Zero maps, components and subobjects

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).

A map can carry extra structure without changing its underlying arrow. Maps into a fixed object, maps out of it and commuting squares give three examples. Their categories let us compare subobjects and quotients without choosing underlying elements. A different question asks whether two objects can be linked by any sequence of arrows. Connected components answer that question, and a zero object gives a particularly strong connection between all objects.

We assume the category, functor and universe conventions of Universes and small categories, and the complete inverse and cancellation arguments in Relations and cancellation in categories. Composition is written \(gf\), with \(f\) applied first. An arrow \(f:X\to Y\) has its specified source and target; two arrows are parallel when both agree. The examples with modules use an associative unital ring and unital left modules, with no commutativity assumption. Basic references are [Stacks, Categories], [Riehl] and those two preceding lessons. We prove the elementary constructions below. The last exercise uses the exact relation-category cancellation criterion already proved in Relations and cancellation, Exercise 4.

Every category here has an ambient set of objects and an ambient set for each Hom collection. We write \(X\in\mathsf C\) for membership in its object set. These ambient sets need not belong to the smaller indexing universe \(\mathcal U\). We distinguish that ambient size from \(\mathcal U\)-smallness throughout.

1. A category whose objects carry an arrow

Let \(F:\mathsf C\to\mathsf D\) be a functor and \(A\) an object of \(\mathsf D\). Define \((F\downarrow A)\) as follows. Its objects are pairs \((X,s)\) with \(s:F(X)\to A\). A morphism from \((X,s)\) to \((Y,t)\) is an arrow \(f:X\to Y\) in \(\mathsf C\) satisfying

\[ s=tF(f). \tag{1.1} \]

The identity of \(X\) satisfies (1.1), because \(F(1_X)=1_{F(X)}\). If \(f:(X,s)\to(Y,t)\) and \(g:(Y,t)\to(Z,u)\), then

\[ \begin{aligned} uF(gf)&=uF(g)F(f),\\ &=tF(f)=s. \end{aligned} \tag{1.2} \]

Thus the original composite is again a morphism. Associativity and both identity laws are the corresponding laws in \(\mathsf C\), since the morphisms are these actual arrows. This proves that \((F\downarrow A)\) is a category. Its forgetful functor sends \((X,s)\) to \(X\) and \(f\) to itself. Each Hom map is inclusion of the subset defined by (1.1), so the functor is faithful. It need not be full: an underlying arrow may fail (1.1).

Define \((A\downarrow F)\) with objects \((X,s:A\to F(X))\) and morphisms satisfying \(t=F(f)s\). Its identity law follows from \(F(1_X)s=s\). For composable morphisms, \(t=F(f)s\) and \(u=F(g)t\) imply \(u=F(g)F(f)s=F(gf)s\). Original composition therefore stays in this Hom subset, and associativity and both units follow from \(\mathsf C\). Forgetting the structure arrow again gives a faithful functor, by inclusion of each Hom subset.

Both constructions depend on \(F\), even when their notation suppresses it. We may name an object by its structure arrow, or just by \(X\) when that arrow is understood. Taking \(F=1_{\mathsf C}\) gives the slice \(\mathsf C/A\) and the coslice \(A/\mathsf C\). No limits or colimits were needed to form them.

There is also an arrow category \(\operatorname{Arr}(\mathsf C)\). An object is an arrow \(f:X\to Y\). A morphism from \(f\) to \(g:X'\to Y'\) is a pair \((u:X\to X',v:Y\to Y')\) such that \(gu=vf\). This is a commuting square: its two directed paths from \(X\) to \(Y'\) have the same composite. More generally, a diagram commutes when the indicated directed paths with the same endpoints have equal composites. A commuting triangle with arrows \(f:X\to Y\), \(g:Y\to Z\) and \(l:X\to Z\) means \(l=gf\).

The identity square at \(f\) is \((1_X,1_Y)\). For a square \((u,v):f\to g\) followed by \((u',v'):g\to h\), define their composite to be \((u'u,v'v)\). The two given square equations yield

\[ \begin{aligned} h(u'u)&=(hu')u,\\ &=(v'g)u=v'(gu),\\ &=v'(vf)=(v'v)f. \end{aligned} \tag{1.3} \]

So composition preserves the square condition. The two component identity equations give both identity laws, and the two component associative laws give associativity for a triple of squares. This completes the category construction.

These constructions have ambient sets of objects and Homs. For \((F\downarrow A)\), the object set is the tagged union of the sets \(\operatorname{Hom}_{\mathsf D}(F(X),A)\), indexed by the ambient object set of \(\mathsf C\); replacement and union give an ambient set. For \((A\downarrow F)\), use the sets \(\operatorname{Hom}_{\mathsf D}(A,F(X))\) instead. For the arrow category, use the tagged union of \(\operatorname{Hom}_{\mathsf C}(X,Y)\), indexed by pairs of objects. Its square Hom sets are subsets of the product of two original Hom sets. These arguments justify all their object and arrow collections in the stated ambient setting.

If \(\mathsf C\) is locally \(\mathcal U\)-small, the two structured-arrow categories are locally \(\mathcal U\)-small because their Homs are subsets of original Homs. The arrow category has the same property because its Homs are subsets of products of two original Homs. The complete product, subset and size-transport results in Universes and small categories, Sections 2–3 apply. None of these arguments requires a \(\mathcal U\)-small object set.

2. Universal endpoints and their zero maps

An object \(P\) is initial if \(\operatorname{Hom}(P,X)\) is a singleton for every \(X\). An object \(T\) is terminal if \(\operatorname{Hom}(X,T)\) is a singleton for every \(X\); equivalently, it is initial in the opposite category. Neither condition asserts that such an object exists in every category.

An endomorphism is an arrow with equal source and target.

Proposition 2.1. Two initial objects have a unique isomorphism between them. The same holds for two terminal objects.

Proof. For initial \(P,Q\), let \(a:P\to Q\) and \(b:Q\to P\) be the unique arrows. There is only one arrow from \(P\) to itself, and its identity is one such arrow. Hence \(ba=1_P\). In the same way \(ab=1_Q\). Thus \(a\) is invertible, and every isomorphism \(P\to Q\) equals \(a\), since every arrow with those endpoints does. For terminal \(T,S\), let \(c:T\to S\) and \(d:S\to T\) be the unique arrows supplied by their terminality. The only endomorphisms of \(T\) and \(S\) are their identities, so \(dc=1_T\) and \(cd=1_S\). This again proves invertibility and uniqueness. \(\square\)

A zero object is both initial and terminal. Fix one, denoted \(0\). For any \(X,Y\), let \(u_X:X\to0\) and \(v_Y:0\to Y\) be the unique arrows. Define the zero map by

\[ 0_{XY}=v_Yu_X:X\longrightarrow Y. \tag{2.1} \]

Proposition 2.2. This map is independent of the chosen zero object. For all \(g:W\to X\) and \(f:Y\to Z\),

\[ f0_{XY}=0_{XZ},\qquad 0_{XY}g=0_{WY}. \tag{2.2} \]

Proof. Let \(0'\) be another zero object, with associated \(u'_X,v'_Y\). Proposition 2.1 gives the unique isomorphism \(k:0\to0'\). Both \(ku_X\) and \(u'_X\) are arrows from \(X\) to the terminal object \(0'\), so they agree. Both \(v'_Yk\) and \(v_Y\) are arrows from the initial object \(0\) to \(Y\), so they agree. Consequently

\[ v_Yu_X=v'_Yku_X=v'_Yu'_X. \tag{2.3} \]

For the first absorption law, \(fv_Y\) is the unique arrow \(0\to Z\), so \(fv_Y=v_Z\). Composing with \(u_X\) gives \(f0_{XY}=v_Zu_X=0_{XZ}\). For the second, \(u_Xg\) is the unique arrow \(W\to0\), so \(u_Xg=u_W\). Composing with \(v_Y\) gives \(0_{XY}g=v_Yu_W=0_{WY}\). \(\square\)

These arguments require no addition on Hom sets. A zero map is a particular existing arrow, and is different from an empty Hom set.

In \(\mathsf{Set}_{\mathcal U}\), the empty set is initial: exactly one empty function goes from it to any set. A singleton is terminal: exactly one function goes to it from any set. There is no zero object. If one existed, Proposition 2.1 would identify it with the empty initial set, whereas terminality would require a function from a singleton to that empty set.

A pointed \(\mathcal U\)-set is a pair \((X,x)\), where \(X\in\mathcal U\) and \(x\in X\); its arrows send distinguished points to distinguished points. Identities do so, and a composite of two point-preserving functions does so, so ordinary function laws define their category. The singleton pointed set is initial, because a map out of it is forced to choose the target point. It is terminal, because the unique function to a singleton preserves every point. It is therefore a zero object.

For an associative unital ring \(R\in\mathcal U\), take unital left \(R\)-modules whose underlying sets belong to \(\mathcal U\). Their \(R\)-linear maps form a category: identity maps preserve addition and the action, and if \(f,g\) do so then \((gf)(m+n)=gf(m)+gf(n)\) and \((gf)(rm)=r(gf)(m)\). Their ordinary composition and identity laws are those of functions. The one-element module is a zero object. There is just one map into it. A linear map out of it must send its sole element \(0\) to \(0\): additivity gives \(h(0)=h(0)+h(0)\), and cancellation in the target group gives \(h(0)=0\). This unique map is linear. Thus both universal endpoint conditions hold for every such \(R\).

For an ordered set regarded as a category, an initial object is a least element and a terminal object is a greatest element, since a Hom set is a singleton exactly when the order relation holds. The ordered set of integers has neither: for each \(n\), there is no arrow \(n\to n-1\) and no arrow \(n+1\to n\).

3. Comparing maps to and from a fixed object

A subcategory restricts the objects and Hom sets, contains the identity of each chosen object, and is closed under the original composition. It is full on its object set if it retains every ambient arrow between those objects. A full subcategory automatically satisfies the category laws, since its Hom sets and their operations are the original ones.

Fix \(X\in\mathsf C\). For monomorphisms \(f:A\to X\) and \(g:B\to X\), an isomorphism between them in \(\mathsf C/X\) is precisely an isomorphism \(a:A\to B\) in \(\mathsf C\) such that \(f=ga\). Indeed an inverse in the slice is an underlying inverse, by the forgetful functor. Conversely, if \(a\) is an underlying isomorphism with that equation, composing it with \(a^{-1}\) gives \(g=fa^{-1}\), so its inverse is also an arrow in the slice. Such \(a\) is unique, because \(g\) is monic and \(ga=ga'\) implies \(a=a'\).

A subobject of \(X\) is one of these isomorphism classes of monomorphisms to \(X\). They form an ambient set. In fact, take the disjoint union of the ambient sets \(\operatorname{Hom}(A,X)\), indexed by the ambient object set of \(\mathsf C\), recording the domain label on each map. This is a set by replacement and union in ambient set theory. Its subset of monomorphisms is a set, and its set of equivalence classes is a subset of its power set. This does not assert \(\mathcal U\)-smallness of all the subobjects of a locally \(\mathcal U\)-small category.

Define a relation on these classes by

\[ \begin{gathered} \bigl[f\bigr]\le\bigl[g\bigr]\quad\Longleftrightarrow\quad f=gh\\ \text{for some }h:A\to B. \end{gathered} \tag{3.1} \]

Theorem 3.1. Relation (3.1) is a well-defined partial order. A factor \(h\), when it exists, is unique and monic.

Proof. Suppose \(f'=f\alpha\) and \(g'=g\beta\), where \(\alpha:A'\to A\) and \(\beta:B'\to B\) are the isomorphisms comparing two choices of representatives. A factor \(f=gh\) then gives

\[ f'=g'(\beta^{-1}h\alpha). \tag{3.2} \]

Conversely, a factor for the new representatives gives one for the old representatives by using \(\alpha^{-1}\) and \(\beta^{-1}\). Thus the relation is independent of both representatives. If \(gh=gh'\), monicity of \(g\) gives \(h=h'\). If \(ha=hb\) for two parallel arrows into \(A\), then \(fa=gha=ghb=fb\); monicity of \(f\) gives \(a=b\). Hence \(h\) is monic.

Identity factors prove reflexivity. If \(f=gh\) and \(g=jl\), then \(f=j(lh)\), proving transitivity. For antisymmetry, suppose \(f=gh\) and \(g=fk\). Then \(f=fkh\) and \(g=ghk\). Cancelling the monomorphisms \(f\) and \(g\) gives

\[ kh=1_A,\qquad hk=1_B. \tag{3.3} \]

Thus \(h\) is an isomorphism over \(X\), and \([f]=[g]\). \(\square\)

There is a corresponding definition for epimorphisms \(e:X\to A\) and \(d:X\to B\). An isomorphism in \(X/\mathsf C\) is an underlying isomorphism \(h:A\to B\) with \(d=he\); its inverse lies in the coslice because \(e=h^{-1}d\). It is unique, since \(he=h'e\) and epicity of \(e\) imply \(h=h'\). A quotient of \(X\) is such an isomorphism class of epimorphisms from it. The tagged union \(\coprod_A\operatorname{Hom}(X,A)\), its subset of epimorphisms and its quotient classes give the ambient-set argument just used for subobjects. We do not identify a quotient class with a chosen underlying quotient set in an arbitrary category.

4. Forgetting every arrow except its component

A category is nonempty when it has an object. A discrete category has only identity arrows: two distinct objects have no arrow between them. It may have many objects, or no objects.

For objects \(X,Y\) of \(\mathsf C\), say that \(X\sim Y\) if there is a finite zigzag of arrows joining them, allowing either direction at each step. A length-zero zigzag gives reflexivity, reversal gives symmetry, and concatenation gives transitivity. Thus \(\sim\) is an equivalence relation. Any arrow gives a length-one zigzag. Conversely, an equivalence relation relating the endpoints of every arrow relates the endpoints of every such zigzag, by symmetry and transitivity. Hence \(\sim\) is exactly the equivalence relation generated by the existence of arrows, rather than merely by isomorphisms.

Write \(\pi_0(\mathsf C)\) for the ambient set of equivalence classes. It is a set because each class is a subset of the ambient object set, and the collection of classes is a subset of its power set. Regard it as a discrete category. There is a functor

\[ q:\mathsf C\longrightarrow\pi_0(\mathsf C) \tag{4.1} \]

sending \(X\) to \([X]\). Every arrow has endpoints in the same class and is sent to that class identity. Every original identity goes to an identity, and a composite goes to the same identity as the composite of its images. These statements prove both functor laws, including for an empty category.

Fix a class \(a\). In \((q\downarrow a)\), a structure arrow \([X]\to a\) exists exactly when \([X]=a\), and is then unique. The same holds for a structure arrow \(a\to[X]\) in \((a\downarrow q)\). In either case all original arrows between two such objects satisfy the structure equation, because their images are identities at \(a\). Forgetting the unique structure arrow therefore identifies each category with the full subcategory \(\mathsf C_a\) on the objects in \(a\). The inverse functor attaches the unique identity structure arrow. Both composites are identities on objects and morphisms, so these are category isomorphisms, in particular equivalences.

Each \(\mathsf C_a\) is nonempty, since \(a\) is an equivalence class of actual objects. Any two of its objects have a zigzag in \(\mathsf C\). Every intermediate object in that zigzag is in the same class \(a\), so the zigzag stays in this full subcategory. Thus \(\mathsf C_a\) is connected. Here connected means nonempty and joined by such finite zigzags. It follows that

\[ \begin{gathered} \mathsf C\text{ is connected}\\ \Longleftrightarrow\quad \pi_0(\mathsf C)\text{ is a singleton}. \end{gathered} \tag{4.2} \]

For the forward direction, every object has the same class and there is at least one object. For the reverse direction, a singleton set of classes gives at least one object; equality of every pair of classes gives a zigzag between every pair of objects. The empty category has empty \(\pi_0\), and is not connected under this convention. An initial or terminal object makes a category connected, since its arrows connect it to every object. Connectedness alone does not supply such an endpoint, as the third exercise shows.

5. Four graded exercises with full solutions

Exercise 1 (introductory: an empty relation is a zero map). Determine the initial, terminal and zero objects in \(\mathsf{Set}_{\mathcal U}\), pointed sets and \(\mathsf{Rel}_{\mathcal U}\). Describe every zero map when it exists. Explain why the empty relation behaves differently from an empty Hom set.

Solution. Section 2 proves that the initial sets are precisely the sets isomorphic to the empty set, hence just the empty set, and that singleton sets are terminal. Conversely a terminal set \(T\) has exactly one map from a singleton, so \(T\) has exactly one element. There is no zero object in \(\mathsf{Set}_{\mathcal U}\).

In pointed sets, the singleton pointed object is both initial and terminal by Section 2. Proposition 2.1 implies that every other initial or terminal pointed object is isomorphic to it, so has exactly one element. For \((X,x)\) and \((Y,y)\), the composite through this zero object sends every element of \(X\) to \(y\). It is the constant point-preserving function.

In \(\mathsf{Rel}_{\mathcal U}\), \(\operatorname{Hom}(\varnothing,X)=\mathcal P(\varnothing\times X)\) and \(\operatorname{Hom}(X,\varnothing)=\mathcal P(X\times\varnothing)\) each contain exactly one relation, the empty relation. Thus \(\varnothing\) is a zero object. The isomorphism criterion in Relations and cancellation, Theorem 4.1 says that isomorphic relation objects are related by a bijection of their underlying sets. Consequently every initial or terminal relation object is itself empty. The composite \(X\to\varnothing\to Y\) is the empty relation: an intermediate element cannot exist. This is the zero arrow \(X\to Y\).

For any \(X,Y\), even nonempty ones, the empty relation is an actual member of \(\mathcal P(X\times Y)\). Hence the Hom set in \(\mathsf{Rel}_{\mathcal U}\) is never empty. An empty Hom set would mean there is no arrow; the zero relation is one specified arrow.

Exercise 2 (intermediate: when an identity becomes zero). In a category with a zero object, prove that \(X\) is a zero object exactly when \(1_X=0_{XX}\). Deduce that a retract of a zero object is a zero object. Use no additive structure.

Solution. If \(X\) is a zero object, every endomorphism of \(X\) is its unique arrow to itself. In particular its identity and the arrow \(0_{XX}\) agree. Conversely, suppose they agree. There is at least one arrow each way between \(X\) and any \(Y\), supplied by the zero maps. For every \(f:X\to Y\), Proposition 2.2 gives \(f=f1_X=f0_{XX}=0_{XY}\). For every \(g:Y\to X\), it gives \(g=1_Xg=0_{XX}g=0_{YX}\). Thus both Hom sets are singletons, proving that \(X\) is initial and terminal.

Suppose \(s:Y\to X\), \(r:X\to Y\) satisfy \(rs=1_Y\), and \(X\) is zero. Then

\[ 1_Y=r1_Xs=r0_{XX}s=0_{YY}, \tag{5.1} \]

where both absorption laws in Proposition 2.2 give the last equality. The equivalence just proved makes \(Y\) a zero object. No subtraction or sum of arrows occurred.

Exercise 3 (advanced: the universal discrete destination). Let \(S\) be a set, regarded as a discrete category. Prove that giving a functor \(H:\mathsf C\to S\) is equivalent to giving a function \(\pi_0(\mathsf C)\to S\), through factorization by (4.1). Prove that a functor \(F:\mathsf C\to\mathsf D\) induces a function on components, and that the correspondence commutes with precomposition by \(F\) and postcomposition by a function of discrete sets. Give a connected category with neither an initial nor a terminal object.

Solution. If \(f:X\to Y\) is an arrow, \(H(f)\) is an arrow in the discrete category \(S\), so \(H(X)=H(Y)\). Reversing and concatenating this equality along a zigzag shows that \(H\) has a constant object value on each class. Therefore \(h([X])=H(X)\) is a well-defined function. Every arrow \(H(f)\) is the unique identity at its object value, so \(H=hq\) on arrows as well as on objects.

Conversely, any function \(h:\pi_0(\mathsf C)\to S\) gives the functor \(hq\), with identities as every arrow image. Recovering the function gives \(h\), since every component has an object representative; recovering a functor gives its original object and arrow assignments. The factorization is unique for the same reason. If \(\mathsf C\) is empty, the component set is empty, and each side has exactly one such assignment. If \(S\) is empty and \(\mathsf C\) is nonempty, neither side has an assignment, so that boundary case also agrees.

A functor \(F\) sends each arrow to an arrow and each finite zigzag to a finite zigzag. Hence \([X]\mapsto[F(X)]\) is well-defined; call it \(\pi_0(F)\). For \(H:\mathsf D\to S\), the function corresponding to \(HF\) sends \([X]\) to \(H(F(X))\), which is \(h(\pi_0(F)([X]))\). Thus precomposition corresponds exactly to composition with \(\pi_0(F)\). For a function \(b:S\to T\), postcomposition sends that value to \(b(h([X]))\), exactly the function \(bh\). Finally, \(\pi_0(1_{\mathsf C})\) fixes every class and \(\pi_0(GF)([X])=[G(F(X))]\), so identities and composition are respected.

The ordered category \(\mathbb Z\) is connected. Given \(m,n\), either \(m\le n\) or \(n\le m\), so an arrow in one of the two directions connects them. Section 2 proves that it has neither an initial nor a terminal object.

Exercise 4 (expert: a five-element subobject lattice). Classify all subobjects of \(Y=\{0,1\}\) in \(\mathsf{Rel}_{\mathcal U}\). Prove that the ordered set has five elements, with three incomparable elements strictly between its least and greatest elements. Compute their joins and meets and show that the lattice is not distributive. Compare it with the subobjects of \(Y\) in \(\mathsf{Set}_{\mathcal U}\).

Solution. Use the complete criterion from Relations and cancellation, Exercise 4: a monic relation \(R:X\to Y\) gives every \(x\) a private output \(y\), which no other input reaches. Private outputs for two distinct inputs are distinct. If \(X\) had three distinct elements, their three private outputs would have to be distinct elements of the two-element set \(Y\), which is impossible. Thus \(X\) has at most two elements.

If \(X\) is empty there is exactly one relation, and it is monic. Denote its class by \(E\). If \(X\) is a singleton, monicity means its row is nonempty, so the three possibilities are \(\{0\}\), \(\{1\}\) and \(\{0,1\}\). Denote their classes by \(A,B,C\), respectively. They are distinct subobjects: any isomorphism between singleton relation objects is the graph of their unique bijection, and precomposing with it preserves the row.

If \(X\) has two elements, their private outputs exhaust \(Y\). Each row contains its own private output and cannot contain the other's, by privacy. Thus both rows are the distinct singletons. The relation is the graph of a bijection. It is isomorphic over \(Y\) to \(1_Y\), so this gives the single class \(T=[1_Y]\). The isomorphism criterion used here is the full bijective-graph result in Relations and cancellation, Theorem 4.1. This proves completeness of the five classes, including all possible larger domains.

The empty relation from \(\varnothing\) to any domain gives a factor \(E\le[f]\) for every subobject. Each monomorphism itself factors through \(1_Y\), so every class is at most \(T\). Between two of \(A,B,C\), a potential factor is a relation from one singleton to another. It is either empty or the bijective graph. The empty factor produces an empty row, while the bijective factor produces the original target row. A nonempty row can therefore factor through a different one of these three rows in neither direction.

None of \(A,B,C\) equals \(E\), because an empty and a singleton domain are not bijective. None equals \(T\), because singleton and two-element domains are not bijective. The order is exactly the following diagram; every edge goes upwards.

\[ \begin{array}{ccccc} &&T&&\\[-2pt] &\nearrow&\uparrow&\nwarrow&\\[-2pt] A&&B&&C\\[-2pt] &\nwarrow&\uparrow&\nearrow&\\[-2pt] &&E&& \end{array} \tag{5.2} \]

Figure 1. The five subobjects of \(\{0,1\}\) in the relation category. The bottom element \(E\) lies below each of \(A,B,C\); each lies below \(T\); there are no comparisons between the three middle elements. Here \(A,B,C\) have the respective singleton rows \(\{0\},\{1\},\{0,1\}\). The preceding classification and factorization arguments prove every element and edge in the diagram.

The meet of two distinct middle elements is \(E\), since there is no other common lower bound. Their join is \(T\), since there is no other common upper bound. Meets and joins involving \(E,T\), or two equal elements, follow from leastness, greatestness and equality. Thus this is a lattice. Its distributive law fails:

\[ \begin{aligned} A\wedge(B\vee C)&=A\wedge T=A,\\ (A\wedge B)\vee(A\wedge C)&=E\vee E=E, \end{aligned} \tag{5.3} \]

and \(A\ne E\).

For sets, the injectivity characterization of monomorphisms in Relations and cancellation, Exercise 1 identifies subobjects with images of injective maps, up to bijection over \(Y\). Every such image is a subset of \(Y\), and every subset has its inclusion. The factorization order is subset containment: a factor sends each element to the unique element of the larger subset with the same image, and conversely any factor makes the smaller image contained in the larger. The four subsets are \(\varnothing,\{0\},\{1\},Y\). Their graphs give \(E,A,B,T\) above, preserving exactly those comparisons. The extra relational subobject \(C\) is not a function graph: its sole input has two outputs. This is the additional middle element responsible for (5.3).

References