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For a CW complex, the qth cellular chain group is the relative homology group of successive skeleta, a free abelian group with one generator for each q-cell. The attaching maps determine the cellular boundary, and the resulting cellular chain complex computes singular homology.

For a compact torus $T\cong U(1{)}^r$, its continuous one-dimensional characters form a free abelian group of rank $r$. They are precisely $z\mapsto z_1^{m_1}\cdots z_r^{m_r}$ with integer exponents. This follows from the classification of representations of the circle group, applied to each factor. These characters are the integral group weights occurring upon restricting a group representation to a maximal torus; global integrality must be distinguished from integrality against coroots for a simply connected semisimple covering group.

Every degree-$n$ number field has a ring of integers that is a free abelian group of rank $n$. To prove this without assuming finite generation, scale a rational basis to algebraic integers ${\alpha }_1,\dots ,{\alpha }_n$ and set $M={\sum }_j\mathbb{Z}{\alpha }_j$. Their trace pairing has an invertible integer matrix $T$, since the trace form of a number field is nondegenerate. For $x\in {\mathcal{O}}_K$, all $\mathrm{Tr}(x{\alpha }_j)$ are integers. The adjugate matrix formula gives $M\subseteq {\mathcal{O}}_K\subseteq (\det T{)}^{-1}M$. The submodule theorem for free modules over a principal ideal domain makes the middle group free of rank $n$, and a group basis is an integral basis.

An integral quadratic lattice is a finite-rank free abelian group with a nondegenerate integral symmetric bilinear form. Its norm convention here is $q(x)=B(x,x)$. A basis gives an integral Gram matrix; basis changes preserve its determinant up to multiplication by the square of a unit.

A toric lattice is a finite-rank free abelian group $N$, embedded as a discrete full-rank subgroup of $N_{\mathbb{R}}=N{\otimes }_{\mathbb{Z}}\mathbb{R}$. Its dual is $M=\mathrm{Hom}(N,\mathbb{Z})$. This lattice of integral vectors is distinct from an order-theoretic lattice. It supplies the integral structure for toric cones and their dual monomials.

For a number field $K$ of degree of a field extension $n$, an integral basis is a basis ${\omega }_1,\dots ,{\omega }_n$ over $\mathbb{Q}$ such thatThus every algebraic integer in $K$ has a unique expression with integer coefficients in this basis. Merely requiring its entries to be algebraic integers would not suffice: for example, $1,2\sqrt{2}$ is a rational basis of $\mathbb{Q}(\sqrt{2})$ but misses the algebraic integer $\sqrt{2}$ in its integer span.

Here is a proof of the existence of an integral basis. Start with any rational basis $v_1,\dots ,v_n$. Each $v_j$ has a nonzero integer multiple ${\alpha }_j$ that is an algebraic integer. Indeed, ifchoose a positive integer $q$ clearing all the denominators. Then $qv$ satisfies the monic polynomialThe ${\alpha }_j$ are still a rational basis. Put $M={\sum }_j\mathbb{Z}{\alpha }_j$ and form the matrixIts entries are integers, because products and sums of algebraic integers are algebraic integers and a rational algebraic integer is an integer. The nondegeneracy of the trace form of a number field gives $D\ne 0$: if $x\ne 0$, then $\mathrm{Tr}(x{x}^{-1})=n\ne 0$.

For $x={\sum }_j{x}_j{\alpha }_j\in {\mathcal{O}}_K$, each $\mathrm{Tr}(x{\alpha }_i)$ is an integer, soThe adjugate matrix formula for $T^{-1}$ gives $D{x}_j\in \mathbb{Z}$. ConsequentlyThe submodule theorem for free modules over a principal ideal domain says that every subgroup of the finite-rank free abelian group $D^{-1}M$ is again a free abelian group of rank at most $n$. Since ${\mathcal{O}}_K$ contains $M$, its rank is exactly $n$. A basis of this free abelian group is the required integral basis.

Write $R={\mathcal{O}}_K$. We prove unique factorization of ideals in a number field, rather than assuming this conclusion as a property of a Dedekind domain. Three preliminary facts about $R$ will be useful. Its additive group is a finite-rank free abelian group, so every ideal is finitely generated and the ascending chain condition holds. It is an integrally closed domain, by integral dependence is transitive. Finally, every nonzero prime ideal is a maximal ideal. For this last fact, choose $0\ne a\in P$. Multiplication by $a$ is an injective endomorphism of the finite-rank free abelian group $R$ with nonzero determinant, so $R/(a)$ is finite. Consequently $R/P$ is a finite integral domain and hence a field.

First, every nonzero proper ideal contains a product of nonzero prime ideals. Otherwise choose a maximal counterexample $J$ under inclusion, using the ascending chain condition. It cannot be a prime ideal. Choose $x,y\notin J$ with $xy\in J$. The strictly larger ideals $J+(x)$ and $J+(y)$ each contain a product of nonzero prime ideals, allowing the empty product $R$ when necessary. Multiplying those products gives a product contained ina contradiction.

Next fix a nonzero prime ideal $P$, choose $0\ne a\in P$, and choose a product of nonzero prime idealswith $r$ minimal. Because $(a)\subseteq P$, primality and maximality imply $P_j=P$ for some $j$; reorder so that $P_1=P$. Minimality provides $b\in P_2\cdots P_r$ with $b\notin (a)$, taking the remaining product to be $R$ if $r=1$. ThusDefine the fractional idealIt contains $R$ and $x$. The ideal $P{P}^{-1}$ lies between $P$ and $R$, so it is $P$ or $R$. If it were $P$, then $xP\subseteq P$. Applying the determinant trick to a finite generating set of an ideal for $P$ gives a monic polynomial over $R$ satisfied by $x$. Since $R$ is an integrally closed domain, this would put $x$ in $R$, a contradiction. ThereforeThis proves that each nonzero prime ideal is an invertible fractional ideal, independently of the desired factorization theorem.

Now take a nonzero proper ideal $A$ and a maximal ideal $P$ containing it. The ideal $B=P^{-1}A$ satisfiesThe first inclusion is strict: if $A=B$, the same $x\in P^{-1}\setminus R$ would stabilize $A$, and the determinant trick would again force $x\in R$. Repeating gives a strictly ascending chain until it reaches $R$; the ascending chain condition ensures this happens after finitely many steps. The identities $A=PB$ then express $A$ as a product of nonzero prime ideals.

For uniqueness, suppose $P_1\cdots P_r=Q_1\cdots Q_s$. Since the right-hand side is contained in $P_1$, some $Q_j$ is contained in $P_1$. Both are nonzero maximal ideals, so $Q_j=P_1$. Multiplication by $P_1^{-1}$ cancels this factor. Repeat. A nonempty product of proper prime ideals cannot equal $R$, so both products finish together. This proves uniqueness up to reordering, including repeated factors.

The group $G(K)$ consists of the nonvanishing continuous functions, and $E(K)$ is a subgroup because multiplication is commutative: $e^g{e}^h=e^{g+h}$. The objective is a logarithm decomposition with finitely many integer obstructions, one for each bounded complementary component. No local connectivity of $K$ is assumed.

We use two elementary finite-neighborhood facts, which make the outline independent of the general Jordan theorem. Given a neighborhood $V$ of $K$, take a finite grid neighborhood $N\subset V$ with $K\subset \mathrm{int}N$. It can be chosen to be a finite union of polygonal surfaces with boundary. Its complement has finitely many bounded holes. Holes that lie in the same component of $\mathbb{C}\setminus K$ can be joined by thin polygonal corridors disjoint from $K$, which are removed from $N$. A hole lying in the unbounded component is similarly joined to the exterior. For the remaining finitely many components, corridors may be continued to include their chosen representatives $a_i$. Small polygonal adjustments preserve $K\subset \mathrm{int}N$ and $N\subset V$. Thus every remaining bounded hole lies in a distinct bounded component of $\mathbb{C}\setminus K$ and contains its representative. One may also arrange this for any prescribed finite list of bounded components by first keeping their representative points out of the grid neighborhood. All paths used are compact and disjoint from $K$, so sufficiently thin corridors remain disjoint from $K$.

For such a finite polygonal $N$, winding indices identify the homology group $H_1(N;\mathbb{Z})$ with the free abelian group on its bounded holes. Here is a finite-chain explanation. Triangulate a large disc with $N$ as a subcomplex. Every integral one-cycle bounds a two-chain in the disc. Its filling coefficient is constant on each component of the complement of $N$, is zero on the exterior, and is the winding index there. If all bounded-hole indices vanish, the filling is supported in $N$, so the cycle already bounds in $N$. Conversely the sum of the oriented faces in one hole has its chain boundary in $N$ and has index one in that hole and zero in the others. Thus cycles ${\gamma }_j$ can be chosen with $\mathrm{wind}({\gamma }_j,a_i)={\delta }_{ij}$. These can be integral sums of boundary curves, which matters when components of $N$ are nested. This proves the needed statement using finite triangulation, not separation of an arbitrary simple closed curve. Together these are finite-hole neighborhoods of a planar compact set.

Now extend $f\in G(K)$ to a continuous $F$ near $K$, using the Tietze extension theorem on its real and imaginary parts. Since ${\min }_K|f|>0$, shrink the neighborhood until $F$ has no zeros, and choose $N$ within it as above. The winding of $F$ on the cycles ${\gamma }_j$ defines integers $k_j$. Setwhere only the finitely many represented holes are used. The winding of $F/r$ on every cycle of $N$ is zero. On each component choose a local complex logarithm at a base point and continue it along paths. Zero winding number around every loop makes the continued value independent of the path; local logarithm charts make it continuous. The continuous logarithm lifting criterion therefore gives $F/r=e^g$ on $N$, and restriction gives $f=e^{g{|}_K}r{|}_K$. Thus $G(K)=E(K)G(\mathbf{a})$, even when $K$ has infinitely many holes: each individual function uses finitely many factors.

To prove uniqueness, suppose a finite product $r=\prod (z-a_i{)}^{k_i}$ equals $e^g$ on $K$. Extend $g$ continuously to a neighborhood. There $r{e}^{-g}$ is close to $1$ after shrinking, so it has its small-branch logarithm. Hence $r$ itself has a continuous function giving a complex logarithm on that neighborhood. Choose $N$ there detecting all the finitely many $a_i$ as separate holes. Its dual cycles give $\mathrm{wind}(r\circ {\gamma }_j,0)=k_j$, while a function with a logarithm has zero winding number. Therefore every $k_j=0$. This proves both freeness of the listed rational functions and $E(K)\cap G(\mathbf{a})=\{1\}$. All subgroups commute, soThis is planar logarithm factorization. If there are no bounded components the second factor is trivial.

For the Jordan curve deduction, let $J$ be homeomorphic to the circle. Parametrize it by $\gamma :[0,1]\to J$ with just the two endpoints identified. For $f\in G(J)$, lift a logarithm of $f\circ \gamma$ along the interval. Its endpoint increment is $2\pi in$. The integer $n$ is additive, every integer is attained by a function with $f(\gamma (t))=e^{2\pi int}$, and $n=0$ precisely when the lift descends to a logarithm on $J$. Hence $G(J)/E(J)\cong \mathbb{Z}$. The factorization says this is free abelian on the bounded complementary components; comparison of ranks, or tensoring with $\mathbb{Q}$, shows that exactly one bounded component exists. There is also the unique unbounded component, so the complement has exactly two components.

We still need the common-boundary assertion. A proper closed arc has trivial exponential quotient because every nonvanishing function on an interval admits a logarithm. Factorization consequently makes the complement of such an arc connected, and since it is open it is path-connected. Fix $p\in J$ and any neighborhood $V$ of $p$, and choose a small open arc $A\subset J\cap V$ containing $p$. The remainder $J\setminus A$ is a proper closed arc. Join a point in each of the two complementary components by a path avoiding $J\setminus A$. This path must meet $J$, and every such meeting lies in $A$. The segment before its first meeting lies in the first region, and the segment after its last meeting lies in the second; continuity makes both regions meet $V$. Thus every point of $J$ lies on both boundaries. Conversely either boundary is contained in the closed set $J$. There are exactly two regions, one bounded and one unbounded, with common boundary $J$, the Jordan curve theorem. This is Jordan separation from the exponential quotient; it does not assume the theorem in the factorization proof.

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.

A proper finite-index subgroup of a torsion-free group cannot be a direct summand when the complement is a subgroup of that same free abelian group. Indeed, if $L=M\oplus K$, every $k\in K$ has $18k\in M$ because the quotient has order eighteen. But also $18k\in K$, so $18k=0$ by the trivial intersection. The ambient group ${\mathbb{Z}}^3$ is torsion-free, forcing $k=0$ and hence $K=0$. This would give $M=L$, contrary to index eighteen. Thus $M$ is not a direct summand.

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.

Let $F=\mathbb{Z}{e}_1\oplus \mathbb{Z}{e}_2$ and $N=\mathbb{Z}(6e_1+9e_2)$. The precise presented group is $A=F/N$, with $x=e_1+N$ and $y=e_2+N$. It is abelian and has exactly the imposed additive relation and its integer multiples.

Change the integral basis to $u=2e_1+3e_2$, $v=e_1+2e_2$. The change-of-basis determinant is $2\cdot 2-3\cdot 1=1$, so these are a basis of the free abelian group. Explicitly, $e_1=2u-3v$ and $e_2=-u+2v$. The relation generator becomes $6e_1+9e_2=3u$. ThereforeThe image of $u$ is nonzero, since $(2,3)$ is not an integer multiple of $(6,9)$, but three times it vanishes. Thus $A$ has an element of exact order three. A free abelian group has no nonzero finite-order element, so $A$ is not free abelian.

Fix a finite-rank free abelian group $H_{\mathbb{Z}}$. A pure integral Hodge structure of weight $w$ is a decompositionIts Hodge numbers are $h^{p,q}={\dim }_{\mathbb{C}}{H}^{p,q}$, and its Hodge filtration is $F^p={⨁}_{r\ge p}{H}^{r,w-r}$. Equivalently, the filtration has these dimensions and satisfiesThe decomposition is recovered as $H^{p,q}=F^p\cap \overline{F^q}$.

A polarized Hodge structure additionally has a nondegenerate integral bilinear form $Q$ satisfying $Q(u,v)=(-1{)}^{w}Q(v,u)$, extended complex-bilinearly, withWe choose the geometric sign convention in which the positivity condition isThe expression is real by the symmetry and conjugation conditions. In particular it is $iQ(v,\bar{v})>0$ on a weight-one $(1,0)$ summand. Other conventions absorb the constant factor $(-1{)}^{w(w-1)/2}$ into $Q$; a consistent choice matters in weight two.

Fix $H_{\mathbb{Z}}$, $Q$ and the Hodge numbers. The compact dual of a period domain is the projective flag varietyThe orthogonality conditions are algebraic. The Griffiths domain $D$ is the subset whose filtrations also satisfy real opposition and the displayed positivity. It is open in the usual complex topology of $\check{D}$; the inequalities are not additional complex algebraic equations. Thus $\check{D}$ retains the complex flag conditions, whereas $D$ retains genuine polarized Hodge structures.

Definitions organize category theory, but theorems give that organization mathematical force. I would therefore argue against treating the subject as an exception in which definitions are intrinsically more important than theorems. The distinction is especially misleading here because good categorical definitions are designed to expose hypotheses from which substantial theorems follow.

A category records objects, composable arrows, identities and associative composition. A functor preserves that structure. This language allows one to compare otherwise different mathematical settings, but it does not itself say which constructions exist. For example, the definition of a product in a category is not a recipe for forming a Cartesian product of underlying sets: it requires a universal factorization of pairs of arrows. If two products in a category $(P,p_1,p_2)$ and $(Q,q_1,q_2)$ exist, the universal properties give unique arrows $f:P\to Q$ and $g:Q\to P$ respecting projections. The composites $gf$ and $fg$ respect the same projections as the identities, so uniqueness forces them to be identities. Thus a definition yields a genuine theorem: products in a category are unique up to unique compatible isomorphism. The proof matters because it licenses replacing a construction by any other realization of the same property.

The Yoneda lemma is a more substantial example. For a covariant representable functor $\mathcal{C}(A,-)$ and $X:\mathcal{C}\to \mathbf{Set}$, a natural transformation $t:\mathcal{C}(A,-)\to X$ is determined by $a=t_A(1_A)$. Naturality forcesConversely this formula defines a natural transformation for every $a\in X(A)$. Hence $\mathrm{Nat}(\mathcal{C}(A,-),X)\cong X(A)$. The definition of naturality is indispensable, but the theorem identifies every possible transformation, proves representables detect arrows, and explains why universal objects can be studied through hom-sets. Merely knowing the definition does not supply this conclusion.

Similarly, an adjunction is defined by natural hom-set bijections. The construction of a free abelian group gives $\mathbf{Ab}({\mathbb{Z}}^{(S)},A)\cong \mathbf{Set}(S,UA)$, because a group homomorphism is determined uniquely by its values on the basis. This makes a universal extension problem transparent. But the Freyd general adjoint functor theorem goes further: it turns categorical limit preservation and a solution set into an actual adjoint. Its smallness hypotheses cannot be discarded as bookkeeping. For instance, the underlying-set functor from finite groups to finite sets has no left adjoint at a singleton. A putative universal finite group with universal element of order $m$ could not map that element to a generator of a finite cyclic group of order greater than $m$. The theorem's hypotheses separate existence from attractive notation.

The monoidal coherence theorem offers another example. A monoidal category has associativity and unit isomorphisms satisfying the pentagon identity for a monoidal category and triangle identity for a monoidal category. Coherence proves that every formal reassociation and unit insertion produces the same structural map. This is what justifies calculating without displaying all parentheses; it is not true merely because one has named an associator. Nor does it identify arbitrary additional maps, such as distinct braidings.

Finally, the opposite category construction shows how a definition can economize on proofs: reversing arrows takes products in a category to coproducts in a category, initial objects to terminal objects and monads to comonads. Yet this economy works because theorems and their proofs are invariant under the reversal. The subject is powerful precisely when definitions isolate reusable structure and theorems establish its consequences. Separating their importance obscures that partnership.

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.

Embed the ring of integers into ${\mathbb{R}}^2$ by its two real embeddings. Its image is a full Euclidean lattice of some covolume $d>0$. Fix $C>d$ and consider the symmetric rectangle $|{\sigma }_1(\alpha )|\le T$, $|{\sigma }_2(\alpha )|\le C/T$. Its area is $4C>4d$, so the Minkowski convex body theorem gives a nonzero algebraic integer ${\alpha }_T$ in it. Its algebraic norm is a nonzero integer with $|N({\alpha }_T)|\le C$.

As $T\to \infty$, infinitely many distinct ${\alpha }_T$ must occur: a finite collection of nonzero elements has a positive minimum of $|{\sigma }_2(\alpha )|$, whereas here that coordinate tends to zero. The principal ideals $({\alpha }_T)$ have norm $|N({\alpha }_T)|\le C$. There are only finitely many ideals of bounded norm: they are subgroups of bounded index in a rank-two free abelian group, and integer Hermite normal forms give finitely many possibilities at each such index. Thus infinitely many of these elements generate one ideal. Ratios of two generators of the same ideal are units, because inclusion in both directions supplies inverse algebraic integers. Among infinitely many distinct generators choose two whose ratio is not $\pm 1$. This gives a non-trivial unit without assuming Dirichlet's theorem.

We use the James reduced product theorem and the Bott–Samelson theorem: if $X$ is a connected based CW complex with free integral homology, then $JX\to \Omega \Sigma X$ is a weak equivalence, and its induced Pontryagin ring isHere $\Sigma$ is the reduced suspension, $T$ is the tensor algebra, and the multiplication is induced by concatenating James words, hence by concatenating loops. Taking $X=S^{2n}$ supplies a single generator $x$ of degree $2n$, sois free, with one generator in every degree $2nk$ and zero in the other degrees.

The diagonal makes this a homology coalgebra. The generator $x$ is a primitive homology class:There are no nontrivial lower positive degrees in which its reduced diagonal could land. Compatibility of the diagonal with loop multiplication givesThe degree of $x$ is even, so the two tensor factors commute without a Koszul sign rule.

The universal coefficient theorem for cohomology has no Ext terms here because homology is free. Let $a_k$ be the cohomology class dual to $x^k$, with $a_0=1$ and $|a_k|=2nk$. The cup product is dual to the diagonal, henceTherefore the integral cohomology ring is the divided power algebraExplicitly $H^{2nk}=\mathbb{Z}{a}_k$ for $k\ge 0$, and all other cohomology groups vanish. In particular $a_1^k=k!a_k$: over the integers this is not a polynomial ring on $a_1$. This is the integral cohomology of an odd-sphere loop space.

For the requested homology multiplication, use the homology cross product followed by concatenation:The constant loop gives the degree-zero unit. Loop concatenation is associative up to homotopy, which suffices for associativity on homology; Moore loops can make the space-level operation strictly associative. The Künneth theorem identifies the tensor-product homology because its groups are free abelian groups. The Bott–Samelson theorem identifies this product with word multiplication, so $x^i\ast x^j=x^{i+j}$. The Pontryagin ring of an odd-sphere loop space has presentationWith only one generator the tensor algebra has the indicated polynomial presentation. The homology product and the divided-power cohomology product are different operations.

$\boxed{{\mathbb{Z}}^4\text{ is not a nontrivial free product}.}$ In a nontrivial free product, nonidentity elements $g,h$ from different factors satisfy $gh\ne hg$, since these are distinct alternating normal forms. Such a group is not an abelian group, whereas the free abelian group ${\mathbb{Z}}^4$ is abelian.

The ordinary Poincare duality statement here is for a compact manifold without boundary, of dimension $d$, oriented over the field $F$. Its fundamental class $[M{]}_F$ induces isomorphismsfor every $q$. For a manifold with boundary the appropriate statement is Poincare-Lefschetz duality with relative groups; the ordinary pairing need not be nondegenerate. Over a field, the universal coefficient theorem for cohomology identifies $H^{d-q}(M;F)$ with the full dual of $H_{d-q}(M;F)$. The cap-cup evaluation identity and duality therefore makea perfect pairing. Explicitly, a nonzero $a$ has a nonzero cap product, and a linear functional $b$ on its homology group takes a nonzero value on that product. The same argument in the other variable proves nonsingularity. On the whole graded cohomology, define $B(a,b)$ by taking the degree-$d$ part of $a⌣b$ before evaluation. For a nonzero component $a_q$ choose $b$ homogeneous of degree $d-q$ to pair nontrivially with it. All other components contribute zero in degree $d$. This proves that the Poincare duality pairing is a nondegenerate bilinear form on $H^{\ast }(M;F)$; it need not be symmetric on all degrees.

Put $r=2n+1$, so $r$ is positive and odd, and orient each $S^r\times S^r$ by its product orientation. The Künneth theorem gives integral cohomology $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^2$ in degree $r$, and zero elsewhere. Its two degree-$r$ generators $\alpha ,\beta$ have ${\alpha }^2={\beta }^2=0$, $\alpha \beta$ equal to its top orientation class, and $\beta \alpha =-\alpha \beta$.

For a connected sum of oriented manifolds, excision and the long exact sequence for deleting a ball show that deleting a ball removes the top homology class and leaves all lower positive homology groups unchanged. The boundary sphere represents zero in the punctured manifold: it is the boundary of its relative fundamental chain. In the Mayer–Vietoris sequence for the two punctured pieces joined along the separating sphere, the orientation class of the connected sum maps onto that sphere's class. The intermediate positive groups are consequently the direct sums of the groups of the two original manifolds. This also covers $r=1$, where the sphere is a circle and the boundary-class observation is necessary in the middle degree. Iterating and applying the universal coefficient theorem for cohomology givesHere $g\ge 1$; under the conventional extension $W_0=S^{2r}$ the same formula holds with a zero middle group.

Let $\omega$ be the top cohomological orientation class. The connected-sum pinch map to the wedge of the $g$ sphere products gives degree-one projections to each summand. Pull back the two factor classes from summand $i$ to obtain ${\alpha }_i,{\beta }_i$. Their product is $\omega$, since the projection has degree one. Classes from different wedge summands have zero positive-degree cup products, and graded commutativity of the cup product supplies the reversed sign. Thus the full cohomology ring of a connected sum of odd-dimensional sphere products is the graded free abelian group just displayed with multiplicationThe unit is $1$, and $\omega$ times any positive-degree class is zero by dimension. This describes all products, including ${\omega }^2=0$.

The smooth involution is a diffeomorphism, since it is its own inverse. Its fixed set is closed; discreteness and compactness therefore make it finite. The supplied positivity of $\det (I-D{f}_x)$ makes every fixed point nondegenerate with local index $+1$. The Lefschetz-Hopf fixed-point theorem then givesIndeed the degree-zero and top-degree traces are both one, because the manifold is connected and $f$ preserves its orientation; the middle degree is odd.

On $V=H^r(W_g;\mathbb{R})$, the form $\Omega (a,b)=⟨a⌣b,[W_g]⟩$ is a nondegenerate alternating bilinear form by Poincare duality and the oddness of $r$. Thus it is a symplectic vector space of dimension $2g$. Naturality and orientation preservation show that $T$ preserves $\Omega$, and $T^2=I$. For the eigenspaces of a symplectic involution, write $V=V_{+}\oplus V_{-}$: the polynomial $(t-1)(t+1)$ has distinct roots over $\mathbb{R}$. If $v_{+}\in V_{+}$ and $v_{-}\in V_{-}$, then $\Omega (v_{+},v_{-})=\Omega (T{v}_{+},T{v}_{-})=-\Omega (v_{+},v_{-})$, so the two vector subspaces are orthogonal. Each restricted form is nondegenerate, since a vector annihilating its own vector subspace also annihilates the other and hence all of $V$. Their dimensions are therefore even, say $\dim V_{+}=2a$, $\dim V_{-}=2b$, with $a+b=g$. ConsequentlyThis proves the fixed-point congruence for an involution on an odd-sphere connected sum:

Each factor is a free abelian group of rank two. Take the infinite symmetric generating setThe Bass-Serre tree has an edge $e_g$ joining $gA$ to $gB$ for every $g\in A\ast B$. Indeed, the normal form theorem for a free product gives $A\cap B=\{1\}$, so a nonempty coset intersection consists of exactly one element. Let $m_g$ be the midpoint of $e_g$ and send the Cayley vertex $g$ to $m_g$.

Two distinct edges $e_g,e_h$ share an endpoint precisely when $g^{-1}h\in A\setminus \{1\}$ or $g^{-1}h\in B\setminus \{1\}$. This is exactly adjacency in the Cayley graph for $S$. A path in that graph therefore gives a path between the midpoints of the same length in the tree. Conversely, the unique tree segment between two edge midpoints passes through a succession of adjacent edges and gives a Cayley path of that length. Hence, on vertices,All edge midpoints occur, and every tree point is within $1/2$ of one of them. Extending along Cayley edges gives a quasi-isometry of the metric graphs; ambiguity from folding the cliques at a tree vertex costs only a bounded additive error. This is the free product with factor generating set construction.

For finite generating sets the answer changes. Use $S_0=\{a,a^{\prime },b,b^{\prime }{\}}^{\pm 1}$. The retraction $A\ast B\to A$ killing $B$ sends every generator to a generator of $A$ or to the identity. Therefore the restriction of $d_{S_0}$ to $A$ is exactly the standard ${\ell }^1$ word metric on ${\mathbb{Z}}^2$.

In this grid, take vertices $(0,0),(2N,0),(0,2N)$. Use the two axis segments as two sides, and the path from $(2N,0)$ through $(2N,2N)$ to $(0,2N)$ as the third. All are metric geodesics in the whole Cayley graph, by the retraction argument. The corner $(2N,2N)$ is at distance $2N$ from either of the other sides. These metric geodesic triangles are arbitrarily thick, so this Cayley graph is not a Gromov-hyperbolic metric space.

A tree is $0$-hyperbolic. Hyperbolicity is invariant under quasi-isometry for metric geodesic spaces, and all finite word metrics on this group are equivalent. Consequently no Cayley graph for a finite generating set is quasi-isometric to any tree, including a locally infinite one.