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Fix a prime number $p$. Take the full subcategory of finitely generated abelian groups with no element of order $p^2$. It is closed under finite direct sums and subgroups. If $t_p(G)$ is the $p$-primary torsion subgroup, the map $G\to R(G)=G/(p{t}_p(G))$ is a reflection into this subcategory: every map to an allowed group kills $p{t}_p(G)$. Therefore ordinary kernels in a category and reflected cokernels in a category supply all finite categorical limits and colimits. No nonzero finitely generated group has zero reflection, so the epimorphisms are precisely surjections and are all conormal epimorphisms. Multiplication by $p^2$ on $\mathbb{Z}$ is a monomorphism but not a normal monomorphism: its cokernel here is $\mathbb{Z}/p\mathbb{Z}$, whose kernel is $p\mathbb{Z}$, not $p^2\mathbb{Z}$. Multiplication by $p$ is normal; composing it with itself gives this nonnormal monomorphism.

Module theory studies abelian groups with scalar multiplication by a ring. It extends linear algebra: a module over a field is a vector space, while ring scalars let the same language describe integer arithmetic and linear operators. The new phenomena are visible before any abstract classification: some nonzero scalars cannot be divided out, some modules have torsion elements of a module, and some have no basis of a module.

The distinction is useful because an integer relation such as $6m=0$ carries information which disappears if one assumes all nonzero scalars have multiplicative inverses. A finitely generated module over a principal ideal domain can be separated into a finite free module and a torsion module; the structure theorem for finitely generated modules over a principal ideal domain unifies the classification of finitely generated abelian groups with the study of linear operators through polynomial rings. See Keith Conrad's Modules over a PID, Theorem 4.1 and the operator applications.

A historical anchor is Emmy Noether's 1921 paper ^[ref]. Its first section studies rings and ideals under the condition that every ideal is a finitely generated module over its ring, and connects this to the ascending chain condition; later sections develop primary decomposition. This is an early source for the structural viewpoint underlying Noetherian rings. ^[ref] makes those statements accessible in English.

For an elliptic curve over an algebraically closed field, ${\mathrm{Pic}}^0(E)\cong E(k)$ via $P\mapsto [P-O]$. This group is infinite and divisible by any prime $\ell$ different from the characteristic, since $[\ell ]:E\to E$ has nonzero derivative and is a surjective morphism of projective curves. A finitely generated abelian group divisible by $\ell$ has zero free rank and is finite. Hence $E(k)$, and therefore $\mathrm{Cl}(E)$, cannot be finitely generated. This proof applies to countable algebraically closed fields as well.

A nilpotent group has lower central series ${\gamma }_{1}G=G$, ${\gamma }_{j+1}G=[G,{\gamma }_{j}G]$ terminating at the identity. A polycyclic group has a finite subnormal series with cyclic factors, which may be finite or infinite.

If a nilpotent group is generated by finitely many elements, then ${\gamma }_{j}G/{\gamma }_{j+1}G$ is generated by the finitely many weight-$j$ iterated commutators in these generators. Modulo the next term, commutator identities make these quotients abelian and permit this collection of words. Each quotient is thus a finitely generated abelian group. Refine each by a series with cyclic factors and splice the refinements through the finite lower central series. This proves that a finitely generated nilpotent group is polycyclic.

For the duality-group assertion, first prove the useful finite-index reduction. Induct on the length of a polycyclic series, taking its last proper term $H◃G$. By induction $H$ has a finite-index poly-infinite cyclic group. Because $H$ is finitely generated, it has only finitely many subgroups of any fixed finite index. Intersect all automorphic images of that subgroup to obtain a characteristic finite-index subgroup $H_0\subset H$ which is still poly-infinite cyclic. It is normal in $G$. The quotient $G/H_0$ is finite-by-cyclic. If it is finite, $H_0$ already suffices; if its cyclic quotient is infinite, the subgroup generated by a lift of its generator is infinite cyclic and has finite index. Its preimage in $G$ is an extension of $H_0$ by $\mathbb{Z}$, hence poly-infinite cyclic. This proves polycyclic groups are virtually poly-infinite cyclic.

Take such a finite-index subgroup $P$ of the given dimension-three Poincare duality group. It is again a dimension-three duality group. The extension rule proved in Question 3, applied successively to its infinite cyclic series, makes $P$ a duality group of dimension equal to its Hirsch length. Hence that length is three. Its last quotient giveswhere $H$ has a poly-infinite cyclic series of length two. Thus $H\cong \mathbb{Z}⋊\mathbb{Z}$, with the last generator acting on the first by $+1$ or $-1$. In the first case $H={\mathbb{Z}}^2$. In the second case $H$ is the Klein-bottle group $⟨a,b\mid ba{b}^{-1}=a^{-1}⟩$. Its characteristic subgroup $⟨a,b^2⟩\cong {\mathbb{Z}}^2$ is the centralizer of its commutator subgroup $⟨a^2⟩$. Replace $H$ by this characteristic index-two subgroup if necessary. It remains normal in $P$, and the quotient is finite-by-infinite-cyclic. Taking the preimage of a finite-index infinite cyclic subgroup givesThis extension splits: any lift $t$ of the cyclic generator supplies a section. Therefore $P_0={\mathbb{Z}}^2{⋊}_B\mathbb{Z}$ for some $B\in G{L}_2(\mathbb{Z})$.

In additive notation on ${\mathbb{Z}}^2$, $[t,v]=(B-I)v$. All its other internal commutators vanish, so induction givesConsequentlyThis is the nilpotence criterion for an abelian-by-cyclic group. If $B=I$ the group is abelian; otherwise it is two-step nilpotent. The criterion concerns this actual semidirect product. A finite extension of it is only known to be virtually nilpotent: for example ${\mathbb{Z}}^2{⋊}_{-I}\mathbb{Z}$ is virtually abelian but its lower central series never terminates.

Write elements of the real Heisenberg group as $(x,y,z)$. Direct matrix multiplication and inversion giveThese formulas show that $N_k=(k\mathbb{Z}{)}^3$ is a subgroup. It is discrete. For the right quotient, multiplying by $(ka,kb,kc)$ changes the coordinates toThus $(x\mathrm{mod}k,y\mathrm{mod}k)$ is well defined on the quotient. Over a small base rectangle choose real lifts $x,y$ of its coordinates. Every coset above that rectangle has a representative $(x,y,z)$, and two such representatives differ precisely by $z\mapsto z+kc$. Henceis a smooth local trivialization. The fibres are circles, proving the locally trivial fibration over ${\mathbb{R}}^2/k{\mathbb{Z}}^2\cong T^2$. Reducing first $x,y$ and then $z$ also gives a compact fundamental region, so $N_k$ is cocompact.

The ambient Lie group is diffeomorphic to ${\mathbb{R}}^3$, hence simply connected and contractible. Its quotient has fundamental group $N_k$, and the first integral homology is its abelianization. Let $a=(k,0,0)$, $b=(0,k,0)$ and $c=(0,0,k)$. Every element is uniquely $a^m{b}^n{c}^{\ell }$, since its third coordinate is $k^{2}mn+k\ell$. The commutator formula isso $c$ is central and $[a,b]=c^k$. Every commutator lies in $(0,0,k^2\mathbb{Z})$, and this subgroup is generated by $[a,b]$. Thus $N_k^{\prime }=⟨c^k⟩$. The homomorphism $(x,y,z)\mapsto (x/k,y/k,z/k\mathrm{mod}k)$ is surjective onto ${\mathbb{Z}}^2\oplus \mathbb{Z}/k$ and has exactly this kernel; the cross term in multiplication is divisible by $k$ after dividing by $k$. ThereforeFor $k=1$ the last factor is trivial. This is the Scaled Heisenberg lattice, whose fibre is central but whose total group need not split as a direct product of groups.

A preadditive category has an abelian group of morphisms between each pair of objects, and composition is additive in both variables. An additive category is a preadditive category with a zero object and finite products in a category; equivalently one may require finite coproducts. The equivalence, and the stronger assertion that these constructions agree canonically, follow from the argument below. An abelian category is an additive category with kernels in a category and cokernels in a category, in which every monomorphism is a categorical kernel and every epimorphism is a categorical cokernel. A monomorphism with the first property is a normal monomorphism; an epimorphism with the second is a conormal epimorphism.

Let $P={\prod }_{i=1}^n{A}_i$, with projections $p_i$. There is a unique $i_j:A_j\to P$ with $p_i{i}_j={\delta }_{ij}{1}_{A_j}$, interpreting the other components as zero. Since every $p_j$ sends ${\sum }_i{i}_i{p}_i$ to $p_j$, the product's universal property givesGiven arrows $f_i:A_i\to B$, put $f={\sum }_i{f}_i{p}_i$. Then $f{i}_j=f_j$. Conversely any arrow with these restrictions equals $f{1}_P={\sum }_{i}f{i}_i{p}_i$, so it is uniquely this sum. Thus $(P,i_i)$ is a coproduct in a category. The empty product is the zero object and also the empty coproduct. The dual argument starts from coproducts and constructs projections; it proves the equivalent definition. Hence finite products and coproducts are the same biproducts, with the displayed identities giving the canonical identification.

For the image factorization in an abelian category, let $h:A\to B$, let $c:B\to Q$ be its cokernel, and let $i:I\to B$ be the kernel of $c$. Since $ch=0$, write $h=ie$. We prove that $e:A\to I$ is epic. Suppose $t:I\to Z$ satisfies $te=0$ and let $l:L\to I$ be its categorical kernel. Then $e=lr$ for some $r:A\to L$. The composite $il$ is monic and hence is a normal monomorphism, say $il=\ker z$ for $z:B\to W$. Since $zh=zilr=0$, the categorical cokernel property makes $z$ factor through $c$, and therefore $zi=0$. The categorical kernel property of $il$ gives $s:I\to L$ with $ils=i$. Cancellation of $i$ yields $ls=1_I$; since $l$ is monic this also gives $sl=1_L$. Thus $l$ is invertible and $t=0$. In a preadditive category, $ue=ve$ now implies $(u-v)e=0$, hence $u=v$. This proves $e$ is an epimorphism and supplies the required factorization.

For uniqueness, an epimorphism that is a categorical cokernel is the categorical cokernel of its own categorical kernel. Indeed, if $e=\mathrm{coker}r$ and $k=\ker e$, then $r$ factors through $k$; an arrow killing $k$ therefore kills $r$ and factors uniquely through $e$. Dually, a normal monomorphism is the categorical kernel of its own categorical cokernel. If $h=mp$ is any epi–mono factorization, monicity of $m$ gives $\ker p=\ker h$. Both $p$ and $e$ are consequently categorical cokernels of this same categorical kernel. Their universal properties give a unique isomorphism $\theta$ between the intermediate objects with $\theta p=e$, and cancellation of $p$ gives $i\theta =m$. Therefore the epi–mono factorization is unique up to the unique compatible isomorphism.

Here is an additive category with truncated primary torsion, with the prime fixed as $2$. Its objects are finitely generated abelian groups having no element of order $4$, and its morphisms are all group homomorphisms. Equivalently their $2$-primary torsion subgroup has exponent at most $2$. It is a full subcategory of abelian groups, closed under finite direct sums and subgroups. Its hom groups and bilinear composition are inherited, and the zero group and finite direct sums belong to it, so it is additive. Categorical kernels are ordinary subgroup categorical kernels, which are still finitely generated and have no element of order $4$.

To construct categorical cokernels, for an arbitrary finitely generated abelian group $G$ let $t_2(G)$ be its $2$-primary torsion subgroup, and defineThe Fundamental theorem of finitely generated abelian groups shows that this quotient lies in the subcategory: it leaves the free and odd-primary parts unchanged and replaces each cyclic $2$-primary summand by either $0$ or $\mathbb{Z}/2\mathbb{Z}$. Every homomorphism $G\to H$ to an allowed group kills $2t_2(G)$, because it sends $2$-primary torsion into a subgroup killed by $2$. Thus $G\to R(G)$ has the universal property of a reflection. For $f:A\to B$, its categorical cokernel in the subcategory isThis has exactly the categorical cokernel universal property by the preceding factorization. Finite categorical limits exist because equalizers are categorical kernels of differences and finite products exist. Finite colimits exist dually because coequalizers are categorical cokernels of differences and finite coproducts exist.

The epimorphisms in this example are precisely the surjective homomorphisms. A surjection is epic already in abelian groups. Conversely, if $f$ is epic in the subcategory, its categorical cokernel is zero, so $R(B/f(A))=0$. A nonzero finitely generated abelian group has nonzero reflection: a nonzero free or odd-primary summand survives, and a nonzero finite $2$-primary summand has a nonzero quotient modulo twice itself. Hence $B/f(A)=0$ and $f$ is surjective. Its ordinary categorical kernel belongs to the subcategory, and its quotient by that categorical kernel is its codomain, which also belongs to the subcategory. It is therefore the categorical cokernel of its categorical kernel. This proves that every epimorphism is conormal.

The monomorphismis not normal. Its categorical cokernel in this category is reduction modulo $2$, since $R(\mathbb{Z}/4\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$. The categorical kernel of that categorical cokernel is the inclusion $2\mathbb{Z}\hookrightarrow \mathbb{Z}$, whereas the image of $m$ is $4\mathbb{Z}$. A normal monomorphism is the categorical kernel of its own categorical cokernel, as proved above, so $m$ cannot be normal. In fact multiplication by $2$ is itself the categorical kernel of reduction modulo $2$, up to the identification $\mathbb{Z}\cong 2\mathbb{Z}$. Its composite with itself is multiplication by $4$. Thus this example already exhibits two normal monomorphisms whose composite is not normal.

For the general assertion, let $m:A\to B$ be a nonnormal monomorphism in any additive category of the specified kind. Put $c=\mathrm{coker}m$ and $k:K\to B$ be its categorical kernel, and factor $m=kg$. Both $m$ and $k$ are monic, so $g$ is monic. Set $d=\mathrm{coker}g$ and $l:L\to K$ be its categorical kernel. Then $k,l$ are normal, and $g=lr$ for some $r$.

Suppose their composite were normal, say $kl=\ker t$. Since $tm=tklr=0$, $t$ factors through $c$, whence $tk=0$. The categorical kernel property of $kl$ supplies $s:K\to L$ with $kls=k$; cancellation shows $ls=1_K$, and monicity makes $l$ an isomorphism. Because $dl=0$, this forces $d=0$. In a preadditive category, vanishing of the categorical cokernel of $g$ implies that $g$ is epic: any $u,v$ with $ug=vg$ have $u-v$ factor through the zero categorical cokernel. By hypothesis $g$ is therefore a categorical cokernel, and it is also monic. If $g=\mathrm{coker}a$, monicity and $ga=0$ force $a=0$; a categorical cokernel of a zero arrow is an isomorphism. Thus $g$ is invertible, making $m=kg$ normal, a contradiction. Consequently normal monomorphisms cannot be closed under composition in any such category; the explicit pair $k,l$ witnesses the failure.

Use additive notation for this finitely generated abelian group. The change of generators $A=a+3d$, $e=c-b$, together with $b,d$, is invertible over the integers: $a=A-3d$ and $c=b+e$. The second relation gives $2e=0$. Subtracting it from the first gives $4A=0$, while the third becomes $4b+2e=0$, hence $4b=0$. The last relation becomes $8A+4b+4e=0$ and is consequently redundant. Conversely $4A=4b=2e=0$ imply every original relation, so no further relation has been introduced or lost. Thus the cyclic group decomposition isThe infinite cyclic group factor contributes only zero to an element killed by two. Each finite factor has two such elements, so the kernel of multiplication by two has $2^3=8$ elements. Removing the identity leavesThe invertible generator change also supplies a direct presentation proof of the decomposition, without relying solely on a Smith normal form computation.

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.

An elliptic curve works over every algebraically closed field, including one algebraic over a finite field. Give it explicitly in ${\mathbb{P}}^2$ byIn the first case the only point at infinity is $O=[0:1:0]$, where the derivative with respect to $z$ is nonzero. In the affine chart $z=1$, a singular point of an algebraic variety would have $y=0$, $x^3-x=0$, and $3x^2-1=0$. The possible roots $x=0,1,-1$ cannot satisfy the last equation in characteristic different from $2$: at $0$ it is $-1$, and at $\pm 1$ it is $2$. In the second case the three homogeneous partial derivatives are $x^2,z^2,y^2$, up to harmless signs, and cannot vanish simultaneously at a projective point. Thus both cubics are smooth varieties. A reducible plane cubic would have intersecting positive-degree components by Bézout theorem, and be singular at an intersection; therefore these smooth cubics are irreducible varieties. The genus-degree formula gives $g=(3-1)(3-2)/2=1$. With the marked point $O$ they are elliptic curves.

For a smooth algebraic curve, its divisor class group is its Picard group. Its degree-zero subgroup isThe map $P\mapsto [P-O]$ gives an isomorphism of abelian groupsTo recall why it is a bijection, Riemann-Roch theorem in geometric genus one says every degree-one divisor class has exactly one independent section: a canonical divisor has degree zero, so the dual degree-$-1$ term has no section. Thus the class contains one effective divisor on an algebraic curve of degree one, namely a point $P$, and that point is unique. Translating by $-O$ gives the claimed bijection; this is precisely the chord-and-tangent group law on the elliptic curve.

Choose a prime integer $\ell$ different from $\mathrm{char}k$. The multiplication-by-n morphism $[\ell ]:E\to E$ has derivative $\ell$ times the identity at $O$, so it is nonconstant. A nonconstant morphism of algebraic varieties between smooth projective curves is surjective: its image is closed by proper morphism and cannot have algebraic dimension zero. Since $k$ is an algebraically closed field, this makes $[\ell ]$ surjective on $k$-points. ThereforeYet $E(k)$ is infinite. For each $x\in k$, the affine equation has a solution for $y$ in the algebraically closed field; since $k$ is infinite, this gives infinitely many points.

By the Fundamental theorem of finitely generated abelian groups, a finitely generated abelian group $A$ satisfying $A=\ell A$ has no free part and is a finite torsion group of order prime to $\ell$. The infinite group $E(k)$ cannot have this property while being finitely generated. Thus ${\mathrm{Pic}}^0(E)$ is not finitely generated. A subgroup of a finitely generated abelian group is finitely generated, soThis proof uses divisibility and infinitude, rather than uncountability, so it remains valid over countable algebraically closed fields.

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.

TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

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.

For connected oriented closed manifolds of dimension $n$, the degree of a map between oriented manifolds is the integer determined bywhere the brackets are the chosen fundamental classes. Functoriality gives $\mathrm{deg}(g\circ f)=\mathrm{deg}g\mathrm{deg}f$.

Give $S^n$ its boundary orientation as the boundary of the unit ball in ${\mathbb{R}}^{n+1}$. The antipodal map is the restriction of $-I$. Its ambient determinant is $(-1{)}^{n+1}$, and it takes the outward normal at $x$ to the outward normal at $-x$. Thus its effect on the boundary orientation has this same sign, givingThe original PDF has $A:S^n\to S^n$, correcting the extra prime on the target in the TeX transcription.

For even dimension $n=2k\ge 2$, cellular homology of real projective space gives $H_n({\mathbb{RP}}^n;\mathbb{Z})=0$. Thus the composite map on top homology factors through zero, andZero is attained by constant maps.

For odd dimension $n=2k+1\ge 3$, Real projective space is orientable: the antipodal deck transformation has degree $+1$. Orient it so that the double covering map $q:S^n\to {\mathbb{RP}}^n$ has degree $2$. Since $S^n$ is simply connected, the lifting criterion for a covering space gives $f=q\circ \tilde{f}$ for a map $\tilde{f}:S^n\to S^n$. Hencewhich is even. Every even integer occurs: collapse the complement of an oriented embedded disk in ${\mathbb{RP}}^n$ to obtain a degree-one map $g:{\mathbb{RP}}^n\to S^n$, and choose $\tilde{f}$ of any prescribed integer degree $m$. Maps of all integer degrees on spheres are obtained, for example, by suspending the circle maps $z\mapsto z^m$. Setting $f=q\circ \tilde{f}$ gives degree $2m$. ThusThese are the degrees of maps factoring through real projective space. If $k=0$ is permitted in the odd-dimensional clause, there is an exception: ${\mathbb{RP}}^1\cong S^1$, so every integer degree occurs. The lifting argument requires dimension at least two. The initial connected-manifold degree definition excludes $S^0$; with the usual reduced-homology definition for self-maps of $S^0$, a factorization through the one-point ${\mathbb{RP}}^0$ has degree zero.

Finally suppose $\mathrm{deg}f=p$ is prime. Fix any prime $q\ne p$ and work over $F={\mathbb{F}}_q$. We claim that $f^{\ast }:H^r(N;F)\to H^r(S^n;F)$ is injective. If $a\ne 0$, Poincare duality supplies $b\in H^{n-r}(N;F)$ with $⟨a⌣b,[N{]}_F⟩\ne 0$. Naturality of the cup product and evaluation givesbecause $p$ is invertible in $F$. Hence $f^{\ast }a\ne 0$, proving cohomological injectivity of a map of invertible degree.

The intermediate cohomology of the sphere is zero, so $H^r(N;F)=0$ for $0<r<n$. Over a field, the universal coefficient theorem for cohomology identifies this with the dual of $H_r(N;F)$, so these homology groups vanish as well. The universal coefficient theorem for homology then givesEach integral group is a finitely generated abelian group, by the permitted finite CW complex model. Its decomposition can have no free summand, since that would survive modulo $q$, and no torsion summand divisible by any prime $q\ne p$. Thus it is a finite $p$-primary group. There are only finitely many intermediate degrees, so the exponents of their finite cyclic summands have a common bound $p^a$. Taking $a\ge 1$ also covers the case in which all the groups vanish. ThereforeThis proves that a prime-degree sphere map forces primary torsion, including one uniform exponent for all the degrees.

Present a finitely generated abelian group $G$ as the cokernel of an integer matrix $A:{\mathbb{Z}}^m\to {\mathbb{Z}}^n$. Changing $A$ by the invertible row and column operations used in its Smith normal form does not change the isomorphism type of the cokernel. Ifthenwhere factors with $d_i=1$ may be omitted. This is the Fundamental theorem of finitely generated abelian groups; the rank of an abelian group $r$ and invariant factors $d_i>1$ are unique.

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.

For a number field $K$ and a finite set $S$ of finite primes, an S-unit is an element of $K^{\times }$ whose discrete valuation is zero at every prime outside $S$. Its group is the unit group of the ring obtained by allowing denominators at $S$. The Dirichlet unit theorem and the valuation homomorphism to ${\mathbb{Z}}^S$ show that this group is a finitely generated abelian group.