= Solution
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 $m=\ker q=\ker(\operatorname{coker}m)$. 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 factorization
$$
A\twoheadrightarrow\operatorname{coim}f
\xrightarrow{\sim}\operatorname{im}f
\hookrightarrow B,
$$
and 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 and
$$
\mathbf Z(n,p)=
\begin{cases}
\mathbb Z,&p=n\text{ or }p=n-1,\\
0,&\text{otherwise}.
\end{cases}
$$
Let 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, and
$$
\partial_{n-1}\partial_n=0
$$
because 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 $Z_n=\ker d_n$, $B_n=\operatorname{im}d_{n+1}$, and let $q:C_n\to Q_n=\operatorname{coker}d_{n+1}$. Since $d_nd_{n+1}=0$, there is a unique $\bar d_n:Q_n\to C_{n-1}$ with $\bar d_nq=d_n$. The image-to-kernel factorization gives a canonical isomorphism
$$
H_n(C_\bullet)
=\operatorname{coker}(B_n\hookrightarrow Z_n)
\cong\ker(\bar d_n:Q_n\to C_{n-1}).
$$
Passing 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>.
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