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An abelian category has enough projectives when every object admits an epimorphism from a projective object in a category. Repeating this on the kernel in a category constructs a projective resolution of every object. Finite direct sums of projective objects remain projective and give the degreewise split horseshoe construction.

In an abelian category, the coimage of a morphism is its domain modulo its kernel in a category, expressed as a cokernel in a category. The canonical arrow from the coimage to the image is an isomorphism, giving the image factorization in an abelian category.

A normal monomorphism is a monomorphism that is the kernel in a category of some morphism. In a pointed category with kernels and cokernels, a monomorphism is normal exactly when it is the kernel of its own cokernel.

Assume $L_{1}F$ is left exact for every additive right exact $F$. For $0\to K\to P\to X\to 0$ with $P$ projective, left exactness makes $L_{1}F(K)\to L_{1}F(P)=0$ a monomorphism. Hence $L_{1}F(K)=0$. Dimension shifting then gives $L_{2}F(X)=0$. As $X,F$ are arbitrary, (iii)$\Rightarrow$(ii), completingThe argument also recovers projectivity of every such kernel in a category directly from the universal first-derived-functor criterion.

Assume subobjects of projectives are projective. For any object $X$, choose $0\to K\to P\to X\to 0$ with $P$ projective. The kernel in a category $K$ is projective by the assumption. The derived long exact sequence gives a dimension-shifting isomorphismbecause $L_{2}F(P)=L_{1}F(P)=0$. The right side is zero because $K$ is projective. Since $X$ and the right exact functor $F$ were arbitrary, $\boxed{\text{(i)}\Rightarrow \text{(ii)}}$.

An abelian category is an additive category with kernels in a category and cokernels in a category in which the canonical comparison from coimage to image of every morphism is an isomorphism. This is the standard image-coimage form of the definition, equivalently expressed by normality of every monomorphism and epimorphism. Additivity means abelian-group hom-sets with bilinear composition, a zero object and finite biproducts. A sequence is exact at an object when the image of the incoming morphism is the kernel in a category of the outgoing one.

For $f:X\to Y$, construct $q:X\to \mathrm{coim}f=\mathrm{coker}(\ker f)$ and $j:\mathrm{im}f=\ker (\mathrm{coker}f)\to Y$. The universal properties give $f=j\bar{f}q$: first $f$ kills its kernel in a category and factors through $q$; then the resulting map is killed by $\mathrm{coker}f$ because $q$ is epic, so it factors through $j$. The comparison $\bar{f}$ is invertible by the abelian axiom. Cokernels in a category are epic and kernels in a category monic, directly by uniqueness in their universal properties, so $e=\bar{f}q$ is epic and $j$ monic. Thus $\boxed{f=je\text{ is an epimorphism-monomorphism factorization}}$.

In a short exact sequence $0\to X\stackrel{i}{\to }Y\stackrel{p}{\to }Z\to 0$, we have $i=\ker p$ and $p=\mathrm{coker}i$, up to the canonical isomorphisms. If $r:Y\to X$ retracts $i$, the map $1_Y-ir$ kills $i$ and factors as $sp$ for some $s:Z\to Y$. Composing with $p$ gives $psp=p$, so epicity gives $ps=1_Z$. Conversely if $ps=1_Z$, then $1_Y-sp$ is killed by $p$ and factors as $ir$. Composing with $i$ gives $iri=i$, so monicity gives $ri=1_X$. ConsequentlyIn that case $Y\cong X\oplus Z$.

For the derived-functor requests, use additive right exact functors, as customary in abelian homological algebra. An abelian category with enough projectives permits an augmented projective resolution $\cdots \to P_2\to P_1\to P_0\to X\to 0$, obtained by repeatedly taking an epimorphism from a projective onto the previous kernel in a category. Apply $F$ and defineProjective lifts construct maps of resolutions; the comparison and chain-homotopy results make the construction independent of choices and functorial. For a short exact sequence, the horseshoe construction produces a degreewise split short exact sequence of resolutions, hence of their images under additive $F$. Its homology objects give the long exact sequenceThis supplies the requested sketch; detailed comparison and functoriality proofs are not needed here.

A projective object $X$ has a resolution concentrated in degree zero, so $L_{1}F(X)=0$ for every such $F$. Conversely choose $0\to K\stackrel{i}{\to }P\to X\to 0$ with $P$ projective, and take $F=\mathcal{A}(-,K)$ with target ${\mathbf{Ab}}^{\mathrm{op}}$. Precomposition makes this covariant to the opposite category. The usual left exactness of contravariant Hom, which follows directly from the cokernel in a category universal property, says exactly that this $F$ is right exact in that opposite target. If $L_{1}F(X)=0$, the long exact sequence and $L_{1}F(P)=0$ make $F(K)\to F(P)$ monic in ${\mathbf{Ab}}^{\mathrm{op}}$. Equivalently,is epic, hence surjective, in abelian groups. The identity of $K$ has a preimage, giving a retraction of $i$. The sequence splits, and $X$ is a direct summand of $P$, hence projective: a lifting problem for $X$ extends along the projection $P\to X$, is solved for $P$, and restricts along the summand inclusion. ThereforeThis is projectivity detected by all first left derived functors. The following parts prove the three requested equivalent formulations of a hereditary abelian category with enough projectives.

Use the kernel squares in an abelian category argument. If $b$ is monic and $r:X\to A$, $s:X\to K^{\prime }$ satisfy $ar=f^{\prime }s$, then $bgr=g^{\prime }ar=g^{\prime }{f}^{\prime }s=0$, so $gr=0$. The kernel in a category property gives a unique $t:X\to K$ with $ft=r$. The equation $f^{\prime }kt=aft=ar=f^{\prime }s$ and monicity of $f^{\prime }$ give $kt=s$. This proves the left square is a pullback in a category.

Now suppose the right square is a pullback, without imposing the earlier monicity hypothesis on $b$. The pair $(f^{\prime },0):K^{\prime }\to A^{\prime }{\times }_{B^{\prime }}B$ gives a unique $r:K^{\prime }\to A$ with $ar=f^{\prime }$ and $gr=0$. Factor $r=f\ell$ through the kernel. Then $f^{\prime }k\ell =af\ell =ar=f^{\prime }$, so $k\ell =1_{K^{\prime }}$. Also $rk$ and $f$ have the same two pullback projections, hence $rk=f$. Thus $f\ell k=f$ and $\ell k=1_K$. Therefore $k$ is an isomorphism. These arguments use only the relevant kernels, zero arrows and pullback properties; the abelian hypothesis supplies them.

An abelian category is an additive category with a kernel in a category and a cokernel in a category for every morphism, in which every monomorphism is a kernel and every epimorphism is a cokernel. In particular, every epimorphism $p$ is the cokernel of its own kernel: if $p$ is initially a cokernel of $u$, then $u$ factors through $\ker p$, and the two cokernel universal properties agree. The dual statement holds for monomorphisms. Finite biproducts and kernels construct finite limits.

For pullback stability of epimorphisms in an abelian category, let $p:B\to C$ be an epimorphism and $k:C^{\prime }\to C$ any morphism. The mapis epic, since its restriction to the $B$ summand is $p$. Its kernel in a category $i=(h,p^{\prime }):B^{\prime }\to B\oplus C^{\prime }$ exhibits the pullback in a category: the equation $d(h,p^{\prime })=0$ means $ph=k{p}^{\prime }$ and has exactly that universal property.

Suppose $v:C^{\prime }\to D$ satisfies $v{p}^{\prime }=0$. The morphism $[0,v]:B\oplus C^{\prime }\to D$ annihilates $i$, so, because $d$ is the cokernel in a category of $i$, it factors as $[0,v]=wd$ for some $w:C\to D$. Restricting to $B$ gives $wp=0$, and epicity gives $w=0$. Hence $v=0$. Applying this to the difference of any two morphisms agreeing after $p^{\prime }$ proves that $p^{\prime }$ is an epimorphism. ThereforeThis proof uses normality from the definition, rather than assuming the stability to be proved.

For the pullback of a short exact sequence in an abelian category, use the displayed square's notation $g,g^{\prime },h,k$. Exactness makes $f:A\to B$ the kernel in a category of $g$. Since $gf=0=k0$, the pullback in a category gives a uniqueIf $u:X\to B^{\prime }$ has $g^{\prime }u=0$, then $g(hu)=k{g}^{\prime }u=0$, so there is a unique $v:X\to A$ with $fv=hu$. Both projections of the pullback give $f^{\prime }v=u$: their composites with $h$ are equal and their composites with $g^{\prime }$ are zero. The uniqueness of $v$ follows from monicity of $f$. Thus $f^{\prime }$ is the kernel in a category of $g^{\prime }$. By the preceding stability result, $g^{\prime }$ is an epimorphism. We obtain the required exact sequence in an abelian category:

Finally form the pullback in a category $E=P{\times }_C{P}^{\prime }$ of the two projective presentations. Pulling back each of their short exact sequences in an abelian category givesBecause each of $P^{\prime }$ and $P$ is a projective object in a category, the respective final epimorphisms have sections: lift their identity morphisms through those epimorphisms. The usual additive category splitting argument identifies each middle object with the biproduct of its kernel and quotient. Explicitly, a kernel inclusion $j$ and section $s$ give the isomorphism $[j,s]$ from that biproduct to the middle object; $1-sq$ factors through $j$ and provides its inverse's kernel component. Henceand we conclude the Schanuel lemma in an abelian 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 short exact sequence in an abelian category has one nonzero morphism entering the middle object and one leaving it: $0\to A\stackrel{i}{\to }B\stackrel{p}{\to }C\to 0$. Equivalently, $i$ is the kernel in a category of the epimorphism $p$. It splits if $p$ has a section; in that case the additive category splitting argument identifies $B$ with the biproduct $A\oplus C$. Categories of modules give familiar examples.