Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 25 (Category Theory)

For the covariant version of the Yoneda embedding, put $Y(A)=\mathcal{C}(A,-)$. A morphism $u:A\to B$ induces the natural transformation $Y(u):Y(B)\to Y(A)$ whose component sends $v:B\to X$ to $vu$. Thus $Y$ is a functor from the opposite category ${\mathcal{C}}^{\mathrm{op}}$ to the functor category $[\mathcal{C},\mathbf{Set}]$. The locally small category hypothesis makes each value a set.

The Yoneda lemma says that, for a functor $F:\mathcal{C}\to \mathbf{Set}$, evaluation at the identity morphism gives a natural bijectionGiven $a\in F(A)$, define ${\theta }_X^a(v)=F(v)(a)$. For $w:X\to X^{\prime }$, the functor law gives $F(w){\theta }_X^a(v)=F(wv)(a)={\theta }_{X^{\prime }}^a(wv)$, proving naturality. Conversely, naturality of $\theta$ at $v:A\to X$ gives ${\theta }_X(v)=F(v){\theta }_A(1_A)$. These two constructions are inverse, because $F(1_A)(a)=a$.

The bijection is natural in both variables. A natural transformation $\beta :F\to H$ sends the element $a$ to ${\beta }_A(a)$, matching postcomposition by $\beta$. For $u:A\to B$, precomposing with $Y(u)$ sends $\theta$ to a transformation from $Y(B)$ and its distinguished element is ${\theta }_B(u)=F(u)(a)$. This is exactly the required dependence on $A$.

Taking $F=Y(B)$ gives $\mathrm{Nat}(Y(A),Y(B))\cong \mathcal{C}(B,A)$. Consequently the covariant Yoneda embedding is full and faithful, with the reversal of arrows accounted for by its opposite category domain.

Suppose every component ${\alpha }_A$ is a monomorphism in the Category of sets, hence an injective function. If $\beta ,\gamma :H\to F$ are natural transformations with $\alpha \beta =\alpha \gamma$, then ${\alpha }_A{\beta }_A={\alpha }_A{\gamma }_A$ at every object. Componentwise cancellation gives ${\beta }_A={\gamma }_A$, so $\beta =\gamma$. Thus $\alpha$ is a monomorphism in the functor category.

Conversely, suppose $\alpha$ is a monomorphism and $x,y\in F(A)$ satisfy ${\alpha }_A(x)={\alpha }_A(y)$. The Yoneda lemma supplies natural transformations ${\theta }^x,{\theta }^y:Y(A)\to F$. Its naturality in $F$ identifies $\alpha {\theta }^x$ and $\alpha {\theta }^y$ with the equal elements ${\alpha }_A(x)$ and ${\alpha }_A(y)$. Cancellation by $\alpha$ gives ${\theta }^x={\theta }^y$; evaluating at $1_A$ gives $x=y$. Therefore a natural transformation is a monomorphism exactly when all its components are injective functions. This is the pointwise monomorphism in a set-valued functor category criterion.

Let $D:J\to \mathcal{C}$ be a finite diagram in a category. Form the finite products in a categoryDefine $s,t:P\rightrightarrows Q$ by ${\pi }_{a}s=D(a){\pi }_j$ and ${\pi }_{a}t={\pi }_k$. Take their equalizer $e:L\to P$. The maps ${\lambda }_j={\pi }_{j}e$ satisfy $D(a){\lambda }_j={\lambda }_k$, so they constitute a categorical cone.

Any other categorical cone $x_j:X\to D(j)$ gives a unique $x:X\to P$ by the product in a category property. Its cone equations say $sx=tx$, so the equalizer gives a unique $\bar{x}:X\to L$ with $e\bar{x}=x$. The projections then give ${\lambda }_j\bar{x}=x_j$, and uniqueness follows from both universal properties. Thus this equalizer of two product maps is the finite limit of the diagram in a category.

The empty indexing category is included: its empty product in a category is a terminal object, which is the empty categorical limit. Hence no nonemptiness restriction has entered the construction.

With a terminal object $1$, the pullback in a category of $A\to 1\leftarrow B$ is the binary product in a category $A\times B$. Repetition gives all nonempty finite products in a category, and $1$ supplies the empty one.

For $f,g:A\rightrightarrows B$, form the diagonal ${\Delta }_B:B\to B\times B$ and pull it back along $⟨f,g⟩:A\to B\times B$. Write the resulting maps as $e:E\to A$ and $h:E\to B$. The equation $⟨f,g⟩e={\Delta }_{B}h$ says $fe=ge=h$. A map $x:X\to A$ with $fx=gx$ pairs with $fx:X\to B$ to give a unique map into this pullback in a category. Thus $e$ is the equalizer of $f,g$.

We now have the finite products in a category and equalizers needed in part (a). A terminal object and pullbacks in a category therefore suffice for all finite limits.

An object of the slice category $\mathcal{C}/A$ is a morphism $x:X\to A$; a map from $x$ to $y:Y\to A$ is $u:X\to Y$ with $yu=x$. The identity morphism $1_A:A\to A$ is a terminal object of this slice category, even when $\mathcal{C}$ itself has no terminal object.

For a cospan $X\stackrel{u}{\to }Z\stackrel{v}{\leftarrow }Y$ in the slice category, form its underlying pullback in a category $P$ in $\mathcal{C}$, with maps $p:P\to X$ and $q:P\to Y$. If $x,y,z$ are the structure maps to $A$, then $xp=zup=zvq=yq$. This common map equips $P$ with its slice structure. An underlying mediating map into $P$ is automatically over $A$, because its composite with $xp$ is the prescribed structure map. Hence the same square is a pullback in a category in $\mathcal{C}/A$.

Part (b) applies to the slice's terminal object and pullbacks in a category. Every slice category has all finite limits when the ambient category has pullbacks in a category. The finite completeness of slice categories result needs no ambient terminal-object hypothesis.

A monomorphism $m:A\to B$ satisfies $mu=mv\Rightarrow u=v$ for every parallel pair into $A$. An epimorphism $e:X\to Y$ satisfies $ue=ve\Rightarrow u=v$ for every parallel pair out of $Y$. A strong monomorphism is a monomorphism with the following lifting property: whenever $e:X\to Y$ is an epimorphism and $mu=ve$, with $u:X\to A$ and $v:Y\to B$, there is a unique $d:Y\to A$ with $de=u$ and $md=v$. A regular monomorphism is a morphism occurring as the equalizer of two parallel maps. A balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism.

Suppose $m:A\to B$ equalizes $p,q:B\rightrightarrows D$. If $mu=mv$, the common composite has two factorizations through the equalizer, so its uniqueness property gives $u=v$. Hence every regular monomorphism is indeed a monomorphism.

Now take the lifting square just described. The equations $pmu=qmu$ imply $pve=qve$. Cancel the epimorphism $e$ to obtain $pv=qv$. The equalizer gives a unique $d$ with $md=v$. Since $mde=ve=mu$ and $m$ is a monomorphism, $de=u$. Thus every regular monomorphism is a strong monomorphism; the existence of the lift uses epic cancellation, and its uniqueness uses the equalizer property.

If every monomorphism is strong, let $m:A\to B$ be both a monomorphism and an epimorphism. Use $m$ as both vertical maps in the lifting square, with horizontal maps $1_A$ and $1_B$. The strong monomorphism property gives $d:B\to A$ satisfying $dm=1_A$ and $md=1_B$. Thus $m$ is an isomorphism, proving that the category is balanced.

For the converse, take $e:X\to Y$ epic, $m:A\to B$ monic, and $mu=ve$. Form the pullback in a category of $m$ along $v$:The equality $mu=ve$ gives $t:X\to P$ with $pt=e$ and $qt=u$. By preservation of monomorphisms under pullback in a category, $p$ is monic. It is also epic: if $hp=kp$, then $he=hpt=kpt=ke$, and cancellation of $e$ gives $h=k$. The balanced category hypothesis therefore makes $p$ an isomorphism.

Set $d=q{p}^{-1}$. Then $md=v$ and $de=qt=u$. Any other such lift agrees with $d$ after composing with the monomorphism $m$, so it is equal to $d$. Consequently in a category with pullbacks in a category, being balanced is equivalent to every monomorphism being a strong monomorphism. This is the balanced categories with pullbacks have strong monomorphisms principle; the proof does not assume pullback stability of epimorphisms.

The limit form of the Special adjoint functor theorem is as follows. Let $\mathcal{C}$ be a locally small category, a complete category, and a well-powered category, and suppose it has a small cogenerating family $(Q_i{)}_{i\in I}$. Let $\mathcal{D}$ be a locally small category. Then a functor $G:\mathcal{C}\to \mathcal{D}$ has a left adjoint if and only if it preserves all small categorical limits. The dual exchanges completeness, well-poweredness and cogenerators for cocompleteness, well-copoweredness and generators, and characterizes functors with a right adjoint.

Here a cogenerating set distinguishes unequal parallel maps by postcomposition into one of the $Q_i$. A well-powered category has a set of subobjects of each fixed object. Completeness means existence of all small categorical limits. We use the initial-object lemma for complete categories with a weakly initial set: a complete category that is locally small has an initial object exactly when it has a weakly initial set. A proof is given in Question 5(c). We also use the fact that when $G$ preserves small categorical limits, the comma category $(B\downarrow G)$ is complete, with its limits constructed in $\mathcal{C}$; local smallness is inherited from $\mathcal{C}$. Finally, having an initial object in $(B\downarrow G)$ for each $B$ is the universal arrow from an object to a functor characterization of a left adjoint.

Assume $G$ preserves small categorical limits, and fix $B\in \mathcal{D}$. We will construct a weakly initial set in $(B\downarrow G)$, rather than assume a solution set. Start with an arbitrary object $(C,x)$, where $x:B\to GC$. Call a subobject $m:S\hookrightarrow C$ supporting if $x=G(m)x_S$ for some $x_S:B\to GS$. There is at least one, namely $1_C$. Because $G$ preserves pullbacks in a category, it preserves monomorphisms: the self-pullback characterization of a monic map is preserved. Consequently each $x_S$ is unique.

There is a set of supporting subobjects by well-poweredness. Their intersection of subobjects exists by completeness; write it $m_0:C_0\hookrightarrow C$. Applying $G$ to this intersection categorical limit gives a unique $x_0:B\to G{C}_0$ lifting all the $x_S$, and hence $x=G(m_0)x_0$. Minimality says that any subobject $n:R\hookrightarrow C_0$ supporting $x_0$ is invertible: $m_{0}n$ supports $x$, so $m_0$ factors through $m_{0}n$, giving a right inverse to the monic $n$.

If $u,v:C_0\to Q_i$ satisfy $G(u)x_0=G(v)x_0$, their equalizer supports $x_0$, since $G$ preserves that equalizer. It is therefore an isomorphism, and $u=v$. We have obtained an injective functionLet $J_i$ be its image, a subset of the fixed set $S_i=\mathcal{D}(B,G{Q}_i)$. The evaluation embedding into cogenerator products is a monomorphismIndeed each $s\in J_i$ corresponds to a unique map $C_0\to Q_i$, and the cogenerating set property makes the resulting family jointly monic. Notice that we use the realized subsets $J_i$, not all of $S_i$: there need not be a map $C_0\to Q_i$ available for an unused index.

There is a set of tuples $J=(J_i)$, since $I$ and every $S_i$ are sets. For each resulting $P_J$, choose representatives of its subobjects; these form a set by well-poweredness. For every representative $R$, include every map $y:B\to GR$, a set by local smallness of $\mathcal{D}$. All resulting pairs $(R,y)$ form a set. Our given $(C,x)$ receives a map from such a pair: transport $x_0$ along the isomorphism between $C_0$ and the chosen representative, then compose its inverse with $m_0$. Thus these pairs form a weakly initial set in $(B\downarrow G)$. This is the cogenerator bound for comma-category solution sets.

The initial-object lemma for complete categories with a weakly initial set gives an initial object $(LB,{\eta }_B)$. For $h:B\to B^{\prime }$, define $Lh$ by the unique equationUniqueness proves the functor laws and gives the natural bijectionsThus $L⊣G$. Conversely, a right adjoint preserves small categorical limits: its adjunction identifies maps into the proposed limiting object with compatible families of maps into the diagram. This proves both directions of the limit form of the special adjoint functor theorem. Applying the argument to the opposite categories proves the dual statement. Choices of representatives and adjoint objects are understood in the usual ambient-universe convention for large categories.

The comma category $(X\downarrow G)$ has objects $(A,x)$ with $A\in \mathcal{C}$ and a function $x:X\to G(A)$. Its morphisms $(A,x)\to (B,y)$ are maps $f:A\to B$ in $\mathcal{C}$ satisfying $G(f)x=y$. The identity morphism and composition in a category are inherited from $\mathcal{C}$; the functor laws make the defining equations stable under both.

For a singleton $1$, its map $1\to G(A)$ is just an element $a\in G(A)$. Thus $(1\downarrow G)$ is the covariant category of elements of $G$, with arrows $f:(A,a)\to (B,b)$ satisfying $G(f)(a)=b$.

Suppose $G$ is a representable functor, with a natural isomorphism $\theta :\mathcal{C}(R,-)\to G$. Put $r={\theta }_R(1_R)$. For any $(A,a)$ in the comma category $(1\downarrow G)$, the representing bijection gives a unique $f:R\to A$ with ${\theta }_A(f)=a$. By naturality, ${\theta }_A(f)=G(f)(r)$, exactly the condition for $f:(R,r)\to (A,a)$. Hence $(R,r)$ is an initial object.

Conversely, if $(R,r)$ is an initial object, thenis a bijection for each $A$: existence and uniqueness are precisely initiality applied to every $(A,a)$. For $h:A\to B$, $G(h){\theta }_A(f)=G(hf)(r)={\theta }_B(hf)$, proving naturality. Thus $r$ is a universal element and $\theta$ is a representation of a functor. Being representable is equivalent to the singleton comma category having an initial object.

A weakly initial set is a set of objects $(W_i{)}_{i\in I}$ such that every object receives at least one map from some $W_i$. An initial object is itself a singleton weakly initial set. We prove the converse without replacing uniqueness by weak initiality.

Let $\mathcal{E}$ be a complete category that is locally small. Form $W={\prod }_i{W}_i$. For every object $A$, choose $W_i\to A$ and compose with the projection $W\to W_i$, so $W$ is weakly initial. Completeness supplies a terminal object, so the given weakly initial set cannot be empty.

The endomorphisms of $W$ form a set. Take a simultaneous equalizer $e:I\to W$ of all of them with $1_W$; equivalently, equalize the two maps from $W$ into ${\prod }_{s\in \mathcal{E}(W,W)}W$ with components $s$ and $1_W$. Thus $se=e$ for every endomorphism $s$. The object $I$ is weakly initial, since $I\stackrel{e}{\to }W\to A$ exists for every $A$. Weak initiality of $W$ gives a map $u:W\to I$. Applying the equalizing identity to $eu$ yields $eue=e$, so monicity of $e$ gives $ue=1_I$.

For any endomorphism $v:I\to I$, the map $evu$ is an endomorphism of $W$. Therefore $evue=e$. Since $ue=1_I$, this becomes $ev=e$, giving $v=1_I$. Now let $f,g:I\rightrightarrows A$ and form their equalizer $j:E\to I$. Weak initiality of $I$ gives $h:I\to E$. The composite $jh$ is an endomorphism of $I$, hence the identity. Consequently $f=fjh=gjh=g$.

There is at least one map from $I$ to every object, and the argument proves that there is at most one. The object $I$ is an initial object. This establishes the initial-object lemma for complete categories with a weakly initial set using only small products in a category and equalizers.

If $G$ is a representable functor, it preserves small categorical limits. Part (b) makes $(1\downarrow G)$ have an initial object, which gives a weakly initial set.

Conversely, assume limit preservation and a weakly initial set in $(1\downarrow G)$. The needed comma-limit fact is explicit: for a small diagram in a category $(A_j,a_j)$, form $L={\lim }_j{A}_j$ in $\mathcal{C}$. The elements $a_j\in G(A_j)$ form a compatible family, hence determine a unique $a\in G(L)$ under $G(L)\cong {\lim }_{j}G(A_j)$. The projections from $(L,a)$ are a categorical limit in the comma category. This includes the empty diagram, since $G$ carries the terminal object to a singleton. Each comma hom-set is a subset of a hom-set in $\mathcal{C}$, so the comma category is also locally small.

Part (c) now gives an initial object of $(1\downarrow G)$, and part (b) supplies a representation of a functor. ThereforeThe size requirement is a genuine solution-set condition; limit preservation alone does not produce it in an arbitrary complete category.

A monad $\mathbb{T}=(T,\eta ,\mu )$ consists of an endofunctor $T:\mathcal{C}\to \mathcal{C}$ and natural transformations $\eta :1_{\mathcal{C}}\Rightarrow T$ and $\mu :T^2\Rightarrow T$ satisfyingAn algebra for a monad is $(A,a)$ with $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $a{\mu }_A=aT(a)$. A morphism of algebras for a monad $h:(A,a)\to (B,b)$ obeys $ha=bT(h)$. The functor laws show that identities and composites obey this equation, giving the Eilenberg-Moore category ${\mathcal{C}}^{\mathbb{T}}$.

For an adjunction $F⊣G$, let $\eta$ be its adjunction unit and $\epsilon$ its adjunction counit. The monad induced by an adjunction isThe two unit laws are $G{\epsilon }_{FA}{\eta }_{GFA}=1_{GFA}$ and $G({\epsilon }_{FA}F{\eta }_A)=1_{GFA}$, respectively the two triangle identities for an adjunction.

For associativity, naturality of the adjunction counit at ${\epsilon }_{FA}:FGFA\to FA$ saysApply $G$ to obtain ${\mu }_A{\mu }_{TA}={\mu }_{A}T{\mu }_A$. The composites defining $\eta$ and $\mu$ are natural transformations, so all monad axioms have been checked, not only their underlying objectwise types.

Define the forgetful functor $G^{\mathbb{T}}(A,a)=A$, acting as the identity on underlying morphisms. Define the free algebra functor byThe monad laws make ${\mu }_X$ an algebra for a monad structure, and naturality of $\mu$ makes $T(h)$ an monad algebra morphism.

For $(A,a)$, define the transpose mapsThe map $\Psi (k)$ is an monad algebra morphism, because naturality of $\mu$ and the algebra associativity law giveFor $k:X\to A$, naturality of $\eta$ gives $\Phi \Psi (k)=aT(k){\eta }_X=a{\eta }_{A}k=k$. Conversely, if $h$ is an monad algebra morphism, thenPrecomposing by a map into $X$ or postcomposing by an monad algebra morphism respects both formulas. Thus the bijection is natural and $F^{\mathbb{T}}⊣G^{\mathbb{T}}$. This is the free-forgetful Eilenberg-Moore adjunction.

Its adjunction unit is the given ${\eta }_X$; its adjunction counit at $(A,a)$ is the monad algebra morphism $a:(TA,{\mu }_A)\to (A,a)$. Hence $G^{\mathbb{T}}{F}^{\mathbb{T}}=T$, and the induced multiplication is the underlying counit at $(TX,{\mu }_X)$, namely ${\mu }_X$. Therefore this adjunction induces exactly the original monad, including its unit and multiplication of a monad.

A semi-additive category has commutative-monoid enrichment: each hom-set is a commutative monoid, and composition in a category preserves addition and zero in each variable. In the finite-biproduct convention, which we use here, it also has finite products in a category and coproducts in a category. A preadditive category has abelian groups as its hom-sets, with composition additive in each variable; this definition alone does not require finite products in a category or a zero object.

Let $P={\prod }_{i=1}^n{A}_i$ with projections $p_i$. Define $i_j:A_j\to P$ by $p_i{i}_j=1_{A_j}$ for $i=j$ and $p_i{i}_j=0$ otherwise. Bilinearity givesso product uniqueness gives ${\sum }_j{i}_j{p}_j=1_P$. For a family $f_j:A_j\to X$, set $f={\sum }_j{f}_j{p}_j$. Then $f{i}_j=f_j$. If $h:P\to X$ has the same restrictions, $h=h{\sum }_j{i}_j{p}_j={\sum }_j{f}_j{p}_j=f$. Thus the product in a category is also the coproduct in a category, with its canonical injections. Conversely, the analogous construction from a finite coproduct in a category gives projections and the same identities, proving that it is a product in a category.

For the empty family, a terminal object $T$ has a singleton endomorphism monoid, so $1_T=0$. Every map $T\to X$ equals its composite with $1_T=0$, hence equals the zero map; this map exists by enrichment. Therefore $T$ is also an initial object, a zero object. The dual argument applies to an initial object. Consequently finite products in a category and coproducts in a category coincide canonically as biproducts, including the nullary case.

Fix an object $C$. For $x:C\to B$, use $rx:C\to A$ as its identity arrow. For $a:C\to A$, its source is $fa$ and target is $ga$. If $ga=fb$, define the composite of $a$ followed by $b$ byThese expressions use addition and subtraction in the hom-set of the preadditive category. The reflexive pair identities $fr=gr=1_B$ giveso the composite has the required endpoints. Also $f(a^{-1})=ga$ and $g(a^{-1})=fa$.

The unit identities follow directly: $a\circ rfa=rfa+a-rfa=a$ and $rga\circ a=a+rga-rga=a$. If $ga=fb$ and $gb=fc$, either parenthesization of three arrows equals $a+b+c-rga-rgb$, so composition is associative. The inverse identities areTherefore these operations define a groupoid on the specified object and arrow sets. Precomposing by $k:C^{\prime }\to C$ respects each expression, by additivity of composition, so the groupoids are natural in $C$.

Every reflexive pair therefore has the required groupoid structure on generalized points. This is the reflexive-pair groupoid formula in a preadditive category. If the composable-arrow pullback in a category $A{\times }_{g,B,f}A$ exists, its projections $p,q$ give the actual internal composition map $p+q-rgp$. The printed generalized-point definition does not assume that this pullback exists, and preadditivity alone would not ensure it.

Take any morphism $h:A\to B$ and let $c:B\to C$ be its categorical cokernel. Let $i:I\to B$ be the categorical kernel of $c$. Since $ch=0$, kernel universality gives $e:A\to I$ with $h=ie$. The map $i$ is a monomorphism. We must establish the pseudo-epimorphism property of $e$, rather than merely name $I$ an image.

Suppose $t:I\to Z$ satisfies $te=0$. Let $k:K\to I$ be its categorical kernel. Then $e=ka$ for some $a:A\to K$. The composite $ik$ is a monomorphism, so the hypothesis that every monomorphism is normal provides a map $q:B\to D$ of which $ik$ is a categorical kernel, up to its unique compatible isomorphism. Now $qh=qika=0$. The categorical cokernel property of $c$ gives $q=\bar{q}c$, and therefore $qi=\bar{q}ci=0$.

Since $ik$ is the categorical kernel of $q$, the map $i$ factors through it: $i=ik\ell$ for some $\ell :I\to K$. Cancel the monomorphism $i$ to get $k\ell =1_I$. As $k$ is also a monomorphism, $\ell k=1_K$, so $k$ is an isomorphism. Finally $t=tk\ell =0$, proving that $e$ is a pseudo-epimorphism.

ThusThe pseudo-epimorphism factorization through the kernel of a cokernel proof uses a pointed category, existence of categorical kernels and categorical cokernels, and normality of every monic map. It does not silently replace pseudo-epic cancellation against zero by the stronger epimorphism condition.

In an abelian category, for $f:A\to B$ set $\mathrm{coim}f=\mathrm{coker}(\ker f)$ and $\mathrm{im}f=\ker (\mathrm{coker}f)$. The canonical map from coimage to image is an isomorphism. Thus the image factorization in an abelian category isunique up to a unique compatible isomorphism.

The short five lemma states that in a commuting diagram of short exact sequences $0\to L\to M\to R\to 0$ and $0\to L^{\prime }\to M^{\prime }\to R^{\prime }\to 0$ in an abelian category, if the vertical maps $L\to L^{\prime }$ and $R\to R^{\prime }$ are isomorphisms, then so is $M\to M^{\prime }$.

For the Five lemma, consider commuting exact rowswith vertical maps $v_i:A_i\to B_i$. We prove the usual stronger version: $v_1$ epic, $v_2$ and $v_4$ invertible, and $v_5$ monic imply that $v_3$ is invertible. In particular the conclusion holds if all four outer vertical maps are isomorphisms.

Put $L=\mathrm{im}{d}_2$ and $R=\mathrm{im}{d}_3$, and define $L^{\prime },R^{\prime }$ similarly. Exactness gives short exact sequencesThe functoriality of abelian image factorization supplies their vertical outer maps. We show that both are isomorphisms.

First, $L\cong \mathrm{coker}{d}_1$ and $L^{\prime }\cong \mathrm{coker}{d}_1^{\prime }$: exactness at $A_2$ identifies $\mathrm{im}{d}_1$ with $\ker d_2$, and the image-coimage isomorphism for $d_2$ gives the claimed cokernel. Write $q:A_2\to L$ and $q^{\prime }:B_2\to L^{\prime }$ for these categorical cokernels. To construct an inverse to the induced map $\ell :L\to L^{\prime }$, observeCancellation of the epimorphism $v_1$ gives $q{v}_2^{-1}{d}_1^{\prime }=0$, so $q{v}_2^{-1}$ factors uniquely as $\bar{\ell }{q}^{\prime }$ with $\bar{\ell }:L^{\prime }\to L$. From $\ell q=q^{\prime }{v}_2$, composing with the epic $q,q^{\prime }$ gives $\bar{\ell }\ell =1_L$ and $\ell \bar{\ell }=1_{L^{\prime }}$.

Second, $R\cong \ker d_4$ and $R^{\prime }\cong \ker d_4^{\prime }$ by exactness at the fourth objects. Write their inclusions as $j:R\to A_4$ and $j^{\prime }:R^{\prime }\to B_4$. The equality $d_4^{\prime }{v}_4=v_5{d}_4$ gives the induced map $\rho :R\to R^{\prime }$. Sinceand $v_5$ is a monomorphism, $d_4{v}_4^{-1}{j}^{\prime }=0$. Kernel universality gives $\bar{\rho }:R^{\prime }\to R$ with $j\bar{\rho }=v_4^{-1}{j}^{\prime }$. Cancelling the monic $j,j^{\prime }$ proves $\bar{\rho }\rho =1_R$ and $\rho \bar{\rho }=1_{R^{\prime }}$.

Apply the short five lemma to the two short exact sequences: their outer maps $\ell ,\rho$ are invertible, so $v_3$ is an isomorphism. This five lemma via image factorization argument uses only universal properties and therefore works in any abelian category, without treating its objects as literal sets of elements.