= Solution
A <preadditive category> has an <abelian group> of <morphisms> between each pair of objects, and composition is additive in both variables. An <additive category> is a <preadditive category> with a <zero object> and finite <products in a category>; equivalently one may require finite coproducts. The equivalence, and the stronger assertion that these constructions agree canonically, follow from the argument below. An <abelian category> is an <additive category> with <kernels in a category> and <cokernels in a category>, in which every <monomorphism> is a <categorical kernel> and every <epimorphism> is a <categorical cokernel>. A <monomorphism> with the first property is a <normal monomorphism>; an <epimorphism> with the second is a <conormal epimorphism>.
Let $P=\prod_{i=1}^n A_i$, with projections $p_i$. There is a unique $i_j:A_j\to P$ with $p_ii_j=\delta_{ij}1_{A_j}$, interpreting the other components as zero. Since every $p_j$ sends $\sum_i i_ip_i$ to $p_j$, the product's <universal property> gives
$$
p_ii_j=\delta_{ij}1_{A_j},\qquad \sum_i i_ip_i=1_P.
$$
Given arrows $f_i:A_i\to B$, put $f=\sum_i f_ip_i$. Then $fi_j=f_j$. Conversely any arrow with these restrictions equals $f1_P=\sum_i fi_ip_i$, so it is uniquely this sum. Thus $(P,i_i)$ is a <coproduct in a category>. The empty product is the <zero object> and also the empty coproduct. The dual argument starts from coproducts and constructs projections; it proves the equivalent definition. Hence \b[finite products and coproducts are the same biproducts], with the displayed identities giving the canonical identification.
For the <image factorization in an abelian category>, let $h:A\to B$, let $c:B\to Q$ be its <cokernel>, and let $i:I\to B$ be the <kernel> of $c$. Since $ch=0$, write $h=ie$. We prove that $e:A\to I$ is epic. Suppose $t:I\to Z$ satisfies $te=0$ and let $l:L\to I$ be its <categorical kernel>. Then $e=lr$ for some $r:A\to L$. The composite $il$ is monic and hence is a <normal monomorphism>, say $il=\ker z$ for $z:B\to W$. Since $zh=zilr=0$, the <categorical cokernel> property makes $z$ factor through $c$, and therefore $zi=0$. The <categorical kernel> property of $il$ gives $s:I\to L$ with $ils=i$. Cancellation of $i$ yields $ls=1_I$; since $l$ is monic this also gives $sl=1_L$. Thus $l$ is invertible and $t=0$. In a <preadditive category>, $ue=ve$ now implies $(u-v)e=0$, hence $u=v$. This proves $e$ is an <epimorphism> and supplies the required factorization.
For uniqueness, an <epimorphism> that is a <categorical cokernel> is the <categorical cokernel> of its own <categorical kernel>. Indeed, if $e=\operatorname{coker}r$ and $k=\ker e$, then $r$ factors through $k$; an arrow killing $k$ therefore kills $r$ and factors uniquely through $e$. Dually, a <normal monomorphism> is the <categorical kernel> of its own <categorical cokernel>. If $h=mp$ is any epi–mono factorization, monicity of $m$ gives $\ker p=\ker h$. Both $p$ and $e$ are consequently <categorical cokernels> of this same <categorical kernel>. Their <universal properties> give a unique <isomorphism> $\theta$ between the intermediate objects with $\theta p=e$, and cancellation of $p$ gives $i\theta=m$. Therefore \b[the epi–mono factorization is unique up to the unique compatible isomorphism].
Here is an <additive category with truncated primary torsion>, with the prime fixed as $2$. Its objects are <finitely generated abelian groups> having no element of order $4$, and its morphisms are all <group homomorphisms>. Equivalently their $2$-primary torsion <subgroup> has exponent at most $2$. It is a full <subcategory> of <abelian groups>, closed under finite <direct sums> and <subgroups>. Its hom groups and bilinear composition are inherited, and the zero group and finite <direct sums> belong to it, so it is additive. <Categorical kernels> are ordinary <subgroup> <categorical kernels>, which are still finitely generated and have no element of order $4$.
To construct <categorical cokernels>, for an arbitrary finitely generated <abelian group> $G$ let $t_2(G)$ be its $2$-primary <torsion subgroup>, and define
$$
R(G)=G/(2t_2(G)).
$$
The <Fundamental theorem of finitely generated abelian groups> shows that this quotient lies in the subcategory: it leaves the free and odd-primary parts unchanged and replaces each cyclic $2$-primary summand by either $0$ or $\mathbb Z/2\mathbb Z$. Every homomorphism $G\to H$ to an allowed group kills $2t_2(G)$, because it sends $2$-primary torsion into a <subgroup> killed by $2$. Thus $G\to R(G)$ has the <universal property> of a reflection. For $f:A\to B$, its <categorical cokernel> in the subcategory is
$$
B\longrightarrow R(B/f(A)).
$$
This has exactly the <categorical cokernel> <universal property> by the preceding factorization. Finite <categorical limits> exist because <equalizers> are <categorical kernels> of differences and finite products exist. Finite <colimits> exist dually because <coequalizers> are <categorical cokernels> of differences and finite coproducts exist.
The <epimorphisms> in this example are precisely the surjective homomorphisms. A surjection is epic already in <abelian groups>. Conversely, if $f$ is epic in the subcategory, its <categorical cokernel> is zero, so $R(B/f(A))=0$. A nonzero finitely generated <abelian group> has nonzero reflection: a nonzero free or odd-primary summand survives, and a nonzero finite $2$-primary summand has a nonzero quotient modulo twice itself. Hence $B/f(A)=0$ and $f$ is surjective. Its ordinary <categorical kernel> belongs to the subcategory, and its quotient by that <categorical kernel> is its codomain, which also belongs to the subcategory. It is therefore the <categorical cokernel> of its <categorical kernel>. This proves that every <epimorphism> is conormal.
The <monomorphism>
$$
m:\mathbb Z\to\mathbb Z,\qquad m(n)=4n
$$
is not normal. Its <categorical cokernel> in this category is reduction modulo $2$, since $R(\mathbb Z/4\mathbb Z)=\mathbb Z/2\mathbb Z$. The <categorical kernel> of that <categorical cokernel> is the inclusion $2\mathbb Z\hookrightarrow\mathbb Z$, whereas the image of $m$ is $4\mathbb Z$. A <normal monomorphism> is the <categorical kernel> of its own <categorical cokernel>, as proved above, so $m$ cannot be normal. In fact multiplication by $2$ is itself the <categorical kernel> of reduction modulo $2$, up to the identification $\mathbb Z\cong2\mathbb Z$. Its composite with itself is multiplication by $4$. Thus this example already exhibits two <normal monomorphisms> whose composite is not normal.
For the general assertion, let $m:A\to B$ be a nonnormal <monomorphism> in any <additive category> of the specified kind. Put $c=\operatorname{coker}m$ and $k:K\to B$ be its <categorical kernel>, and factor $m=kg$. Both $m$ and $k$ are monic, so $g$ is monic. Set $d=\operatorname{coker}g$ and $l:L\to K$ be its <categorical kernel>. Then $k,l$ are normal, and $g=lr$ for some $r$.
Suppose their composite were normal, say $kl=\ker t$. Since $tm=tklr=0$, $t$ factors through $c$, whence $tk=0$. The <categorical kernel> property of $kl$ supplies $s:K\to L$ with $kls=k$; cancellation shows $ls=1_K$, and monicity makes $l$ an <isomorphism>. Because $dl=0$, this forces $d=0$. In a <preadditive category>, vanishing of the <categorical cokernel> of $g$ implies that $g$ is epic: any $u,v$ with $ug=vg$ have $u-v$ factor through the zero <categorical cokernel>. By hypothesis $g$ is therefore a <categorical cokernel>, and it is also monic. If $g=\operatorname{coker}a$, monicity and $ga=0$ force $a=0$; a <categorical cokernel> of a zero arrow is an <isomorphism>. Thus $g$ is invertible, making $m=kg$ normal, a contradiction. Consequently \b[normal monomorphisms cannot be closed under composition in any such category]; the explicit pair $k,l$ witnesses the failure.
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