= Solution
An <abelian category> is an additive category with all kernels and cokernels in which every monomorphism is a kernel and every epimorphism is a cokernel. If $m:A\rightarrowtail B$ is monic and $c=\operatorname{coker}m$, then $m$ factors through $k=\ker c$. The canonical coimage-to-image morphism is an isomorphism in an abelian category, so this factorization identifies $A$ with $\ker c$; hence $m$ is the kernel of its own cokernel. Dually, every epimorphism is the cokernel of its own kernel. The assignments $m\mapsto\operatorname{coker}m$ and $q\mapsto\ker q$ therefore give inverse bijections between subobjects and quotient objects of a fixed object. Well-poweredness is consequently equivalent to well-copoweredness.
The category $\mathcal C^{\mathcal A}$ of <complex in an abelian category>[complexes] has chain complexes as objects and chain maps as morphisms. Write $Z_n=\ker d_n$ and $Q_n=\operatorname{coker}d_{n+1}$. The equation $d_nd_{n+1}=0$ gives both an induced map $\operatorname{im}d_{n+1}\to Z_n$ and a map $Q_n\to\operatorname{im}d_n$. The <homology object> has the two canonically isomorphic descriptions
$$
H_n=\operatorname{coker}(\operatorname{im}d_{n+1}\to Z_n)
=\ker(Q_n\to\operatorname{im}d_n).
$$
Passing to the opposite category exchanges these descriptions, proving self-duality.
Let $S_n(A)=A[n]$. A chain map $A[n]\to C_\bullet$ is exactly a map $A\to Z_n(C)$, while a chain map $C_\bullet\to A[n]$ is exactly a map $Q_n(C)\to A$. Therefore
$$
Q_n\dashv S_n\dashv Z_n.
$$
The <Snake lemma> states that a commutative diagram with exact rows yields the exact sequence of the three kernels, followed by its connecting morphism and the three cokernels. Apply it degree by degree to a short exact sequence of complexes $0\to A_\bullet\to B_\bullet\to C_\bullet\to0$, using the diagrams of cycles, boundaries, and degree objects. The connecting map sends a cycle of $C_n$ to a lift in $B_n$, takes its boundary in $B_{n-1}$, and identifies that boundary with a class in $H_{n-1}(A)$. The Snake lemma gives exactness and produces
$$
\cdots\to H_n(A)\to H_n(B)\to H_n(C)\xrightarrow{\partial}H_{n-1}(A)\to H_{n-1}(B)\to\cdots,
$$
the algebraic <Mayer-Vietoris theorem>.
Back to article page
