$. The Euler characteristic isThe face-boundary map is injective over the integers: coefficients of $a,b,c$ force the first three face coefficients to vanish, after which the diagonal coefficients force the fourth to vanish. Its cokernel eliminates $d,e,f$ and leaves generators $a,b,c$ with the single relation $2(a+b+c)=0$. Since $a+b+c$ is primitive in ${\mathbb{Z}}^3$, this cokernel is ${\mathbb{Z}}^2\oplus \mathbb{Z}/2$. ThereforeModulo two the first three face-boundary columns are $d,e,f$, and the last is their sum. The rank is three, the kernel is one-dimensional, and the edge quotient has dimension three. ConsequentlyThe top mod-two class is the sum of all four faces. Integral homology has no such top class, reflecting nonorientability.

For an open cover $X=U\cup V$, the Mayer–Vietoris theorem gives the long exact sequenceIt ends in $H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to H_0(X)\to 0$. To prove it, let $C_{\ast }^{U,V}(X)$ be the subcomplex generated by singular simplices lying wholly in $U$ or wholly in $V$. There is a short exact sequence of chain complexesExactness follows from the singular-simplex bases: the intersection of the two chain subgroups consists exactly of simplices lying in both sets. The small-chain locality theorem for singular homology says the inclusion of this small-chain complex into $C_{\ast }(X)$ induces an isomorphism on homology; barycentric subdivision is its standard justification. The long exact sequence of a short exact sequence of chain complexes now proves the theorem. For a homological cycle represented as $a+b$, the connecting map is $[a+b]\mapsto [\partial a]$, since $\partial a=-\partial b$ lies in the intersection. This also identifies the displayed maps and their signs.

For the glued solid tori, use open collar enlargements of the two halves. Each retracts to a circle; the overlap retracts to their boundary torus. Thus the halves have $H_0=H_1=\mathbb{Z}$ and higher homology zero, while the overlap has $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^2$. In the overlap choose longitude and meridian generators. The meridian bounds a disk in each solid torus and the longitude generates each first homology, so the map on $H_1$ isIts kernel is the meridian copy of $\mathbb{Z}$ and its cokernel is $\mathbb{Z}$. The map on $H_0$ is the injective map $z\mapsto (z,-z)$. Exactness therefore gives $H_2(X)=\mathbb{Z}$ and $H_1(X)=\mathbb{Z}$. The preceding segment is $0\to H_3(X)\to H_2(U\cap V)\to 0$, giving $H_3(X)=\mathbb{Z}$. Connectedness gives $H_0(X)=\mathbb{Z}$, and all higher groups vanish. HenceAs a geometric check, the identity boundary gluing identifies the product with $S^1\times (D^2{\cup }_{S^1}{D}^2)=S^1\times S^2$.

Yes: for $n\ge 1$, the sphere $S^n$ is connected, closed and oriented, with $H_0=H_n=\mathbb{Z}$ and no intermediate homology. These groups do not require the manifold to be homeomorphic to a sphere, but the sphere provides the requested example.

No. The dimension would again be three. The universal coefficient theorem gives $H^1=0$, since both $\mathrm{Hom}(H_1,\mathbb{Z})$ and $\mathrm{Ext}(H_0,\mathbb{Z})$ vanish. Poincare duality then forces $H_2=0$, contradicting the proposed torsion group. In general the second integral homology of a connected closed oriented three-manifold is free abelian, being dual to its first homology.

Yes: real projective three-space ${\mathbb{RP}}^3$ has these groups. Its usual cell structure has one cell in each dimension zero through three, with boundary maps ${\partial }_1=0$, ${\partial }_2=2$ and ${\partial }_3=0$. Hence its integral homology is $\mathbb{Z}$ in degrees zero and three, $\mathbb{Z}/2$ in degree one, and zero in degree two. It is a connected closed oriented manifold: the antipodal map on $S^3$ preserves orientation, so the sphere's orientation descends to the quotient.

A Morse function is perfect over a specified field when its numbers of critical points equal the Betti numbers over that field. Over the rationals this forces every integer matrix in its Morse chain complex to have rank zero, hence to vanish. Consequently its integral homology is free of rank $c_i$ in every degree; a manifold with integral homology torsion cannot admit such an integral-critical-count perfect function.

For a closed oriented $d$-dimensional manifold over a field $F$, the cup product pairingis a perfect pairing between complementary degrees. The cap product with the fundamental class is an isomorphism by Poincare duality, and the universal coefficient theorem for cohomology over $F$ identifies the complementary cohomology with the full dual space of the resulting homology. Evaluation is therefore a perfect pairing. Taking the top-degree part of the same formula defines a nondegenerate bilinear form on the full graded cohomology. A boundary requires relative groups and Poincare-Lefschetz duality instead.

For a space with finite-dimensional homology over a specified field and only finitely many nonzero Betti numbers, this polynomial records their graded dimensions. Its value at one is the total Betti number, and its value at minus one is the Euler characteristic. The Künneth theorem gives $P_{M\times N}(t)=P_M(t)P_N(t)$ over a field. For integer homology ranks, use the rational Betti numbers.

Apply the homology cross product and then the map induced by loop concatenation. This degree-additive bilinear product supplies the Pontryagin ring. The Bott–Samelson theorem computes it for looped suspensions with free homology.

The homology ring formed from loop concatenation and the homology cross product. Homotopy associativity is enough for associativity in homology, and the constant loop is the unit. More generally the same construction applies to a homotopy-associative H-space.

For $n\ge 1$, the Bott–Samelson theorem gives a single homology generator of degree $2n$ with no word relations. Its Pontryagin product satisfies $x^i\ast x^j=x^{i+j}$. Unlike this polynomial homology product, the cohomology cup product has divided-power coefficients.

A positive-degree class whose reduced homology diagonal vanishes. In a torsion-free homology Hopf algebra, powers of an even primitive class have binomial diagonals. Dualizing that diagonal produces the multiplication of a divided power algebra.

Study homotopy after killing torsion, especially for simply connected spaces of finite type. The rational Whitehead theorem detects rational homotopy equivalences by rational homology, while Sullivan minimal models encode rational homotopy generators and cohomological relations.

For maps between simply connected spaces, a rational homology equivalence is a rational homotopy equivalence. It follows from the relative Hurewicz theorem modulo torsion, or from rationalization and the homological Whitehead theorem.

For a based finite acyclic chain complex, Reidemeister torsion is an alternating product of determinants comparing each given chain basis with bases formed from boundaries and lifts of boundaries. With suitable local coefficients this gives an invariant of a space finer than its homology. Analytic torsion constructs a corresponding invariant from spectra of Laplacian operators, while Turaev torsion refines combinatorial sign and normalization choices.