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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 $c=\mathrm{coker}m$, 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 $m\mapsto \mathrm{coker}m$ 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 complexes has chain complexes as objects and chain maps as morphisms. Write $Z_n=\ker d_n$ and $Q_n=\mathrm{coker}{d}_{n+1}$. The equation $d_n{d}_{n+1}=0$ gives both an induced map $\mathrm{im}{d}_{n+1}\to Z_n$ and a map $Q_n\to \mathrm{im}{d}_n$. The homology object has the two canonically isomorphic descriptionsPassing 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

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 $0\to A_{\bullet }\to B_{\bullet }\to C_{\bullet }\to 0$, 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 producesthe algebraic Mayer–Vietoris theorem.

A Lawvere theory is a small category with finite products generated by one object $[1]$, so every object is a finite power $[n]$. A model in a finite-product category $\mathcal{C}$ is a finite-product-preserving functor from the theory to $\mathcal{C}$. For a finitary monad $T$ on sets, take the opposite of the full subcategory of its Eilenberg-Moore category on the finitely generated free algebras; the resulting finite-power category is its Lawvere theory, and its set-valued models are $T$-algebras.

If ${\mathbf{Set}}^T$ is additive, its full subcategory of finite free algebras is additive, and passing to the opposite preserves biproducts and abelian-group enrichment. Thus the associated theory is an additive category. A product-preserving model sends the abelian-group object $[1]$ and its addition, zero, and inverse maps to an internal abelian group in $\mathcal{C}$.

In an additive theory the product $[n]=[1{]}^n$ is also a biproduct. If ${\iota }_i:[1]\to [n]$ and ${\pi }_i:[n]\to [1]$ are its injections and projections, every $n$-ary operation decomposes uniquely aswhich in a model reads $\alpha (x_1,\dots ,x_n)={\sum }_i{\alpha }_i(x_i)$. Put $R=\mathrm{End}([1])$, with addition from the enrichment and multiplication from composition. The unary operations give every model an $R$-module structure, and the displayed decomposition says that all operations are exactly $R$-linear combinations. Conversely every $R$-module interprets them this way. Hence ${\mathbf{Set}}^T\simeq R\text{-}\mathbf{Mod}$, up to the conventional choice of left versus right modules.

For an adjunction $F⊣U:\mathcal{D}\to \mathcal{C}$ with monad $T=UF$, the comparison functor sends $D$ to the $T$-algebra $(UD,U{\epsilon }_D)$. The adjunction is monadic when this comparison is an equivalence. The comparison-left-adjoint lemma says that if $\mathcal{D}$ has coequalizers of the reflexive pairs used to present $T$-algebras and $U$ preserves them, then the comparison has a left adjoint. Applying the unit and counit criteria yields the Beck monadicity theorem: $U$ is monadic exactly when it reflects isomorphisms and creates coequalizers of all parallel pairs whose images under $U$ admit split coequalizers.

Let $A_n\subseteq A$ be the fixed-point set of ${\alpha }_n$ and identify $A$ with $A\times \{0\}$. Define $F_{n}A$ to have underlying setKeep the old operations on the first summand, put ${\alpha }_{n+1}(a,0)=(a,1)$ for $a\in A_n$, and put ${\alpha }_1(a,m)=(a,m+1)$ for $m>0$; the remaining higher operations on these new points are undefined. Given $A\to U_{n}B$, the only possible extension sends $(a,m)$ along the iterates of ${\alpha }_1$ beginning at ${\alpha }_{n+1}(a)$. This proves $F_n⊣U_n$.

The forgetful functor $U_n:{\mathcal{C}}_{n+1}\to {\mathcal{C}}_n$ reflects isomorphisms. A $U_n$-split coequalizer carries a unique descended partial operation: splitness prevents any new fixed point of ${\alpha }_n$ from appearing without a representative on which ${\alpha }_{n+1}$ is already prescribed. Hence $U_n$ creates these coequalizers, and Beck's theorem proves the adjunction monadic.

The composite $U_n{U}_{n+1}:{\mathcal{C}}_{n+2}\to {\mathcal{C}}_n$ is not monadic. To see the Beck obstruction explicitly, take $B=\{x,y,z\}$, let all operations through ${\alpha }_n$ be the identity, let ${\alpha }_{n+1}$ swap $x,y$ and fix $z$, and define ${\alpha }_{n+2}(z)=z$. Let $q:B\to Q=\{w,z\}$ identify $x$ and $y$, choose the section $s(w)=x$, $s(z)=z$, and put $h=sq$. Form the kernel pair $A=B{\times }_{Q}B$ with coordinatewise operations and projections $f,g:A\rightrightarrows B$. The map $t:B\to A$, $t(b)=(b,h(b))$, satisfiesso $q$ is a split coequalizer of the underlying pair in ${\mathcal{C}}_n$.

Any lifted structure on $Q$ must have ${\alpha }_{n+1}(w)=w$, so ${\alpha }_{n+2}(w)$ must be defined. If it is $w$, map $x,y$ to a fixed point $d$ and $z$ to a fixed point $e\ne d$ in a target with ${\alpha }_{n+2}(d)=e$ and ${\alpha }_{n+2}(e)=e$; this equalizes $f,g$ but its set-theoretic factor through $Q$ does not preserve ${\alpha }_{n+2}$ at $w$. If instead ${\alpha }_{n+2}(w)=z$, use a target with ${\alpha }_{n+2}(d)=d$ and ${\alpha }_{n+2}(e)=e$ to obtain the same failure. Thus the underlying split coequalizer cannot be created in ${\mathcal{C}}_{n+2}$, and Beck's theorem proves that the composite is not monadic.

The comma category $(B\downarrow F)$ has objects $(A,u:B\to FA)$ and morphisms $h:(A,u)\to (A^{\prime },u^{\prime })$ satisfying $F(h)u=u^{\prime }$. An initial object $(LB,{\eta }_B)$ is precisely a universal arrow from $B$ to $F$. Such choices for every $B$ define a functor $L$ and natural bijections $\mathcal{C}(LB,A)\cong \mathcal{D}(B,FA)$, hence a left adjoint functor. The unit of an existing adjunction supplies the initial objects in the reverse direction.

Suppose $F:\mathcal{C}\to \mathcal{D}$ is final and ${\alpha }_A:D(FA)\to X$ is a cocone. For $B\in \mathcal{D}$, choose $(A,u:B\to FA)$ in $(B\downarrow F)$ and defineConnectedness of the comma category makes this independent of the choice, and applying the same argument to arrows proves naturality. Any extension must have this value, so it is unique. Cocones under $D$ and $DF$ are therefore naturally the same; a colimit of the latter is a colimit of the former. Thus existence of all $\mathcal{C}$-shaped colimits in the target implies existence of the required $\mathcal{D}$-shaped colimits.

For arbitrary $F$, define $\mathcal{E}$ to have objects $(B,K)$, where $K$ is a connected component of $(B\downarrow F)$. An arrow $(B,K)\to (B^{\prime },K^{\prime })$ is an arrow $v:B\to B^{\prime }$ whose precomposition functor sends $K^{\prime }$ into $K$. Let $H(B,K)=B$, and let $G(A)=(FA,K_A)$ where $K_A$ contains $(A,1_{FA})$. Then $F=HG$. Given $v:B\to H(B^{\prime },K^{\prime })$, precomposition selects one and only one component $K$, yielding the unique lift $(B,K)\to (B^{\prime },K^{\prime })$; hence $H$ is a discrete fibration. Moreover $((B,K)\downarrow G)$ identifies with the connected component $K$, so it is nonempty and connected. Thus $G$ is a final functor.