Self-checked by the writing AI.

Coimages, images and composition of quotients

A map can identify elements in its source and place the resulting object inside its target. In sets and modules these two operations meet at the same object. In topological spaces they can give the same underlying set with different topologies. Universal properties let us identify exactly which conclusions survive in a general category.

We work in a locally small category $\mathcal C$ with finite limits and finite colimits. These hypotheses supply pullbacks, pushouts, equalizers and coequalizers. The intrinsic definition of a strict epimorphism, including its image-presheaf interpretation, is in Generators and small quotient families, §4. We will compare that definition with the constructions here. No assumption about stability of quotients under pullback is made.

1. Two constructions attached to a map

Fix $f:X\to Y$. Write its kernel pair and its self-pushout as $R_f=X\times_YX,\qquad S_f=Y\amalg_XY.$ The projections are $p_1,p_2:R_f\to X$; the pushout maps are $j_1,j_2:Y\to S_f$. Define $\begin{gathered} q_f:X\longrightarrow C_f,\qquad i_f:I_f\longrightarrow Y,\\ C_f=\operatorname{Coim}f=\operatorname{coeq}(p_1,p_2),\\ I_f=\operatorname{Im}f=\operatorname{eq}(j_1,j_2). \end{gathered} \tag{1.1}$ A coequalizer is epic: two maps out of its target agreeing after the coequalizer are the same factorization. Dually an equalizer is monic. Thus $q_f$ is epic and $i_f$ is monic.

The equation $fp_1=fp_2$ gives a unique $v_f:C_f\to Y$ with $v_fq_f=f$. Since $j_1f=j_2f$ and $q_f$ is epic, $j_1v_f=j_2v_f$. Hence there is a unique comparison $\begin{gathered} u_f:C_f\longrightarrow I_f,\\ f=i_fu_fq_f,\qquad v_f=i_fu_f. \end{gathered} \tag{1.2}$ Uniqueness follows by canceling the epic $q_f$ and then the monic $i_f$. We call $f$ strict if this specified comparison is an isomorphism. An abstract isomorphism between $C_f$ and $I_f$, without compatibility with (1.2), is insufficient.

Two alternate constructions are useful: $\begin{gathered} C_f\simeq X\amalg_{R_f}X,\\ I_f\simeq Y\times_{S_f}Y. \end{gathered} \tag{1.3}$ Here the maps $R_f\to X$ in the pushout are $p_1,p_2$, and the maps $Y\to S_f$ in the pullback are $j_1,j_2$. We check their structural maps.

A map from the pushout on the left to $Z$ is a pair $a,b:X\to Z$ with $ap_1=bp_2$. The diagonal $d:X\to R_f$ satisfies $p_1d=p_2d=\operatorname{id}_X$, so this equation forces $a=b$. The remaining condition is $ap_1=ap_2$, precisely the condition for a map $C_f\to Z$. Thus both maps $X\to X\amalg_{R_f}X$ identify with $q_f$.

For the other isomorphism, the pushout has a codiagonal $s:S_f\to Y$ with $sj_1=sj_2=\operatorname{id}_Y$. If maps $a,b:W\to Y$ satisfy $j_1a=j_2b$, applying $s$ gives $a=b$. The equation then says $j_1a=j_2a$, precisely the equalizer condition. Both projections $Y\times_{S_f}Y\to Y$ identify with $i_f$. These universal-property identifications hold for all test objects and determine the canonical isomorphisms in (1.3).

2. Cancellation and the comparison

We retain Stacks, Lemma 4.13.3, Tag 08LR: a map is monic exactly when its diagonal exhibits its source as its kernel pair, and epic exactly when its codiagonal exhibits its target as its self-pushout. The specified projections and inclusions are part of these assertions. Applied to (1.3), this gives $\begin{gathered} f\text{ monic}\ \Longleftrightarrow\ q_f\text{ invertible},\\ f\text{ epic}\ \Longleftrightarrow\ i_f\text{ invertible}. \end{gathered} \tag{2.1}$ Here is the interface with our constructions. If $f$ is monic, $R_f$ is $X$ with both projections the identity, whose coequalizer is $\operatorname{id}_X$. Conversely, if $q_f$ is invertible, $q_fp_1=q_fp_2$ gives $p_1=p_2$. Every pair $a,b:W\to X$ with $fa=fb$ factors through this kernel pair, so $a=b$.

If $f$ is epic, $j_1f=j_2f$ gives $j_1=j_2$, whose equalizer is $\operatorname{id}_Y$. Conversely, if $i_f$ is invertible, then $j_1=j_2$. Given $a,b:Y\to Z$ with $af=bf$, the pushout produces $t:S_f\to Z$ with $tj_1=a$ and $tj_2=b$; thus $a=b$.

A useful strengthening is $\begin{gathered} f\text{ epic}\ \Longleftrightarrow\ i_f\text{ invertible},\\ i_f\text{ invertible}\ \Longleftrightarrow\ i_f\text{ epic}. \end{gathered} \tag{2.2}$ Only the last implication needs checking. If $i_f$ is epic, $j_1i_f=j_2i_f$ forces $j_1=j_2$, so $i_f$ is an isomorphism. The dual statement says that $q_f$ is monic exactly when it is invertible, exactly when $f$ is monic. These conclusions use the particular equalizer and coequalizer in (1.1), without assuming that an arbitrary morphism which is both monic and epic is invertible.

In particular, if $f$ is epic, strictness is equivalent to invertibility of $v_f:C_f\to Y$. If $f$ is monic, strictness is equivalent to invertibility of $u_f:X\simeq C_f\to I_f$. Strict monomorphisms can therefore be recovered by reversing the strict-epimorphism results below.

3. Strict epimorphisms and descent

A regular epimorphism is a coequalizer of some parallel pair. This is the canonical convention in Adámek–Rosický–Vitale, §0.16. Their Example 3.4 records that, when kernel pairs exist, a regular epimorphism coequalizes its own kernel pair. We retain this result and make its exact interface with strictness explicit.

Theorem 3.1. For $f:X\to Y$, the following five conditions are equivalent:

$f$ is strict and epic.
$v_f:C_f\to Y$ is an isomorphism.
$f$ is a coequalizer of $p_1,p_2:X\times_YX\to X$.
$f$ is a regular epimorphism.
For every $Z$, precomposition with $f$ bijects $\operatorname{Hom}(Y,Z)$ with the maps $r:X\to Z$ such that $ra=rb$ for every pair $a,b:W\to X$ satisfying $fa=fb$.

Proof. Conditions 1 and 2 agree by (2.2). Conditions 2 and 3 agree by uniqueness of a coequalizer with its specified map from $X$. Condition 3 implies 4.

For $4\Rightarrow3$, suppose $f$ is a coequalizer of $a,b:W\to X$. The equation $fa=fb$ gives a map $W\to R_f$ with projections $a,b$. Thus $q_fa=q_fb$, and the coequalizer property of $f$ gives $t:Y\to C_f$ with $tf=q_f$. Now $tv_fq_f=tf=q_f,\qquad v_ftf=v_fq_f=f.$ Cancel the epimorphisms $q_f$ and $f$ to get $tv_f=\operatorname{id}_{C_f}$ and $v_ft=\operatorname{id}_Y$. Hence condition 2 holds.

A map $r:X\to Z$ satisfies the condition in 5 exactly when $rp_1=rp_2$. Necessity uses the particular equalized pair $p_1,p_2$. Sufficiency uses the map $W\to R_f$ associated to each pair $a,b$. Condition 5 is therefore precisely the universal property in 3. $\square$

Thus the intrinsic strict epimorphisms defined in the prerequisite lesson are exactly the regular epimorphisms under the present hypotheses. The intrinsic definition itself makes sense in categories without these finite constructions.

Every $q_f:X\to C_f$ is a strict epimorphism, because it is a coequalizer. A strict epimorphism which is monic is an isomorphism: by monicity its kernel pair has both projections the identity, and the identity of its source is their coequalizer.

We also retain the stronger canonical fact from Adámek–Rosický–Vitale, §0.17: every regular epimorphism is strong, and hence extremal. The checked lifting argument is as follows. In a commutative square $mq'=bf$, where $f:X\to Y$ is regular epic, $q':X\to A$, $b:Y\to B$, and $m:A\to B$ is monic, the equations $fp_1=fp_2$ imply $mq'p_1=mq'p_2$. Cancel $m$; then Theorem 3.1 gives a unique $d:Y\to A$ with $df=q'$. Since $mdf=mq'=bf$, cancel $f$ to get $md=b$. This is the required unique diagonal. If $f=mq'$, take $b=\operatorname{id}_Y$; then $md=\operatorname{id}_Y$, and $mdm=m$ gives $dm=\operatorname{id}_A$ by monicity. Thus $m$ is invertible. These implications do not say that every epimorphism is regular.

Descent lemma 3.2. If $f=hg$, where $g:X\to Z$ is a strict epimorphism and $h:Z\to Y$, there is a unique $a:Z\to C_f$ with $ag=q_f,\qquad v_fa=h. \tag{3.1}$

Proof. If $gs=gt$ for maps $s,t:W\to X$, then $fs=ft$, and the kernel-pair property gives $q_fs=q_ft$. Intrinsic strictness of $g$ therefore descends $q_f$ uniquely to $a$ with $ag=q_f$. Then $v_fag=f=hg$; cancel the epic $g$ to get $v_fa=h$. $\square$

This lemma controls factorizations through a quotient even when $v_f$ is not known to be monic.

4. When quotients compose

Call $\mathcal C$ balanced if every morphism which is both monic and epic is invertible. Consider these properties:

(A) Every morphism is strict.
(B) Every epimorphism is strict.
(C) For every $f$, the canonical $v_f:C_f\to Y$ is monic.
(D) The category is balanced.
(E) Composites of strict epimorphisms are strict epimorphisms.

Theorem 4.1. With only finite limits and finite colimits, $\begin{gathered} \text{(A)}\ \Longrightarrow\ \text{(B)},\\ \text{(B)}\ \Longleftrightarrow\ \bigl(\text{(C) and (D)}\bigr),\\ \text{(C)}\ \Longleftrightarrow\ \text{(E)}. \end{gathered} \tag{4.1}$

Proof of $(E)\Rightarrow(C)$. Write $f=vq$, where $q=q_f:X\to I=C_f$, and form $t=q_v:I\to J=C_v$. Both $q$ and $t$ are strict epimorphisms. For every pair $a,b:W\to X$, $\begin{gathered} fa=fb\ \Longleftrightarrow\ qa=qb,\\ qa=qb\ \Longleftrightarrow\ tqa=tqb. \end{gathered} \tag{4.2}$ For the first equivalence, the forward direction follows by factoring $(a,b)$ through $R_f$; the reverse follows from $f=vq$. For the second, $qa=qb$ implies the displayed equality after $t$. Conversely, $v$ factors through $t$, so $tqa=tqb$ implies $fa=fb$, hence $qa=qb$.

In particular, $q$ and $tq$ have the same kernel pair, with the same projections to $X$: their universal cones are exactly the same pairs by (4.2). The map $q$ coequalizes this pair, since $R_q=R_f$ by the first equivalence. By (E), $tq$ also coequalizes this pair. Consequently the comparison $t:I\to J$ between these two coequalizers is invertible. Explicitly, the coequalizer property of $tq$ gives $r:J\to I$ with $rtq=q$. Cancel $q$ to get $rt=\operatorname{id}_I$, and cancel $tq$ in $trtq=tq$ to get $tr=\operatorname{id}_J$. By (2.1) applied to $v$, invertibility of $q_v=t$ makes $v$ monic. This proves (C).

Proof of $(C)\Rightarrow(E)$. Let $g:X\to Y$ and $h:Y\to Z$ be strict epimorphisms. Factor $\begin{gathered} k=hg=mq,\\ q=q_k:X\to W=C_k,\qquad m=v_k:W\to Z. \end{gathered}$ Property (C) makes $m$ monic. Lemma 3.2 gives $a:Y\to W$ with $ag=q$ and $ma=h$.

For any $s,t:V\to Y$, $hs=ht$ implies $mas=mat$, so $as=at$ by monicity; the reverse implication is immediate. Thus $a$ equalizes the kernel pair of $h$. Theorem 3.1 gives $b:Z\to W$ with $bh=a$. Now $\begin{gathered} mbh=ma=h,\\ bmq=bmag=bhg=ag=q. \end{gathered}$ Cancel $h$ and $q$ to obtain $mb=\operatorname{id}_Z$ and $bm=\operatorname{id}_W$. Hence $m$ is invertible, and $k$ is a strict epimorphism by Theorem 3.1. This proves (E).

Proof of the other implications. Property (A) implies (B). If (B) holds, composites of epimorphisms are epic, so every composite of strict epimorphisms is strict by (B). The equivalence just proved gives (C). A map which is both epic and monic is then a strict epimorphism and monic, hence invertible by §3; this is (D).

Conversely assume (C) and (D), and let $f:X\to Y$ be epic. Its factor $v_f$ is monic by (C), and epic because $v_fq_f=f$ is epic: agreement after $v_f$ implies agreement after $f$. Balancedness makes $v_f$ invertible. Theorem 3.1 now makes $f$ strict, proving (B). $\square$

Corollary 4.2. If every epimorphism is strict, each factorization $f=me$ with $e:X\to Z$ epic and $m:Z\to Y$ monic has a unique isomorphism $a:Z\to C_f$ satisfying $ae=q_f$ and $v_fa=m$.

Proof. Theorem 4.1 makes $v_f$ monic, and the assumption makes $e$ strict. Monicity of $m$ says $es=et$ exactly when $fs=ft$; therefore $e$ and $q_f$ are coequalizers of the same kernel pair. Their unique compatible comparison is an isomorphism. Its compatibility with $m$ follows by canceling $e$. $\square$

The isomorphism conclusion also holds whenever this particular $e$ is strict and $m$ is monic, without assuming that every epimorphism is strict. The same kernel-pair argument proves it. Ordinary epimorphisms alone are insufficient, as Exercise 2 shows.

5. Sets, modules, presheaves and spaces

For a function $f:X\to Y$, its kernel pair identifies exactly the elements with the same value. The coequalizer is the set of fibres, identified with the subset $f(X)\subset Y$ by $[x]\mapsto f(x)$. This identification is bijective, including the empty-source case. The self-pushout consists of two copies of $Y$, with the two copies of each $f(x)$ identified. No point outside $f(X)$ changes copies. Thus $j_1(y)=j_2(y)$ exactly for $y\in f(X)$, and its equalizer is $f(X)$. The comparison $u_f$ is the identity under these identifications, so every map of sets is strict.

For modules, we retain the abelian-category interface in Sheaves of modules on a ringed space, Section 2, Theorem 2.1. Its statement includes sheaves of left modules over noncommutative rings. On a one-point space these are exactly left modules over the given ring: a sheaf is determined by its module on the point, with the zero module on the empty open set, and sheaf maps are module maps. The result therefore gives the ordinary coimage–image isomorphism for every unital ring. The following calculation checks its interface with our kernel-pair and self-pushout definitions.

For a homomorphism $f:M\to N$ of left $R$-modules, its kernel pair consists of pairs $(x,x')$ with $x-x'\in\ker f$. A linear map $r:M\to L$ equalizes the two projections exactly when it kills $\ker f$: use pairs $(k,0)$ for necessity and differences for sufficiency. Hence $C_f=M/\ker f$, with its specified quotient map.

The self-pushout is $\begin{gathered} L_f=\{(f(x),-f(x)):x\in M\},\\ S_f=(N\oplus N)/L_f. \end{gathered} \tag{5.1}$ Indeed maps from it to $L$ correspond to pairs $a,b:N\to L$ with $af=bf$. For $y\in N$, its two images agree exactly when $(y,-y)=(f(x),-f(x))$ for some $x$, which means $y\in f(M)$. Thus $I_f=f(M)$, with its induced module structure, and $u_f$ is $M/\ker f\longrightarrow f(M),\qquad [x]\longmapsto f(x).$ It is well-defined, injective, surjective and linear. Its inverse is linear by applying the bijection to the addition and scalar equations. This checks strictness for all left modules, with no commutativity or finiteness restriction on the ring or modules.

For a small category $\mathcal A$, let $F,G:\mathcal A^{op}\to\operatorname{Set}$ and $f:F\to G$ be natural. Their kernel pair, self-pushout, coequalizer and equalizer are computed objectwise. For example, restrictions send a pair of elements with equal $f$-values to another such pair, and send a fibre-equivalence class to the class of its restriction. The pointwise universal factors are natural because their defining equations commute with restrictions and because the universal factors are unique. The same reasoning supplies all four finite constructions and their universal properties.

Consequently $\begin{gathered} I_f(a)=f_a(F(a))\subseteq G(a),\\ C_f(a)=F(a)/{\sim_f}. \end{gathered} \tag{5.2}$ where $x\sim_f x'$ means $f_a(x)=f_a(x')$. Naturality sends the first subset into the corresponding subset at any restricted object. The bijections $[x]\mapsto f_a(x)$ commute with restrictions, giving a natural isomorphism $C_f\simeq I_f$. Thus every map of set-valued presheaves is strict, including presheaves on the empty category.

For the category $\operatorname{Top}$ of all topological spaces, limits have the usual subspace topology in products, and pushouts are quotient spaces of disjoint unions. To check the self-pushout, a map out of its quotient is continuous exactly when its restrictions to the two copies of $Y$ are continuous and agree after $f$. Its underlying set is therefore the same pushout set described above. Its equalizer has underlying set $f(X)$ and the subspace topology inherited from $Y$.

A coequalizer of a parallel pair of continuous maps is the set quotient by the generated equivalence relation, with the quotient topology: a map out is continuous exactly when its composite with the quotient map is continuous. For the kernel pair of $f$, this relation is precisely equality of $f$-values. Hence $\begin{gathered} C_f=(f(X),\tau_{\mathrm{quot}}),\\ I_f=(f(X),\tau_{\mathrm{sub}}). \end{gathered} \tag{5.3}$ Here $\tau_{\mathrm{quot}}$ is the quotient topology from $X$, and $\tau_{\mathrm{sub}}$ is the subspace topology from $Y$. The canonical comparison is the identity of the underlying set. It is continuous since every subspace-open $V\subseteq f(X)$ has open inverse image in $X$. It is an isomorphism exactly when the two topologies agree. Thus a continuous map is strict exactly when it is a quotient map onto its image equipped with the subspace topology.

Monomorphisms in $\operatorname{Top}$ are injective: test distinct points with a singleton; conversely an injective map cancels underlying functions. Epimorphisms are surjective. Surjectivity implies cancellation; if a point is missing from the image, two distinct functions to the indiscrete two-point space can agree on the image, and both are continuous. Thus strict epimorphisms are exactly quotient maps, and strict monomorphisms are exactly embeddings.

Every $v_f$ in $\operatorname{Top}$ is injective, so property (C) holds. Quotient maps also compose directly: a subset of the final target is open exactly when its inverse image under the second quotient is open, exactly when its inverse image under the composite is open. Property (E) holds. Balancedness fails: the identity from a discrete two-point space to an indiscrete two-point space is continuous, monic and epic, with discontinuous inverse. Consequently (B) and (A) fail. These examples concern all spaces; imposing a Hausdorff condition would change the colimit constructions.

6. The image presheaf remembers the descent equations

The categorical image $I_f$ need not represent the pointwise image of the Yoneda map. Keep these two objects distinct. Set $\begin{gathered} Q_f(W)=\{f\circ a\mid a: W\to X\}\\ \subseteq\operatorname{Hom}(W,Y). \end{gathered} \tag{6.1}$ Precomposition makes $Q_f$ a subpresheaf of $h_Y=\operatorname{Hom}(-,Y)$. There is a natural bijection $\operatorname{Hom}(C_f,Z) \simeq \operatorname{Nat}(Q_f,h_Z). \tag{6.2}$ This is an identification with maps into a representable presheaf, without asserting that $Q_f$ is representable.

Explicitly, $t:C_f\to Z$ gives $\beta_W(fa)=tq_fa,\qquad a:W\to X. \tag{6.3}$ If $fa=fb$, then $q_fa=q_fb$, so the formula is well-defined. For $c:V\to W$, it sends $fac$ to $tq_fac=\beta_W(fa)c$, proving naturality.

Conversely let $\beta:Q_f\to h_Z$, and set $r=\beta_X(f):X\to Z$. Naturality along $a:W\to X$ gives $\beta_W(fa)=ra$. Hence $fa=fb$ implies $ra=rb$. In particular $rp_1=rp_2$, so $r=tq_f$ for a unique $t:C_f\to Z$. Starting with $t$, this procedure recovers $t$ by canceling $q_f$. Starting with $\beta$, it recovers every component by its displayed formula. Postcomposition with $Z\to Z'$ commutes with (6.3), proving naturality in $Z$.

For example, a surjective module map need not split. For $q:\mathbb Z\to\mathbb Z/2$, $Q_q(\mathbb Z/2)$ consists only of the zero map, whereas $h_{\mathbb Z/2}(\mathbb Z/2)$ contains the identity. Indeed a homomorphism from $\mathbb Z/2$ to $\mathbb Z$ sends its generator to an element killed by $2$, hence to zero. Nevertheless $C_q=I_q=\mathbb Z/2$, and (6.2) holds. The descent equations captured by $Q_q$ determine maps out of the quotient, even when its individual components omit many maps into that quotient.

7. Graded exercises with complete solutions

Exercise 1 (warm-up: an ordered category). Let $S$ be a nonempty set and regard its power set as a category: there is one arrow $A\to B$ exactly when $A\subseteq B$. Compute the coimage, image and comparison of every arrow. Determine which of (A)–(E) hold.

Solution. Finite products are intersections, finite coproducts are unions, and the initial and terminal objects are $\varnothing$ and $S$. Every parallel pair consists of the same arrow, so its equalizer and coequalizer are the appropriate identity arrows. Pullbacks and pushouts are the corresponding intersections and unions over the stated objects, so the required finite constructions exist.

For $f:A\subseteq B$, its kernel pair is $A$, with both projections the identity, and its self-pushout is $B$, with both inclusions the identity. Thus $C_f=A$, $I_f=B$, $q_f=\operatorname{id}_A$, $i_f=\operatorname{id}_B$, and $u_f=f$. The arrow is strict exactly when $A=B$. Every arrow is both monic and epic, since each relevant Hom set has at most one element. In particular $\varnothing\to S$ refutes (A), (B) and (D). Every $v_f$ is monic, so (C) holds. Strict epimorphisms are the identity arrows; their composites are identities, so (E) holds. This example keeps the equivalence (C)$\Leftrightarrow$(E) separate from balancedness.

Exercise 2 (standard: the same image with two topologies). Give $X=\{0,1,2\}$ the topology $\{\varnothing,\{0\},\{0,1\},X\}$, and $Y=\{a,b,c\}$ the topology $\{\varnothing,\{a,b\},Y\}$. Define $f(0)=f(1)=a$, $f(2)=b$. Compute $R_f$, $S_f$, $C_f$, $I_f$, and $u_f$. Exhibit two factorizations of $f$ as an epimorphism followed by a monomorphism whose middle spaces are not isomorphic.

Solution. The only proper nonempty open set of $Y$ has inverse image $X$, so $f$ is continuous. The kernel pair is $\{(0,0),(0,1),(1,0),(1,1),(2,2)\}$ with the subspace topology from $X\times X$ and coordinate projections.

The self-pushout has points $a,b,c_1,c_2$; the first two points have their two copies identified, and $c_1,c_2$ retain their copies. A subset is open exactly when its inverse images in both copies of $Y$ are open. This gives precisely $\begin{gathered} \varnothing,\quad \{a,b\},\quad \{a,b,c_1\},\\ \{a,b,c_2\},\quad \{a,b,c_1,c_2\}. \end{gathered}$ For completeness, an open subset containing a $c_i$ must contain $a,b$, since its restriction to that copy must be all of $Y$. A nonempty open subset without either $c_i$ must likewise be $\{a,b\}$, by the restrictions. All the listed subsets satisfy these tests.

The coimage has points $a,b$. Their inverse images in $X$ are $\{0,1\}$ and $\{2\}$, respectively, so its topology is $\{\varnothing,\{a\},\{a,b\}\}$. The image has the same points with the indiscrete topology induced by $Y$. Thus $u_f$ is a continuous bijection from the first space to the second with discontinuous inverse; $f$ is not strict.

One factorization uses $C_f$: its first map is a quotient map, hence strict epic; its second map $v_f:C_f\to Y$ is continuous and injective, hence monic. Another uses $I_f$: the first map $X\to I_f$ is continuous and surjective, hence epic, and the second is the subspace inclusion, hence monic. These middle spaces have different numbers of open subsets and cannot be isomorphic. The first map in the second factorization is not strict: the subset $\{a\}$ is not open although its inverse image is open. Thus the uniqueness result requiring a strict first factor applies only to the first factorization.

Exercise 3 (advanced: functorial comparisons). An object of the arrow category is a map $f:X\to Y$; a morphism $f\to f'$ is a square $(a,b)$ with $bf=f'a$. After choosing the finite constructions in (1.1), prove that coimage and image define functors on the arrow category and that $u_f$ is a natural transformation between them.

Solution. Let $p_1,p_2:R_f\to X$. The equation $f'ap_1=bfp_1=bfp_2=f'ap_2$ gives a map $R_f\to R_{f'}$ with projections $ap_1,ap_2$. Consequently $q_{f'}ap_1=q_{f'}ap_2$, so there is a unique $\bar a:C_f\to C_{f'},\qquad \bar a q_f=q_{f'}a.$ For images, the square induces $s:S_f\to S_{f'}$ with $sj_1=j'_1b$ and $sj_2=j'_2b$: these maps from the two copies of $Y$ agree on $X$ by $bf=f'a$. Hence $j'_1bi_f=sj_1i_f=sj_2i_f=j'_2bi_f,$ so there is a unique $\bar b:I_f\to I_{f'}$ with $i_{f'}\bar b=bi_f$.

For an identity square, both induced maps satisfy the equations of identity maps, so uniqueness makes them identities. For two composable squares, the composites of induced coimage maps satisfy the defining equation for the composite square; canceling $q_f$ proves equality with its induced map. The same conclusion for images follows by canceling the final equalizer inclusion. Thus these assignments are functors.

Finally, $\begin{aligned} i_{f'}\bar b u_fq_f &=bi_fu_fq_f=bf=f'a\\ &=i_{f'}u_{f'}q_{f'}a =i_{f'}u_{f'}\bar a q_f. \end{aligned}$ Cancel $i_{f'}$ and $q_f$ to obtain $\bar b u_f=u_{f'}\bar a$. This proves naturality. On the full subcategory of strict arrows, these comparisons form a natural isomorphism. Every cancellation involved the specified universal maps, so the result also tracks the choices of finite constructions.

Exercise 4 (expert: pushouts preserve quotients, pullbacks may fail). Prove that a pushout of a strict epimorphism is a strict epimorphism. Then construct an explicit quotient map of finite spaces whose pullback along a subspace inclusion is not a strict epimorphism. Conclude that properties (C) and (E) do not imply stability under pullback.

Solution. Let $q:X\to Y$ be strict epic, hence a coequalizer of a pair $d_1,d_2:W\to X$. Push it out along $r:X\to X'$, obtaining maps $q':X'\to Y'$ and $r':Y\to Y'$ with $q'r=r'q$. We prove that $q'$ coequalizes $rd_1,rd_2$.

It equalizes them by the pushout equation. If $t:X'\to Z$ equalizes them, then $tr$ equalizes $d_1,d_2$. There is a unique $v:Y\to Z$ with $vq=tr$. The pushout gives $w:Y'\to Z$ with $wq'=t$ and $wr'=v$. If $w'$ also satisfies $w'q'=t$, then $w'r'q=w'q'r=tr=vq.$ Cancel $q$, so $w'r'=v$; the pushout property now gives $w'=w$. Hence $q'$ is regular epic and therefore strict epic.

For the counterexample, let $X$ be the disjoint union of two two-point spaces. Name their points $a_1,b_1$ and $b_2,c_2$. Their open sets are, respectively, $\varnothing,\{b_1\},\{a_1,b_1\}, \qquad \varnothing,\{c_2\},\{b_2,c_2\}.$ Let $Y=\{a,b,c\}$, with open sets $\varnothing,\{c\},\{b,c\},Y$, and define $q(a_1)=a$, $q(b_1)=q(b_2)=b$, $q(c_2)=c$. For a subset $U\subseteq Y$, its inverse image is open exactly when $a\in U\Rightarrow b\in U,\qquad b\in U\Rightarrow c\in U.$ These are precisely the conditions defining the listed open subsets of $Y$. The surjection $q$ is therefore a quotient map, hence strict epic.

Take the subspace $T=\{a,c\}\subset Y$, with topology $\{\varnothing,\{c\},T\}$. Its inverse image is $\{a_1,c_2\}$, a discrete subspace of $X$: the open component $\{a_1,b_1\}$ isolates $a_1$, and the open set $\{c_2\}$ isolates $c_2$. The pullback $X\times_YT$ identifies with this inverse-image subspace. Indeed the map $x\mapsto(x,q(x))$ is continuous into the product and has continuous inverse given by the first projection, and the pullback universal property identifies exactly these pairs.

The pulled-back map is the continuous bijection from this discrete two-point space to $T$. It is not a quotient map, because $\{a\}\subset T$ has open inverse image but is not open. Hence it is epic and monic but not strict epic. Property (C), and therefore (E), holds in all of $\operatorname{Top}$ by §5, while this quotient fails stability under pullback. That stability is an additional hypothesis.

Original lesson text: CC0 1.0. Named canonical results are cited with their checked interfaces; no source prose is reproduced. Self-checked; no independent review has occurred.