Abelian Categories

If $h:X\to A$ satisfies $mh=0$, then also $m{0}_{X,A}=0$, so [L2] gives $h=0_{X,A}$. Therefore the unique map $0\to A$ from [L1] has the kernel universal property for $m$, because every morphism killed by $m$ factors uniquely through $0$.

Kernels are unique up to a unique compatible isomorphism, so the displayed kernel $k:K\to A$ is isomorphic to the zero morphism $0\to A$ from step 1.1. Hence $K$ is a zero object and $k$ is the zero map into $A$. The cokernel claim is the formal dual of the same argument with right cancellation in [L2].

Let $u:0\to A$ be the unique map from [L1]. Since every composite $h\circ u:0\to X$ equals the unique map $0\to X$, every morphism $h:A\to X$ annihilates $u$. Each such $h$ factors uniquely through $1_A$, namely as $h=h\circ 1_A$. Therefore $1_A$ is a cokernel of $u$.

The dual argument with the unique map $A\to 0$ shows that $1_A$ is also a kernel of $A\to 0$. So both displayed identifications hold up to the unique compatible isomorphism of kernels and cokernels.

Because $k$ is a kernel of $f$, one has $f\circ k=0$.

The morphism $q$ is a cokernel of $k$, so step 1.1 gives a unique $\tilde{f}:\mathrm{coim}(f)\to B$ with $\tilde{f}q=f$. That is exactly the claimed factorization through the coimage.

Because $c$ is a cokernel of $f$, one has $c\circ f=0$.

The morphism $i$ is a kernel of $c$, so step 1.1 gives a unique $\widehat{f}:A\to \mathrm{im}(f)$ with $i\widehat{f}=f$. That is the claimed factorization through the image.

By [L1], there is a unique morphism $\tilde{f}:\mathrm{coim}(f)\to B$ with $\tilde{f}{q}_f=f$. Let $c_f:B\to C_f$ be a cokernel of $f$. Then $c_f\tilde{f}{q}_f=c_{f}f=0$, and [L3] makes $q_f$ epic, so $c_f\tilde{f}=0$. Because $i_f:\mathrm{im}(f)\to B$ is a kernel of $c_f$, there is a unique map $\bar{f}:\mathrm{coim}(f)\to \mathrm{im}(f)$ with $i_f\bar{f}=\tilde{f}$.

Composing the identity of step 1.1 with $q_f$ gives $i_f\bar{f}{q}_f=\tilde{f}{q}_f=f$, so $\bar{f}$ has the required property. If another map $u:\mathrm{coim}(f)\to \mathrm{im}(f)$ also satisfies $i_{f}u{q}_f=f$, then $i_{f}u=\tilde{f}$ by the epicity of $q_f$ from [L3], and the uniqueness of the kernel factorization in step 1.1 gives $u=\bar{f}$.

By [L1], the map $q_f$ is a cokernel of a kernel of $f$. Therefore [L2] makes $q_f$ epic.

Again by [L1], the map $i_f$ is a kernel of a cokernel of $f$. Therefore [L2] makes $i_f$ monic.

Because $f$ is monic and epic, [L1] identifies $\ker (f)$ and $\mathrm{coker}(f)$ with zero objects. Hence [L2] gives isomorphisms and for the coimage projection and image inclusion of $f$.

By [L4], the middle map $\bar{f}:\mathrm{coim}(f)\to \mathrm{im}(f)$ is an isomorphism. Since $f=i_f\bar{f}{q}_f$, step 1.1 shows that $f$ is a composite of three isomorphisms, so $f$ itself is an isomorphism.

By [L1] and [L2], the opposite category is additive. Under [L3], a kernel in $\mathcal{A}$ becomes a cokernel in ${\mathcal{A}}^{\mathrm{o}\mathrm{p}}$, and a cokernel becomes a kernel, so every morphism of ${\mathcal{A}}^{\mathrm{o}\mathrm{p}}$ also has both.

The image of $f^{\mathrm{o}\mathrm{p}}$ in the opposite category is the opposite of the coimage of $f$, and the coimage of $f^{\mathrm{o}\mathrm{p}}$ is the opposite of the image of $f$. Therefore the canonical comparison for $f^{\mathrm{o}\mathrm{p}}$ is the opposite of the canonical comparison for $f$, which is an isomorphism by [L1]. So ${\mathcal{A}}^{\mathrm{o}\mathrm{p}}$ satisfies the same AB2 clause and is abelian.

By [L1], the kernel of $m$ is zero. Therefore [L2] identifies the coimage projection $q_m:A\to \mathrm{coim}(m)$ with an isomorphism.

By definition, the image inclusion $i_m:\mathrm{im}(m)\to B$ is a kernel of the cokernel $c$. Since $m=i_m\bar{m}{q}_m$ by [L4], and both $q_m$ and $\bar{m}$ are isomorphisms by step 1.1 and [L5], there is an isomorphism with $i_{m}u=m$. So $m$ is itself a kernel of $c$ up to the unique compatible isomorphism of kernel objects.

By [L6], the opposite of an abelian category is again abelian. Applying step 2.1 there to the opposite of an epimorphism gives that every epimorphism in the original category is the cokernel of its kernel.

Let $q_f:A\to \mathrm{coim}(f)$ and $i_f:\mathrm{im}(f)\to B$ be the defining maps. By [L1], $f$ factors as $f=i_f\bar{f}{q}_f$, and [L2] makes $q_f$ epic and $i_f$ monic. Since $\bar{f}$ is an isomorphism by [L3], the composite $e_f:=\bar{f}{q}_f:A\to \mathrm{im}(f)$ is epic, so $f=i_f{e}_f$ is an epic-monic factorization.

Suppose also $f=m^{\mathrm{\prime }}{e}^{\mathrm{\prime }}$ with $e^{\mathrm{\prime }}$ epic and $m^{\mathrm{\prime }}$ monic. Then $\mathrm{coker}(m^{\mathrm{\prime }})f=0$, so $\mathrm{coker}(m^{\mathrm{\prime }})i_f{e}_f=0$, and the epicity of $e_f$ gives $\mathrm{coker}(m^{\mathrm{\prime }})i_f=0$. Since $m^{\mathrm{\prime }}$ is a kernel of its cokernel by [L5], $i_f$ factors uniquely through $m^{\mathrm{\prime }}$ as $i_f=m^{\mathrm{\prime }}u$. Reversing the roles of $i_f$ and $m^{\mathrm{\prime }}$ gives $m^{\mathrm{\prime }}=i_{f}v$ for a unique $v$.

From step 2.1 one gets $m^{\mathrm{\prime }}=m^{\mathrm{\prime }}uv$ and $i_f=i_{f}vu$, so monicity gives $uv=1$ and $vu=1$. Thus $u$ is an isomorphism. Finally, $m^{\mathrm{\prime }}u{e}_f=i_f{e}_f=f=m^{\mathrm{\prime }}{e}^{\mathrm{\prime }}$, and monicity of $m^{\mathrm{\prime }}$ gives $u{e}_f=e^{\mathrm{\prime }}$. Any other comparison map with the same property equals $u$ by the uniqueness in step 2.1.

By [L1], $f$ admits an epic-monic factorization $f=i_f\circ e_f$, so it factors through its image.

Because $f=ng$, the composite $\mathrm{coker}(n)\circ f$ is zero. Using step 1.1, this becomes $\mathrm{coker}(n)i_f{e}_f=0$. Since $e_f$ is epic, $\mathrm{coker}(n)i_f=0$. Now [L3] says that $n$ is a kernel of $\mathrm{coker}(n)$, so $i_f$ factors uniquely through $n$.

The factorization in step 2.1 is exactly the order relation $[\mathrm{im}(f)]\le [n]$ from [L2]. Hence the image is the least subobject of $B$ through which $f$ factors.

The monomorphism claim is exactly [L2], because [L1] supplies the preadditive and zero-object hypotheses that [L2] needs.

The epimorphism claim is exactly [L3], for the same reason.

Freyd's axioms already make $\mathcal{A}$ balanced: if $m:A\to B$ is monic and epic, the normality clause writes $m$ as a kernel of some $g:B\to C$, so $gm=0$; because $m$ is epic, $g=0$, and the identity $1_B$ is a kernel of $0_{B,C}$, so $m$ is isomorphic to $1_B$.

Let $S=A\bigsqcup B$ and $P=A\times B$ be the coproduct and product supplied by [L1]. The split epics $[1_A,0]:S\to A$ and $[0,1_B]:S\to B$ are cokernels of the opposite injections because maps out of a coproduct are determined by the injections; dually the split monics $⟨1_A,0⟩:A\to P$ and $⟨0,1_B⟩:B\to P$ are kernels of the opposite projections because maps into a product are determined by the projections. Therefore the canonical comparison $c:S\to P$ is both monic and epic, so step 1.1 makes it an isomorphism. Thus binary biproducts exist, and together with the zero-object clause this gives finite biproducts.

By [L3], the finite biproducts from step 2.1 give a canonical commutative-monoid law on each hom-set. On $A\oplus A$, Mitchell's shear $\theta =\left(\begin{array}{cc}{1}_A& 1_A\\ 0& 1_A\end{array}\right)$ is monic and epic by the same kernel-cokernel argument used in step 2.1, hence invertible by step 1.1. Writing ${\theta }^{-1}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the matrix identity $\theta {\theta }^{-1}=1$ yields $1_A+b=0$ in the monoid law of [L3]. For every $x:A\to B$, the morphism $xb$ is therefore an additive inverse of $x$, so the hom-monoids are abelian groups. Hence $\mathcal{A}$ is preadditive, and with finite biproducts it is additive.

Let $f:A\to B$, let $k:K\to A$ be its kernel, let $p:A\to Q=\mathrm{coim}(f)$ be the cokernel of $k$, let $m:I=\mathrm{im}(f)\to B$ be the kernel of a cokernel of $f$, and let $u:Q\to I$ be the canonical morphism from [L4]. Put $i:=mu:Q\to B$. To show that $i$ is monic, let $x:X\to Q$ satisfy $ix=0$, let $q:Q\to Q^{\mathrm{\prime }}$ be a cokernel of $x$, and write $i=jq$. Since $qp$ is a composite of cokernels, [L5] makes it epic, so the conormality clause gives some $h:H\to A$ with $qp=\mathrm{coker}(h)$. Now $fh=iph=jqph=0$, so $h$ factors through $k$; hence $ph=0$. Because $qp$ is a cokernel of $h$, the map $p$ factors through $qp$ as $p=p^{\mathrm{\prime }}qp$. Since $p$ is epic by [L5], $p^{\mathrm{\prime }}q=1_Q$, so $q$ is monic. Then $qx=0$ forces $x=0$, and $i$ is monic.

The dual argument shows that the factorization map $e:=up:A\to I$ is epic: starting from a kernel of a map out of $I$, one passes to the opposite category and repeats step 4.1. Since $m$ is monic and $p$ is epic by [L5], the equalities $mu=i$ and $up=e$ imply that $u$ itself is monic and epic. Step 1.1 then makes $u$ an isomorphism.

Thus every morphism has kernels and cokernels and an invertible canonical coimage-to-image comparison. Together with the additivity from step 3.1, this is exactly the working abelian definition of Abelian category.

If clause 1 holds, then [L1] gives additivity, kernels, and cokernels. The zero object and binary biproducts in [L1] supply the zero-object, product, and coproduct clauses, while [L2] supplies the normality and conormality clauses. So clause 1 implies clause 2.

Clause 2 implies clause 1 by [L3].

If clause 2 holds, then step 1.2 and [L4] give all finite limits and finite colimits, hence in particular pullbacks and pushouts. So clause 2 implies clause 3. Conversely, if clause 3 holds, then the pullback of $A\to 0\leftarrow B$ is a product of $A$ and $B$, the pushout of $A\leftarrow 0\to B$ is a coproduct, the pullback of $A\stackrel{f}{\to }B\stackrel{}{\leftarrow }0$ is a kernel of $f$, and the pushout of $0\stackrel{}{\leftarrow }A\stackrel{f}{\to }B$ is a cokernel of $f$. Together with the stated normal and conormal clauses, that is exactly Freyd's list.

Steps 1.1, 1.2, and 2.1 prove the three-way equivalence.

By [L1], an abelian category satisfies the hypotheses of [L2].

Therefore [L2] applies directly and yields all finite limits and all finite colimits.

By [L1] there is a product of $A$ and $B$, and by [L4] that product is the biproduct $A\oplus B$. In the preadditive structure of [L4], a morphism $x:X\to A\oplus B$ satisfies $(f{p}_A-g{p}_B)x=0$ exactly when $f{p}_{A}x=g{p}_{B}x$. So by [L3], a kernel of $f{p}_A-g{p}_B$ is an equalizer of the parallel pair $f{p}_A,g{p}_B$.

Giving $x:X\to A\oplus B$ is the same as giving its two composites to $A$ and $B$, and the equality in step 1.1 is exactly the pullback compatibility condition. Therefore the equalizer in step 1.1 is a pullback of $f$ and $g$.

Reversing all arrows gives the pushout statement: in an abelian category a pushout is the cokernel of the corresponding difference map.

By [L1], there is a monomorphism $k:P\to A\oplus B$ with $\alpha =p_{A}k$, $\beta =p_{B}k$, and $k$ the kernel of $d:=f{p}_A-e{p}_B$. Since $d{i}_B=-e$, the epicity of $e$ implies that $d$ is epic as well.

Let $q:A\to Q$ be a cokernel of $\alpha$. Because $d$ is epic by step 1.1, [L4] says that $d$ is a cokernel of its kernel $k$. Since $q\alpha =q{p}_{A}k=0$, the map $q{p}_A:A\oplus B\to Q$ kills $k$, so it factors through $d$ as $\bar{q}d=q{p}_A$. Composing with $i_B$ gives $\bar{q}(-e)=0$, and since $e$ is epic, $\bar{q}=0$. Thus $q{p}_A=0$, and composing with $i_A$ yields $q=0$. So $\alpha$ has zero cokernel and is epic by [L3].

Passing to the opposite category turns the given pushout square into a pullback square, and the given monomorphism into an epimorphism. By [L1], the opposite category is still abelian.

Apply [L2] in the opposite category. The opposite of the resulting epimorphism is exactly the pushout leg in the original square, so that leg is monic.

Let $k_{\beta }:K_{\beta }\to P$ be a kernel of $\beta$ and let $k_f:K_f\to A$ be a kernel of $f$. Since $g\beta k_{\beta }=f\alpha k_{\beta }=0$, the kernel property of $k_f$ gives a unique map $u:K_{\beta }\to K_f$ with $k_{f}u=\alpha k_{\beta }$.

Because $f{k}_f=0=g0$, the pullback universal property gives a unique map $s:K_f\to P$ with $\alpha s=k_f$ and $\beta s=0$. Since $\beta s=0$, the kernel property of $k_{\beta }$ gives a unique map $v:K_f\to K_{\beta }$ with $k_{\beta }v=s$. Then $k_{f}uv=\alpha k_{\beta }v=\alpha s=k_f$, so $uv=1$ by monicity of $k_f$; similarly $k_{\beta }vu=s=k_{\beta }$, so $vu=1$ by monicity of $k_{\beta }$. Thus $u$ is an isomorphism.

By [L1], the pullback $P\to D$ of the two monics $m$ and $n$ exists. Because the square defining $P$ is a common lower bound of $[m]$ and $[n]$, and every other common lower bound factors uniquely through that pullback, [L2] says that $P\to D$ represents their intersection subobject.

If the displayed square is a pullback, then its source $X\to D$ is another representative of the same greatest lower bound from step 1.1. Therefore the induced morphism $X\to I$ is an isomorphism.

Conversely, if the induced map $X\to I$ is an isomorphism, then composing the pullback square representing $I$ from step 1.1 with that isomorphism yields the displayed square. Pullbackness is invariant under replacing the corner object by an isomorphic one, so the displayed square is a pullback.

An abelian category is still a category, and the statement only concerns pullback squares that already exist in that ambient category.

Therefore the general theorem [L1] applies verbatim to the abelian setting.

Let $m:M\to A$ be monic, let $c_m:A\to Q_m$ be its cokernel, and let $k_m:K_m\to A$ be the kernel of $c_m$. Since $c_{m}m=0$, the monomorphism $m$ factors through $k_m$. Conversely, [L3] says that $m$ is itself a kernel of $c_m$, so $k_m$ factors through $m$. Thus $[k_m]=[m]$. The dual argument shows that for every epic $q$ one has $[\mathrm{coker}(\ker q)]=[q]$.

If $[m]\le [n]$, then $m=nu$ for some $u$. Since $\mathrm{coker}(n)n=0$, one has $\mathrm{coker}(n)m=0$, so the cokernel universal property of $m$ makes $\mathrm{coker}(n)$ factor through $\mathrm{coker}(m)$. By the quotient-order convention in [L1], this is exactly $[\mathrm{coker}(m)]\le [\mathrm{coker}(n)]$. The kernel assignment preserves the order dually.

Step 1.1 proves that the two assignments are mutually inverse on classes, and step 1.2 proves that both preserve the stated orders. So kernel and cokernel are mutually inverse order isomorphisms between subobjects and quotient objects.

By [L2], the condition that $m$ and $n$ represent the same subobject is exactly $[m]=[n]$. Applying [L3] gives $[\mathrm{coker}(m)]=[\mathrm{coker}(n)]$ as quotient-object classes.

Equality of quotient-object classes means the two cokernels are joined by a unique compatible isomorphism. By [L1], that is exactly the claim that the quotient $A\mathrm{/}B$ is independent of the chosen representing monomorphism.

By [L1] and [L2], the quotient $A\mathrm{/}\ker (f)$ is exactly the coimage of $f$.

The map from step 1.1 to $\mathrm{im}(f)$ is the canonical coimage-to-image morphism of [L3], and [L4] makes it an isomorphism. Therefore $A\mathrm{/}\ker (f)\cong \mathrm{im}(f)$ canonically.

Let $q_C:A\to A\mathrm{/}C$ and $q_B:A\to A\mathrm{/}B$ be the quotient maps from [L1]. Since $q_{B}bc=0$, the morphism $q_B$ kills $C$, so the universal property of $q_C$ gives a unique map $\bar{q}:A\mathrm{/}C\to A\mathrm{/}B$ with $\bar{q}{q}_C=q_B$.

The composite $bc$ is monic: equality $bcu=bcv$ implies $cu=cv$ and then $u=v$. By [L4], $bc$ is therefore a kernel of its cokernel $q_C$. If $h:X\to B$ satisfies $q_{C}bh=0$, there is a unique $t:X\to C$ with $bct=bh$. Monicity of $b$ gives $ct=h$, and monicity of $c$ gives uniqueness. Since $q_{C}bc=0$, this proves that $c$ is a kernel of $q_{C}b$.

Let $p:B\to B\mathrm{/}C$ be the cokernel of $c$. By [L2], the canonical image factorization of $q_{C}b$ gives a monomorphism $\tilde{b}:B\mathrm{/}C\to A\mathrm{/}C$ satisfying $\tilde{b}p=q_{C}b$. Since $p$ is epic by [L3] and $\bar{q}\tilde{b}p=q_{B}b=0$, we have $\bar{q}\tilde{b}=0$.

If $r:A\mathrm{/}C\to Y$ satisfies $r\tilde{b}=0$, then $r{q}_{C}b=r\tilde{b}p=0$. The cokernel property of $q_B$ gives a unique $s:A\mathrm{/}B\to Y$ with $s{q}_B=r{q}_C$. Since $q_B=\bar{q}{q}_C$ and $q_C$ is epic, $s\bar{q}=r$. Conversely, any $s$ satisfying this last equality satisfies $s{q}_B=r{q}_C$, proving uniqueness. Thus $\bar{q}$ is a cokernel of $\tilde{b}$, and [L1] identifies $(A\mathrm{/}C)\mathrm{/}(B\mathrm{/}C)$ canonically with $A\mathrm{/}B$.

By [L1], there is an isomorphism $\varphi :A\mathrm{/}\ker (f)\to \mathrm{im}(f)$ such that the composite of the quotient map $q:A\to A\mathrm{/}\ker (f)$, the isomorphism $\varphi$, and the image inclusion $i_f:\mathrm{im}(f)\to B$ equals $f$.