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To add two extensions of $M$ by $N$, take their direct sum, form a pullback in a category along the diagonal $M\to M\oplus M$, and form a pushout in a category along addition $N\oplus N\to N$. This produces another extension of $M$ by $N$ and defines addition in ${\mathrm{Ext}}_R^1(M,N)$. In a quiver representation, vertexwise splittings express the two extensions by off-diagonal arrow matrices, and the Baer sum adds those matrices modulo the extension complex of quiver representations coboundaries.

For a reflexive pair $f,g:A\rightrightarrows B$ with common section $r$, form the pushout in a category of $f$ and $g$, with maps $j_1,j_2:B\to P$. The equation $j_{1}f=j_{2}g$, composed with $r$, gives $j_1=j_2=q$. This common map is a coequalizer of $f,g$: every map equalizing them defines a commuting pair into the pushout and hence factors uniquely through $q$.

In a pointed category, a pushout in a category of $f$ and $f^{\prime }$ induces an isomorphism between their categorical cokernels, when those cokernels exist. Push out the cokernel map of $f$ together with the zero map, then factor through the cokernel of $f^{\prime }$. Cokernel and pushout uniqueness prove that this map is inverse to the induced comparison. No abelian hypothesis is needed.

In a pointed category, a zero object defines zero morphisms between all objects. A categorical cokernel of $f:A\to B$ is a map $g:B\to C$ with $gf=0$ such that every $u:B\to X$ with $uf=0$ factors uniquely as $u=\bar{u}g$. Equivalently it is the coequalizer of $f$ and the zero map, so $g$ is an epimorphism.

Write $a:A\to A^{\prime }$, $b:B\to B^{\prime }$, and $c:C\to C^{\prime }$ for the vertical arrows, with $g^{\prime }=\mathrm{coker}{f}^{\prime }$ and $cg=g^{\prime }b$. The left pushout in a category applied to the compatible pair $g:B\to C$ and $0:A^{\prime }\to C$ gives $h:B^{\prime }\to C$ satisfying $hb=g$ and $h{f}^{\prime }=0$. The categorical cokernel property of $g^{\prime }$ then gives $d:C^{\prime }\to C$ with $d{g}^{\prime }=h$.

Now $dcg=d{g}^{\prime }b=hb=g$. Cancel the epimorphism $g$ to obtain $dc=1_C$. To prove the other identity, the two arrows $cd{g}^{\prime },g^{\prime }:B^{\prime }\to C^{\prime }$ agree after $b$, because $cd{g}^{\prime }b=cg=g^{\prime }b$, and after $f^{\prime }$, because both composites are zero. The pushout in a category uniqueness clause gives $cd{g}^{\prime }=g^{\prime }$. Cancelling the epimorphism $g^{\prime }$ gives $cd=1_{C^{\prime }}$.

Thus $c$ is an isomorphism: cokernel invariance under pushout holds already in pointed categories with the indicated cokernels.

Let $q:B\to Q$ be the categorical cokernel of $f$ in an abelian category. If $f$ is an epimorphism, the equality $qf=0f$ implies $q=0$. Since $q$ is also an epimorphism, $1_{Q}q=0q$ implies $1_Q=0$. An object with zero identity is a zero object: every morphism to or from it is zero. Hence the cokernel object is zero.

Conversely, suppose the cokernel object is zero. If $uf=vf$, additivity gives $(u-v)f=0$. The categorical cokernel property makes $u-v$ factor through the zero object, so $u-v=0$ and $u=v$. ThusThis is the zero-cokernel criterion for epimorphisms. In a pushout in a category, the two horizontal morphisms have isomorphic cokernels by part (a). Therefore the lower morphism is epic if and only if the upper morphism is epic. In particular pushouts reflect epimorphisms in an abelian category; the same argument also proves preservation.

Use the displayed square's notation $f:A\to B$, $h:A\to C$, $g:B\to D$, $k:C\to D$, with $gf=kh$ and $g$ epic. Form the biproduct and the morphismsThe map $q$ is an epimorphism, since its restriction to $B$ is $g$: equality after $q$ implies equality after $g$. The pullback in a category property says exactly that $j$ is a categorical kernel of $q$. Indeed $qj=gf-kh=0$, and a map into $B\oplus C$ killed by $q$ is a pair $(x,y)$ with $gx=ky$, which factors uniquely through $(f,h)$.

Use the standard abelian category property that every epimorphism is the categorical cokernel of its categorical kernel. If $u:B\to X$ and $v:C\to X$ satisfy $uf=vh$, then $[u,-v]j=0$. Hence there is a unique $w:D\to X$ with $wq=[u,-v]$. Restriction to the two summands gives $wg=u$ and $wk=v$. This is the pushout in a category universal property. Thus a pullback of an epimorphism is a pushout in an abelian category.

In this pushout, $g$ is the pushout of $h$ along $f$. The reflection result of part (b) therefore makes $h$ epic. This proves pullback stability of epimorphisms in an abelian category.

Finally, let $e:E\to B$ be an epimorphism and let $p_1,p_2:R\rightrightarrows E$ be its kernel pair. Their pullback square is a pushout by the result just proved. If $u:E\to X$ satisfies $u{p}_1=u{p}_2$, the two copies of $u$ form a pushout cocone. There is a unique $v:B\to X$ with $ve=u$. Hence epimorphisms in an abelian category are coequalizers of their kernel pairs, so they are regular epimorphisms.

Assume (c), and take an epimorphism $p:M\twoheadrightarrow h_A$ with $M$ a monofunctor. Since $h_A$ is a projective object in a category, there is a natural transformation $s:h_A\to M$ with $ps=1_{h_A}$. A retract in a category of a monofunctor is a monofunctor: if $h_A(f)(u)=h_A(f)(v)$, then naturality of $s$ gives $M(f)s(u)=M(f)s(v)$, injectivity of $M(f)$ gives $s(u)=s(v)$, and applying $p$ gives $u=v$. Thus (c) implies (b), completing

The stronger assertion that every set-valued functor is a monofunctor holds exactly when $\mathcal{C}$ is a groupoid. Sufficiency follows because a functor preserves inverses, so sends every arrow to a bijection.

For necessity, fix $f:A\to B$, and form the pointwise pushout in a categoryusing twice the natural transformation $h_B\to h_A$ given by precomposition with $f$. At an object $X$, $P(X)$ consists of two copies of $\mathcal{C}(A,X)$, with the two copies of $kf$ identified for every $k:B\to X$. No different underlying elements are identified: each generating relation simply joins the two copies of one element. In $P(B)$, the two copies of $f$ coincide, since $f=1_{B}f$. Hence $P(f)$ sends the two copies of $1_A$ in $P(A)$ to the same element. If $P$ is a monofunctor, those copies of $1_A$ coincide. The description of the pushout implies $rf=1_A$ for some $r:B\to A$. Thus every morphism is a split monomorphism.

Apply the same conclusion to $r$: there is $t:A\to B$ with $tr=1_B$. Then $t=trf=f$, and so $fr=1_B$ as well. Every arrow is invertible, proving

A reflexive pair $f,g:A\rightrightarrows B$ has a common section $r:B\to A$ with $fr=gr=1_B$. Form the pushout in a category of $f$ and $g$, and write its two maps from $B$ as $j_1,j_2$. Its relation $j_{1}f=j_{2}g$, composed with $r$, gives $j_1=j_2=q$. Thus $qf=qg$. Any $h:B\to C$ with $hf=hg$ gives the compatible pair $(h,h)$ for this pushout and therefore factors uniquely through $q$. This proves the coequalizer of a reflexive pair from a pushout construction.

There is a genuine transcription difference: the original PDF says finite products in the finite-colimit assertion, while the TeX says finite coproducts. The PDF assertion is false with products. Even retaining a pushout-existence assumption, the full subcategory of the Category of sets on nonempty sets has finite products, pushouts, and all coequalizers, but no initial object, hence no empty colimit. Regard the poset with elements $0,a,b,u_0,u_1,\dots$, orderand incomparable $a,b$ as a category. It has finite meets and top $u_0$, hence finite products in a category. Every reflexive pair in a poset is an equal pair and has its identity as a coequalizer. But $a,b$ have no least upper bound, so they have no coproduct in a category.

For the corrected construction using finite coproducts, let $D:\mathcal{J}\to \mathcal{C}$ be a finite diagram in a category. PutDefine $s,t:Y\rightrightarrows X$ on the summand indexed by $a:i\to j$ as ${\nu }_i$ and ${\nu }_{j}D(a)$. Maps $h:X\to Z$ with $hs=ht$ are exactly cocone under a diagram data on $D$. Hence their universal coequalizer is its colimit. Replace $s,t$ by the reflexive pairwhose common section is the second injection and whose coequalizer is unchanged. This proves construction of finite colimits from coproducts and reflexive coequalizers, including the empty diagram via the empty coproduct:

In an abelian category, suppose a commutative square $gf=kh$ is a pullback in a category and $g$ is an epimorphism. Then $(f,h):A\to B\oplus C$ is the categorical kernel of the epimorphism $[g,-k]$. Since an epimorphism is the cokernel of its kernel, any compatible pair $u:B\to X$, $v:C\to X$ induces a unique map $D\to X$ by factoring $[u,-v]$. Hence the square is also a pushout in a category.

A universal property characterizes an object by a prescribed family of morphisms and a unique factorization of every competing family. It determines that object up to a unique compatible isomorphism. Examples include the pushout in a category and the universal property of the tensor product of modules.