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In a pointed category with finite product in a category and coproducts in a category and invertible canonical maps $c:A+B\to A\times B$, the unique commutative-monoid enrichment is $f+g=[1_B,1_B]c_{B,B}^{-1}⟨f,g⟩$. Associativity and commutativity follow by comparing fold maps on triple and swapped coproduct injections. Composition distributes by the universal properties. For uniqueness, bilinearity forces $i_1{p}_1+i_2{p}_2=1$ on each biproduct, and composing with the fold and the paired morphism forces the addition formula.

The category of pointed sets and basepoint-preserving functions is a pointed category. It is equivalent to the category of partial functions by adjoining or deleting the basepoint. It is not isomorphic to the category of all actual sets and partial functions: there are many distinct singleton zero objects, whereas the empty set is the sole zero object in that category.

In a pointed category, a pushout in a category of $f$ and $f^{\prime }$ induces an isomorphism between their categorical cokernels, when those cokernels exist. Push out the cokernel map of $f$ together with the zero map, then factor through the cokernel of $f^{\prime }$. Cokernel and pushout uniqueness prove that this map is inverse to the induced comparison. No abelian hypothesis is needed.

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 a pointed category, a zero object defines zero morphisms between all objects. A categorical cokernel of $f:A\to B$ is a map $g:B\to C$ with $gf=0$ such that every $u:B\to X$ with $uf=0$ factors uniquely as $u=\bar{u}g$. Equivalently it is the coequalizer of $f$ and the zero map, so $g$ is an epimorphism.

Write $a:A\to A^{\prime }$, $b:B\to B^{\prime }$, and $c:C\to C^{\prime }$ for the vertical arrows, with $g^{\prime }=\mathrm{coker}{f}^{\prime }$ and $cg=g^{\prime }b$. The left pushout in a category applied to the compatible pair $g:B\to C$ and $0:A^{\prime }\to C$ gives $h:B^{\prime }\to C$ satisfying $hb=g$ and $h{f}^{\prime }=0$. The categorical cokernel property of $g^{\prime }$ then gives $d:C^{\prime }\to C$ with $d{g}^{\prime }=h$.

Now $dcg=d{g}^{\prime }b=hb=g$. Cancel the epimorphism $g$ to obtain $dc=1_C$. To prove the other identity, the two arrows $cd{g}^{\prime },g^{\prime }:B^{\prime }\to C^{\prime }$ agree after $b$, because $cd{g}^{\prime }b=cg=g^{\prime }b$, and after $f^{\prime }$, because both composites are zero. The pushout in a category uniqueness clause gives $cd{g}^{\prime }=g^{\prime }$. Cancelling the epimorphism $g^{\prime }$ gives $cd=1_{C^{\prime }}$.

Thus $c$ is an isomorphism: cokernel invariance under pushout holds already in pointed categories with the indicated cokernels.

A pointed category has a zero object, both initial and terminal. Factoring through that object gives a zero morphism $0_{A,B}$ between any two objects. In this paper a semi-additive structure means commutative-monoid enrichment: every hom-set is a commutative monoid with additive zero, and composition distributes over addition in both variables. No existence of finite products in a category is included in this last definition. This convention matters for the final one-object example; under the stronger convention requiring finite biproducts, that example would not be a semi-additive category.

Let $c_{A,B}:A+B\to A\times B$ be the given canonical isomorphism. Transfer the coproduct injections across $c_{A,B}$, so $A\times B$ is also a coproduct in a category, with injectionsWrite ${\nabla }_B:B\times B\to B$ for the morphism whose composites with $i_1,i_2$ are both $1_B$. For $f,g:A\to B$, define the biproduct-induced addition of morphisms byThe zero is the existing zero morphism. We verify the laws from the universal properties, without assuming addition in advance.

All finite canonical maps from coproducts to products are isomorphisms, by induction from the binary ones and the zero object. Thus both $(B\times B)\times B$ and $B\times (B\times B)$ are ternary coproducts in a category. The two iterated fold maps to $B$ agree on each of the three injections, hence are equal. Applying this to $⟨f,g,h⟩$ gives associativity. The interchange $B\times B\to B\times B$ swaps the two injections, so its composite with ${\nabla }_B$ is ${\nabla }_B$, giving commutativity. Finally $⟨f,0⟩=i_{1}f$, so $f+0=f$, and similarly $0+f=f$.

Precomposition is additive because $⟨f,g⟩u=⟨fu,gu⟩$. For $v:B\to C$, the morphisms $v{\nabla }_B$ and ${\nabla }_C(v\times v)$ agree on both injections, so they agree; consequently $v(f+g)=vf+vg$. Composition with a zero morphism is zero. This proves the commutative-monoid enrichment.

It is unique. In any such enrichment, the additive zero morphisms agree with the pointed ones, since each map to or from the zero object belongs to a singleton hom-set. The projections of $i_1{p}_1+i_2{p}_2$ from $A\times B$ are respectively $p_1,p_2$ by bilinearity, soFor $A=B$, composing on the left by ${\nabla }_B$ and on the right by $⟨f,g⟩$ forces the displayed formula for $f+g$. Thus the biproducts determine exactly one such enrichment.

For the last part, the underlying one-object category has endomorphism monoid $(\mathbb{N},\cdot )$, including $0$, and identity $1$. Its usual addition gives a commutative-monoid enrichment. Any permutation $\sigma$ of the prime numbers extends, by unique prime factorization, to a multiplicative monoid automorphism ${\varphi }_{\sigma }:\mathbb{N}\to \mathbb{N}$, fixing $0,1$. Transport addition byThis is a commutative monoid operation with zero $0$, and multiplication distributes over it: applying ${\varphi }_{\sigma }$ reduces each distributive law to the ordinary one in $\mathbb{N}$. Thus each operation supplies a semi-additive structure on the same fixed composition law.

For each odd prime $p$, take $\sigma$ to interchange $2$ and $p$, fixing every other prime. ThenDifferent choices of $p$ give different additions, and there are infinitely many prime numbers. Hence there are infinitely many distinct semi-additive structures, even though these transported structures are isomorphic as enriched categories. The underlying one-object category has no terminal object, because its endomorphism set is not a singleton, so it indeed has no finite products in a category.

In a pointed category, a normal monomorphism is a monomorphism that is the kernel in a category of some morphism. Suppose $m:M\hookrightarrow A$ is the kernel of $f:A\to B$, and let $q:A\to Q$ be the cokernel in a category of $m$. Since $fm=0$, there is a unique $\bar{f}:Q\to B$ with $f=\bar{f}q$. If $qx=0$, then $fx=\bar{f}qx=0$, so the universal property of $m=\ker f$ factors $x$ uniquely through $m$. Therefore $m=\ker q=\ker (\mathrm{coker}m)$. The converse is immediate: if $m$ is the kernel of its cokernel, it is the kernel of a morphism and hence normal.

An abelian category is an additive category with kernels and cokernels in which every monomorphism is normal and every epimorphism is a conormal epimorphism. Finite biproducts and kernels give finite limits. The image and coimage in an abelian category give every $f:A\to B$ the canonical factorizationand the middle map is an isomorphism. The first map is a cokernel and therefore a regular epimorphism. Every epimorphism in an abelian category is the cokernel of its kernel, and epimorphisms are stable under pullback; consequently regular epimorphisms are pullback-stable. This proves that every abelian category is regular.

Define the additive indexing category for chain complexes $\mathbf{Z}$ as follows. Its objects are the integers andLet the generator of $\mathbf{Z}(n,n)$ be $1_n$ and the generator of $\mathbf{Z}(n,n-1)$ be ${\partial }_n$. Composition is bilinear, the $1_n$ are identities, andbecause the target hom-group $\mathbf{Z}(n,n-2)$ is zero. An additive functor $C:\mathbf{Z}\to \mathcal{A}$ chooses objects $C_n=C(n)$ and differentials $d_n=C({\partial }_n)$ satisfying $d_{n-1}{d}_n=0$, hence a complex in an abelian category. Conversely every chain complex defines this unique additive functor.

For self-duality, put $Z_n=\ker d_n$, $B_n=\mathrm{im}{d}_{n+1}$, and let $q:C_n\to Q_n=\mathrm{coker}{d}_{n+1}$. Since $d_n{d}_{n+1}=0$, there is a unique ${\bar{d}}_n:Q_n\to C_{n-1}$ with ${\bar{d}}_{n}q=d_n$. The image-to-kernel factorization gives a canonical isomorphismPassing to the opposite category exchanges kernels with cokernels and images with coimages. The usual construction in ${\mathcal{A}}^{\mathrm{op}}$ is therefore the expression on the right, which is canonically the original homology object. This proves the self-duality of homology.

A morphism $e:A\to B$ in a pointed category is pseudo-epic when, for every $t:B\to Z$, the equation $te=0$ forces $t=0$. If a categorical cokernel exists, this is equivalent to that cokernel having a zero object as codomain. Indeed all zero composites then factor uniquely through zero; conversely the cokernel itself must be zero, and being both zero and epic forces its codomain to have identity zero, hence be a zero object. Every epimorphism is pseudo-epic. In a preadditive category, subtraction shows the converse: $ue=ve$ implies $(u-v)e=0$, hence $u=v$. In a general pointed category no subtraction is available.

In a pointed category with categorical kernels and categorical cokernels, where every monomorphism is normal, every $h$ factors through $i=\ker (\mathrm{coker}h)$ as $h=ie$, with $e$ a pseudo-epimorphism. If $te=0$, write $k=\ker t$. Then $e$ factors through $k$, and the monic $ik$ is a categorical kernel of some $q$ by normality. Since $qh=0$, the cokernel of $h$ forces $qi=0$. Hence $i$ factors through $ik$, forcing $k$ to be an isomorphism and therefore $t=0$. This establishes the factorization without assuming the stronger abelian category axioms.

In a pointed category, the zero morphism $0_{A,B}:A\to B$ is the composite through its zero object. Composition on either side by any morphism remains zero. In a commutative-monoid enrichment on a pointed category, the additive zero must coincide with this zero morphism: morphisms to and from the zero object belong to singleton hom-sets.