Building categories from familiar structures

Written by GPT-6.1 Sol (OpenAI), October 2026. Self-checked by the writing AI (GPT-6.1 Sol, Ultra). Original text public domain (CC0).

An arrow can be a function, a continuous map, a comparison in an order, or an element of a monoid. To turn any of these descriptions into a category, we must specify how arrows compose and check the laws. Small categories then become templates: a functor from a particular shape selects an arrow, a parallel pair, or a projector in another category.

We assume elementary sets, abelian groups, rings and the definition of a category. The complete size convention and closure proofs are in Universes and small categories. The inverse, opposite, cancellation and relation proofs are retained from Relations and cancellation in categories. The full structured-arrow, subcategory and component constructions supply the other categorical prerequisites. We use ambient ZFC and a Grothendieck universe \(\mathcal U\) containing the natural numbers.

1. Restrict functions and check composition

An arrow \(f:X\to Y\) has the specified source \(X\) and target \(Y\). The notation \(X\in\mathsf C\) means that \(X\) belongs to the object set of \(\mathsf C\). An endomorphism has equal source and target; an automorphism is an invertible endomorphism. Arrows are parallel when both endpoints agree. Juxtaposition \(gf\) means \(g\circ f\), with \(f\) applied first.

Proposition 1.1. The members of \(\mathcal U\), with functions as arrows, form a locally \(\mathcal U\)-small category \(\mathsf{Set}_{\mathcal U}\). Its full subcategory on finite members is denoted \(\mathsf{FinSet}_{\mathcal U}\).

Proof. A composite of functions is a function with the indicated endpoints. For \(f:X\to Y\), \(g:Y\to Z\) and \(h:Z\to W\), both \((hg)f\) and \(h(gf)\) send \(x\) to \(h(g(f(x)))\); extensional equality gives associativity. Composing on either side with the appropriate identity function leaves each value unchanged. This also works for empty domains. Each function set belongs to \(\mathcal U\) by the complete function-set closure in the prerequisite. Its objects form the ambient set \(\mathcal U\). Restricting to finite objects keeps every function between those objects, so the inherited operations give a full subcategory. \(\square\)

The collection of every ambient set cannot itself be a set. If it were a set \(V\), separation would form \(D=\{x\in V:x\notin x\}\). Since \(V\) contains every set, \(D\in V\), and its definition would give \(D\in D\) if and only if \(D\notin D\). Thus the convention requiring an ambient object set needs a size bound. The complete universe lesson also proves that \(\mathsf{Set}_{\mathcal U}\) is not essentially \(\mathcal U\)-small. Stacks permits certain named categories with a proper class of objects; that convention does not change our object-set requirement.

A topology on \(X\) is a collection \(\tau\subseteq\mathcal P(X)\) containing \(\varnothing,X\), closed under arbitrary unions and finite intersections, including the empty intersection \(X\). A function \(f:(X,\tau_X)\to(Y,\tau_Y)\) is continuous when \(f^{-1}(V)\in\tau_X\) for each \(V\in\tau_Y\).

Proposition 1.2. Topological spaces with underlying sets in \(\mathcal U\), and continuous maps, form a locally \(\mathcal U\)-small category \(\mathsf{Top}_{\mathcal U}\).

Proof. The identity is continuous because its inverse image of each open set is that same open set. For continuous \(f:X\to Y\) and \(g:Y\to Z\), an open \(W\subseteq Z\) has open \(g^{-1}(W)\subseteq Y\), whose inverse image under \(f\) is open. Since

\[ (gf)^{-1}(W)=f^{-1}(g^{-1}(W)), \tag{1.1} \]

the composite is continuous. Associativity and the two identity equations hold as equations of functions, by Proposition 1.1. Continuous maps form a subset of the function set \(Y^X\), so their Hom set belongs to \(\mathcal U\).

For each \(X\in\mathcal U\), the possible topologies form a subset of \(\mathcal P(\mathcal P(X))\). The pairs \((X,\tau)\) therefore form an ambient set by replacement and union over the set \(\mathcal U\). Universe closure puts each topology and each pair in \(\mathcal U\) as well. Consequently this agrees with taking topological spaces encoded as universe members. \(\square\)

Two more restrictions are already fully checked in Zero maps, Section 2: pointed sets retain functions sending the designated point to the designated point, and unital left \(R\)-modules retain additive functions commuting with the \(R\)-action. That section proves preservation by identity and composition for both restrictions. We use those exact proofs, for every associative unital \(R\in\mathcal U\). Relations, Section 4 gives the full relation composition and diagonal laws, and the faithful graph inclusion of functions, including its failure to be full.

2. Orders describe which arrows exist

A partial order on \(I\) is a reflexive, transitive and antisymmetric relation \(\leq\). It is directed, also called filtering, when \(I\) is nonempty and every pair \(i,j\) has a common upper bound in \(I\). It is total when every pair is comparable. Write \(i<j\) for \(i\leq j\) and \(i\ne j\); write \(i\geq j\) for \(j\leq i\), and \(i>j\) for \(j<i\).

An ordered set is inductively ordered when every totally ordered subset has an upper bound in the ambient ordered set. The empty subset is included: its upper bound is any ambient element, so this condition forces nonemptiness. A maximal element \(m\) means that \(m\leq x\) implies \(x=m\); a greatest element means that every \(x\leq m\). These are different conditions.

We use Zorn's lemma in the ambient foundational system: an inductively ordered set has a maximal element. It is the standard choice-equivalent foundational principle in ZFC. The course assumes this foundation; the definition of an inductively ordered set alone is not a proof of Zorn's lemma.

The category of an order has one arrow \(i\to j\) when \(i\leq j\) and none otherwise. Retain the complete laws and reversed-order calculation in Relations, Section 2. Reflexivity supplies the identity and transitivity supplies the composite; uniqueness of each existing arrow proves every required equation. Antisymmetry makes isomorphic objects equal. The empty order still gives a category, although it is neither directed nor inductively ordered.

For example, a directed order may have no greatest element: \(\mathbb N\) is directed because \(\max(i,j)\) bounds both inputs, and \(n+1>n\) excludes a greatest element. It is not inductively ordered, since its entire totally ordered subset has no upper bound in \(\mathbb N\).

3. Endomorphisms describe a multiplication

A monoid is a set \(M\) with an associative multiplication and a two-sided unit \(1\). Its one-object category \(BM\) has object \(*\), arrows \(M\), identity \(1\), and composition

\[ b\circ a=ba. \tag{3.1} \]

The two unit equations in \(M\) are precisely the two category identity equations, and the monoid associative equation is precisely the category associative equation. Conversely, the endomorphisms of the sole object of a one-object category form a monoid by these same equations. The descriptions recover all the original data in either direction.

A groupoid is a category in which every arrow is invertible. If \(M=G\) is a group, its group inverses give both inverse-arrow equations, so \(BG\) is a groupoid. Conversely, if \(BM\) is a groupoid, every element of \(M\) has a two-sided inverse; the monoid is a group. More generally, the exact inverse-closure proof in Relations, Section 1 shows that the automorphisms of any object form a group.

When counting arrows, retain their endpoints. The complete arrow set is the tagged union of the Hom sets, with a typical member \((X,Y,f)\). The map

\[ X\longmapsto(X,X,1_X) \tag{3.2} \]

is injective because equality of the triples forces equality of their first entries. Thus a finite category, meaning one with finitely many typed arrows, has finitely many objects. The converse fails: \(B\mathbb N\), using addition with identity \(0\), has one object and infinitely many arrows. A discrete category has only identities; it may have any ambient set of objects. It is finite exactly when that object set is finite.

A subcategory must contain its chosen objects' identities and be closed under the inherited composition. A full subcategory keeps all ambient arrows between its objects. These laws are retained in Zero maps, Section 3. A full subcategory is replete, or saturated, if it also contains every ambient object isomorphic to one of its objects. The finite-set subcategory is replete: a bijection with a finite set transports a finite enumeration.

Fullness alone does not imply repleteness. In \(\mathsf{Set}_{\mathcal U}\), the full subcategory on the one chosen singleton \(\{\varnothing\}\) omits the different singleton \(\{\{\varnothing\}\}\), although a bijection joins them. Both encodings belong to \(\mathcal U\).

For \(r:X\to Y\) and \(s:Y\to X\) with \(rs=1_Y\), \(r\) is a left inverse of \(s\) and \(s\) a right inverse of \(r\). The arrow \(s\) is a section of \(r\), while \(r\) is a retraction or cosection of \(s\). Retain the complete monic-section and epic-retraction proofs in Relations, Section 1. A monomorphism may be drawn \(X\hookrightarrow Y\) or \(X\rightarrowtail Y\), and an epimorphism \(X\twoheadrightarrow Y\); the drawing records the cancellation property, not an underlying subset or quotient in every category. The exact Hom-injectivity tests and all three composition closure proofs are in the same lesson.

4. Small shapes specify diagram data

The point category \(\mathsf{Pt}\) has one object and its identity only. Its single possible composite is that identity, so both unit laws and associativity hold. The empty category has no objects or arrows; it has no composable pair or triple and no identity to supply, so every category axiom holds vacuously.

The walking arrow \(\mathsf{Arr}\) has objects \(a,b\), their identities, and one arrow \(u:a\to b\). The walking parallel pair \(\mathsf{Par}\) has the same objects and identities and two distinct arrows \(u,v:a\to b\). In either case, the only compositions involving a nonidentity arrow are its composition with an endpoint identity. Define those to be that same arrow. A composable triple contains at most one nonidentity arrow: there is no nonidentity arrow leaving \(b\) or entering \(a\). Either bracketing consequently returns that arrow, or the identity if all three arrows are identities. This verifies associativity and both units.

The walking idempotent \(\mathsf{Pr}\) has one object \(c\) and arrows \(1,p\), with \(p\ne1\). Its multiplication is given in Figure 4.1.

Left factor \(\backslash\) right factor \(1\) \(p\)
\(1\) \(1\) \(p\)
\(p\) \(p\) \(p\)

Figure 4.1. The complete multiplication table of the walking idempotent. Left factor means the arrow applied after the right factor. Every product containing \(p\) is \(p\). This editable table gives every composite and the associative mechanism used below.

The table has two-sided unit \(1\). For any triple, either every factor is \(1\), in which case either bracketing is \(1\), or at least one is \(p\), in which case either bracketing is \(p\). Thus it is an associative monoid, and Section 3 gives the category. All five shapes have finite encoded object and arrow sets in \(\mathcal U\).

Proposition 4.1. For any category \(\mathsf C\), functors from these shapes select exactly the following data:

Proof. A functor must send each identity to the identity of its chosen object. For the walking arrow and parallel pair, this leaves exactly the indicated nonidentity-arrow images to choose. Since all other products use identities, the target's identity laws give every functor composition equation. The source arrows \(u,v\) are distinct, but a functor is not required to be injective on arrows.

For \(\mathsf{Pr}\), the source equation \(p^2=p\) forces \(e^2=e\). Conversely, choosing such an \(e\), sending \(1\) to \(1_X\) and \(p\) to \(e\), respects all four entries of Figure 4.1. Hence it defines a functor. The point and empty assignments have only the forced identity equations or no equations, respectively. These constructions recover a functor's assignments, proving both directions. \(\square\)

A natural transformation between two walking-idempotent diagrams \((X,e)\) and \((Y,d)\) is precisely an arrow \(h:X\to Y\) such that

\[ he=dh. \tag{4.1} \]

Indeed naturality for \(p\) is (4.1), and naturality for \(1\) is automatic. For walking arrows, the analogous two-component condition is the commuting square proved in Zero maps, Section 1. For a parallel pair, the two components must commute with both chosen arrows.

A diagram names objects and arrows; the indicated directed paths commute when their composites agree. In particular, for \(g_1,g_2:Z\to X\) and \(f:X\to Y\), saying that the two resulting compositions coincide means

\[ fg_1=fg_2. \tag{4.2} \]

For a square with \(f:X\to Y\), \(h:X\to V\), \(g:Y\to Z\), \(k:V\to Z\), the equation is \(gf=kh\); for a triangle with \(l:X\to Z\), it is \(gf=l\). These are equations between arrows with the same specified endpoints. When the vertices are categories and the edges are functors, distinguish literal equality of route functors from a specified natural isomorphism between them. The latter is the quasi-commutative convention, retained with its compatibility example in Natural transformations, Section 4.

5. Modules and their two sides

All rings here are associative and unital; their homomorphisms preserve the unit. The zero ring is allowed. A module action preserves addition in both variables, satisfies the associative action law, and is unital. The zero module has one element. A field is a nonzero commutative ring in which every nonzero element is invertible.

Proposition 5.1. Right \(R\)-modules and left \(R^{\mathrm{op}}\)-modules describe the same structures and the same linear maps.

Proof. The opposite ring has the same addition and unit, with product \(r\star s=sr\). Associativity follows since \((r\star s)\star t=t(sr)=(ts)r=r\star(s\star t)\); its two unit laws and two distributive laws are the original laws with multiplication reversed.

For a right \(R\)-module, define \(r\cdot m=mr\). Both distributive laws become the two distributive laws of the right action, and \(1\cdot m=m1=m\). Its left associative action law is

\[ \begin{aligned} r\cdot(s\cdot m)&=(ms)r\\ &=m(sr)=(r\star s)\cdot m. \end{aligned} \tag{5.1} \]

Conversely, for a left \(R^{\mathrm{op}}\)-module define \(mr=r\cdot m\). Its right associative law is

\[ \begin{aligned} (mr)s&=s\cdot(r\cdot m)\\ &=(s\star r)\cdot m\\ &=(rs)\cdot m=m(rs) \end{aligned} \]

and the unit and both distributive equations reverse in the same way. The two conversions literally recover the original actions. An additive map commutes with \(mr\) exactly when it commutes with \(r\cdot m\), so the linear maps agree. Their identities and composites are the same functions. \(\square\)

Proposition 5.2. An abelian group has a unique unital \(\mathbb Z\)-module structure. Its group homomorphisms are precisely its \(\mathbb Z\)-linear maps.

Proof. For \(n\geq0\), define \(na\) to be the sum of \(n\) copies of \(a\), with \(0a=0\). Define \((-n)a=-(na)\). Counting summands gives \((m+n)a=ma+na\) for nonnegative integers. For arbitrary integers, cancel positive and negative copies in the abelian group; the same equality follows, including opposite signs. Also \(n(a+b)=na+nb\) for nonnegative \(n\) by reordering a finite sum; negation gives it for negative \(n\). Repeating an \(n\)-fold sum \(m\) times gives \(m(na)=(mn)a\) for nonnegative \(m,n\). The identity \((-m)b=-(mb)\) and additivity of \(m(-)\), which gives \(m(-b)=-(mb)\), extend this equation to all signs. Finally \(1a=a\). These are all the module equations.

For uniqueness, a unital action has \(1a=a\), and additivity in the integer variable forces \(na\) to be the repeated sum for positive \(n\). It also forces \(0a=0\) and \((-n)a=-(na)\), so there is no other action. An additive map preserves finite sums, zero and negatives, hence all integer multiples. Conversely a module map is additive by definition. Thus \(\mathsf{Mod}(\mathbb Z)\) is exactly the category of abelian groups, with these structures identified. \(\square\)

For a left module \(M\), its endomorphisms form the ring \(\operatorname{End}_R(M)\), with pointwise addition and multiplication \(uv=u\circ v\). Here is the full elementary verification. Sums, negatives and zero remain \(R\)-linear: evaluate on \(m+n\) and on \(rm\), and distribute in the abelian target group. The addition laws hold after evaluation on every element of \(M\). Composite linear maps remain linear by the complete module-category laws in Zero maps, Section 2. Multiplication is associative with unit \(1_M\), by function laws. Its two distributive equations are

\[ \begin{gathered} ((u+v)w)(m)\\ =u(wm)+v(wm),\\ (u(v+w))(m)\\ =u(vm)+u(wm). \end{gathered} \tag{5.2} \]

The second uses additivity of \(u\). This proves every unital ring law. For \(M=0\) the ring has one element and its unit equals zero. Its units are exactly \(\operatorname{Aut}_R(M)\): a two-sided inverse in this ring is exactly a two-sided inverse linear arrow. The complete automorphism-group proof in Relations, Section 1 supplies closure, identity and inverses.

6. Finite generators and relations

Let \(\mathsf{Mod}^f(R)\) be the full subcategory of finitely generated left modules, and \(\mathsf{Mod}^{\mathrm{fp}}(R)\) the full subcategory of finitely presented ones.

A module is finitely generated precisely when it admits a surjection \(R^n\to M\) for a nonnegative integer \(n\). To check the terminology, the images of the standard basis vectors generate the quotient. Conversely a finite generating list \(m_1,\ldots,m_n\) defines the surjection \((r_i)\mapsto\sum_i r_im_i\). The case \(n=0\) supplies the zero module.

A finite free surjection \(q:R^n\to M\) is a finite presentation when its kernel has a finite generating list. Such a list is the image of a linear map \(R^m\to R^n\), so equivalently there is an exact sequence

\[ \begin{gathered} R^m\longrightarrow R^n \xrightarrow{q}M\longrightarrow0,\\ m,n<\infty. \end{gathered} \tag{6.1} \]

The first map need not be injective. We must also check that changing the finite free surjection does not change this property.

Proposition 6.1. If one finite free surjection onto \(M\) has finitely generated kernel, every finite free surjection onto \(M\) does.

Proof interface and specialization. Retain the complete arbitrary-ring kernel identity (4.3), including both containment proofs, in Formal colimits and compact presentations, Section 4. In its summand argument take its \(P=M\), its complementary \(N=0\), and its \(F_M\oplus F_N=F=R^n\), choosing \(F_N=0\). Its \(v:G\to M\) is the given finite free surjection with finitely generated kernel \(H\), and its \(q:F\to M\) is any other finite free surjection. Freeness lifts the finitely many basis images and gives \(h:G\to F\), \(k:F\to G\) with \(qh=v\), \(vk=q\). The exact retained identity becomes

\[ \ker q=h(H)+\operatorname{im}(1_F-hk). \tag{6.2} \]

The first summand is generated by the images of a finite generating list of \(H\). The second is generated by the images of the finite basis of \(F\). Their union generates the sum. Thus the retained complete proof gives the assertion for an arbitrary associative unital \(R\), with no exchange of coefficients and no coherence or Noetherian hypothesis. This also handles \(F=0\), when its image summand is zero. \(\square\)

These full module subcategories are replete. If \(\alpha:M\to N\) is an isomorphism and \(q:R^n\to M\) is surjective, then \(\alpha q\) is surjective and has the same kernel, since \(\alpha\) is injective. A finite generating list or presentation therefore transports to \(N\). Fullness retains all linear maps between their objects.

For later calculations, fix some elementary notation. The symbols \(\mathbb Z,\mathbb Q,\mathbb R,\mathbb C\) denote the usual integers, rationals, reals and complex numbers, with their usual ring or field structures; \(\mathbb N=\{0,1,2,\ldots\}\). The singleton \(\{\mathrm{pt}\}\) has one distinguished element \(\mathrm{pt}\), whereas \(\varnothing\) has none.

For a commutative unital ring \(k\), a \(k\)-algebra is a unital ring \(R\) with a unital map \(\phi:k\to R\) whose image is central: \(\phi(a)r=r\phi(a)\) for every \(a,r\). This permits \(R\) to be noncommutative. The notation \(k^\times\) means its group of units. Products of units are units because \((ab)^{-1}=b^{-1}a^{-1}\); the identity and inverse equations give its group laws. The notation \(k[x_1,\ldots,x_n]\) denotes the polynomial ring in commuting variables. A polynomial is a finitely supported coefficient family indexed by \(\mathbb N^n\); addition is coefficientwise and multiplication uses the finite convolution of coefficients whose multi-indices add to the requested exponent. For \(n=0\) this is \(k\).

Finally \(\delta_{ij}\) is the Kronecker symbol: it is \(1\) when \(i=j\) and \(0\) otherwise. Matrix identities use these values times the appropriate ring unit or object identity. The notation does not put a scalar multiplication on Hom sets in an arbitrary category.

7. Four graded exercises with full solutions

Exercise 1: Upper bounds and empty shapes

Exercise 1 (introductory). Let \(P=\{o,a,b\}\), with \(o<a\), \(o<b\) and no comparison between \(a,b\). Decide whether \(P\) is total, directed or inductively ordered. List its maximal and greatest elements. Do the same for the empty ordered set. Count the arrows of each associated category, and the functors from \(\mathsf{Pt}\) and the empty category into it.

Solution. The pair \(a,b\) is incomparable, so \(P\) is not total. It has no common upper bound, so \(P\) is not directed. Each nonempty chain is a finite totally ordered subset, hence has a largest member: in this example the largest is \(a\), \(b\), or \(o\). That member is an upper bound in \(P\). The empty chain is bounded by \(o\). Thus \(P\) is inductively ordered. Its maximal elements are \(a,b\); it has no greatest element since one would have to bound both. Its order category has the three identities and the two arrows \(o\to a,o\to b\), so has five arrows. There are three functors from \(\mathsf{Pt}\), one for each object, and one functor from the empty category.

The empty order is total vacuously. It is not directed by the nonemptiness condition. It is not inductively ordered: its empty chain has no ambient upper bound. It has no maximal or greatest elements and its category has no arrows. There is no functor from \(\mathsf{Pt}\) into it, since there is no image object. There is exactly one functor from the empty category into it, the empty assignment. Zorn's conclusion applies to \(P\), and its hypothesis fails for the empty order.

Exercise 2: Three projector diagrams in a topological space

Exercise 2 (intermediate). Give \(S=\{0,1\}\) the topology \(\{\varnothing,\{1\},S\}\). Determine its continuous endomorphisms and its idempotent endomorphisms. List the functors \(\mathsf{Pr}\to\mathsf{Top}_{\mathcal U}\) sending \(c\) to \(S\). For every pair of these functors, count the natural transformations.

Solution. There are four functions \(S\to S\). The two constant functions \(c_0,c_1\) are continuous: the preimage of \(\{1\}\) is respectively \(\varnothing,S\), both open. The identity is continuous. The interchange has preimage \(\{0\}\) for \(\{1\}\), which is not open, so it is not continuous. The three continuous maps are all idempotent: repeating a constant changes nothing, and \(1_S^2=1_S\).

Proposition 4.1 therefore gives exactly three functors with the specified object image, indexed by \(e=c_0,c_1,1_S\). A transformation from \(e\) to \(d\) is a continuous \(h:S\to S\) with \(he=dh\). The complete counts, with source diagrams in rows, are:

Source \(\backslash\) target \(c_0\) \(c_1\) \(1_S\)
\(c_0\) 2 1 2
\(c_1\) 1 2 2
\(1_S\) 1 1 3

For \(e=c_i,d=c_j\), the condition says \(h(i)=j\). If \(i=j\), \(h\) can be \(c_i\) or \(1_S\), giving two; if \(i\ne j\), only \(c_j\) works. For \(e=c_i,d=1_S\), the condition \(hc_i=h\) forces \(h\) to be constant, and both constants work. For \(e=1_S,d=c_j\), the condition \(h=c_jh\) forces \(h=c_j\). For both identities every continuous map works, giving three. These cases give all nine entries, for a total of fifteen transformations.

Exercise 3: The opposite multiplication is visible

Exercise 3 (advanced). Let \(R=M_2(\mathbb F_2)\), and regard \(R\) as its regular left module. Describe every left-linear endomorphism, identify its endomorphism ring, and explain why left multiplication by a fixed matrix need not be left-linear. Translate the regular right module into a left \(R^{\mathrm{op}}\)-module.

Solution. If \(f:R\to R\) is left-linear, then \(f(x)=f(x1)=xf(1)\). Conversely, for any \(b\in R\), the map \(r_b(x)=xb\) is additive and satisfies \(r_b(ax)=(ax)b=a(xb)\). Thus \(b\mapsto r_b\) is bijective and additive. Its multiplication obeys

\[ \begin{aligned} (r_b r_c)(x)&=(xc)b\\ &=x(cb)=r_{cb}(x). \end{aligned} \tag{7.1} \]

Hence it is a unital ring isomorphism \(R^{\mathrm{op}}\to\operatorname{End}_R(R)\): in the opposite ring \(b\star c=cb\), exactly as in (7.1).

For the matrix units \(b=E_{12}\) and \(a=E_{21}\), left multiplication \(\ell_b\) has

\[ \begin{aligned} \ell_b(a1)&=ba=E_{11},\\ a\ell_b(1)&=ab=E_{22}. \end{aligned} \tag{7.2} \]

These matrices differ over \(\mathbb F_2\), so \(\ell_b\) is not left-linear. On the regular right module, the left opposite action is \(b\cdot x=xb\). Its law is \(b\cdot(c\cdot x)=(xc)b=x(cb)=(b\star c)\cdot x\), retaining the same reversed order. The isomorphism and action statements use every matrix \(b\), not just the two selected units.

Exercise 4: Change a finite presentation explicitly

Exercise 4 (advanced). Let \(M=\mathbb Z/6\mathbb Z\). Compare \(v:\mathbb Z\to M\), \(v(t)=[t]\), with \(q:\mathbb Z^2\to M\), \(q(x,y)=[x+2y]\). Give lifts \(h,k\) as in Proposition 6.1, determine both summands in (6.2), and exhibit a complete finite presentation using \(q\). Explain the rank-zero case.

Solution. Both maps are additive and commute with integer multiplication, so are linear. They are surjective since \(v(1)=[1]\) and \(q(1,0)=[1]\). The first kernel is \(H=6\mathbb Z\). Choose \(h(t)=(t,0)\) and \(k(x,y)=x+2y\). Then \(qh=v\) and \(vk=q\), as required. Their two kernel summands are

\[ \begin{aligned} h(H)&=\mathbb Z(6,0),\\ (1-hk)(x,y)&=(-2y,y),\\ \operatorname{im}(1-hk)&=\mathbb Z(-2,1). \end{aligned} \tag{7.3} \]

Directly, \(q(x,y)=0\) means \(x+2y=6t\) for some integer \(t\), hence \((x,y)=t(6,0)+y(-2,1)\). Conversely each displayed generator maps to zero. Thus they generate exactly the kernel. The map \(j:\mathbb Z^2\to\mathbb Z^2\), \(j(t,s)=(6t-2s,s)\), has that image. It is also injective: if \(j(t,s)=0\), the second coordinate gives \(s=0\), and then \(6t=0\) gives \(t=0\). Therefore

\[ 0\longrightarrow\mathbb Z^2 \xrightarrow{j}\mathbb Z^2 \xrightarrow{q}M\longrightarrow0 \tag{7.4} \]

is exact. For a rank-zero free source, \(\mathbb Z^0=0\). A surjection \(0\to N\) forces \(N=0\), and its kernel is zero, generated by the empty list. This is the rank-zero finite presentation; it cannot present the nonzero \(M\) above.

8. References and proof interfaces