[proofplan] We construct $p':E'\to C$ from the universal property of the pushout, using the compatible pair $p:E\to C$ and $0:A'\to C$. We then prove that $p'$ is the cokernel of $i'$ by testing morphisms out of $E'$ that kill $i'$. Finally, we prove that $i'$ is the kernel of $p'$ by using the kernel property of $i=\ker p$ and then applying the pushout universal property. [/proofplan]

[step:Construct the induced morphism from the pushout to $C$] Let $0_{A',C}:A'\to C$ denote the zero morphism. Since the original sequence is exact at $E$, we have \begin{align*} p i = 0_{A,C}. \end{align*} Hence the two morphisms $p:E\to C$ and $0_{A',C}:A'\to C$ are compatible with the span $E \xleftarrow{i} A \xrightarrow{v} A'$, because \begin{align*} p i = 0_{A,C}=0_{A',C}v. \end{align*} By the universal property of the pushout square, there exists a unique morphism $p':E'\to C$ such that \begin{align*} p'\tilde v=p, \qquad p'i'=0_{A',C}. \end{align*} [/step]

[step:Show that $p'$ is the cokernel of $i'$] Let $X$ be an object of $\mathcal A$, and let $q:E'\to X$ be a morphism satisfying \begin{align*} q i'=0_{A',X}. \end{align*} Then \begin{align*} (q\tilde v)i = q(\tilde v i) = q(i'v) = (qi')v = 0_{A',X}v = 0_{A,X}. \end{align*} Since $p=\operatorname{coker} i$, there exists a unique morphism $r:C\to X$ such that \begin{align*} r p=q\tilde v. \end{align*} We claim that $q=rp'$. Indeed, the two morphisms $q:E'\to X$ and $rp':E'\to X$ agree after precomposition with the pushout structure maps: \begin{align*} (rp')\tilde v=r(p'\tilde v)=rp=q\tilde v, \end{align*} and \begin{align*} (rp')i'=r(p'i')=r0_{A',C}=0_{A',X}=qi'. \end{align*} By uniqueness in the pushout universal property, $q=rp'$. The morphism $r$ is unique with this property. If $s:C\to X$ also satisfies $sp'=q$, then \begin{align*} sp=s(p'\tilde v)=(sp')\tilde v=q\tilde v=rp. \end{align*} Since $p=\operatorname{coker} i$ is an epimorphism, $s=r$. Therefore $p'=\operatorname{coker} i'$. [/step]

[step:Show that $i'$ is the kernel of $p'$]We use the standard biproduct-cokernel construction of pushouts in an abelian category. Define \begin{align*} j:A\to E\oplus A', \qquad j=\begin{pmatrix} i \\ -v \end{pmatrix}. \end{align*} Since $i$ is monic, $j$ is monic: if $t:Y\to A$ satisfies $jt=0$, then $it=0$, hence $t=0$. The pushout object $E'$ is represented as the cokernel of $j$, with quotient morphism \begin{align*} \pi:E\oplus A'\to E' \end{align*} and structure maps \begin{align*} \tilde v=\pi\begin{pmatrix}1_E\\0\end{pmatrix}, \qquad i'=\pi\begin{pmatrix}0\\1_{A'}\end{pmatrix}. \end{align*} Indeed, the equality $\pi j=0$ is equivalent to $\tilde v i=i'v$, and the cokernel universal property of $\pi$ is exactly the pushout universal property for pairs of morphisms out of $E$ and $A'$ that agree after precomposition with $i$ and $v$. Define \begin{align*} \bar p:E\oplus A'\to C, \qquad \bar p=\begin{pmatrix}p&0_{A',C}\end{pmatrix}. \end{align*} Since \begin{align*} \bar p j = \begin{pmatrix}p&0_{A',C}\end{pmatrix} \begin{pmatrix}i\\-v\end{pmatrix} = pi-0_{A',C}v = 0_{A,C}, \end{align*} the cokernel property of $\pi$ gives the unique morphism $p':E'\to C$ with \begin{align*} p'\pi=\bar p. \end{align*} This morphism is the same $p'$ constructed from the pushout, because it satisfies $p'\tilde v=p$ and $p'i'=0_{A',C}$. We first prove that $i'$ is monic. Let $b:Y\to A'$ be a morphism with $i'b=0$. Then \begin{align*} \pi\begin{pmatrix}0\\b\end{pmatrix}=0. \end{align*} Because $j$ is monic and $\pi=\operatorname{coker}j$, the normality axiom in an abelian category gives $j=\ker\pi$. Hence there is a unique morphism $t:Y\to A$ such that \begin{align*} \begin{pmatrix}0\\b\end{pmatrix} = \begin{pmatrix}i\\-v\end{pmatrix}t. \end{align*} Taking the $E$-component gives $it=0$, so $t=0$ because $i$ is monic. Taking the $A'$-component then gives $b=-vt=0$. Thus $i'$ is monic. Now let $Y$ be an object of $\mathcal A$, and let $a:Y\to E'$ be a morphism satisfying \begin{align*} p'a=0_{Y,C}. \end{align*} Form the pullback of $a$ and the epimorphism $\pi$: \begin{align*} \begin{array}{ccc} P & \xrightarrow{\ell} & E\oplus A' \\ \downarrow \rho & & \downarrow \pi \\ Y & \xrightarrow{a} & E'. \end{array} \end{align*} Since cokernels are epimorphisms and epimorphisms are stable under pullback in an abelian category, $\rho:P\to Y$ is an epimorphism. The equality $\pi\ell=a\rho$ gives \begin{align*} \bar p\ell = p'\pi\ell = p'a\rho = 0_{P,C}. \end{align*} Since $i=\ker p$, the kernel of $\bar p=\begin{pmatrix}p&0_{A',C}\end{pmatrix}$ is \begin{align*} m:A\oplus A'\to E\oplus A', \qquad m= \begin{pmatrix} i&0\\ 0&1_{A'} \end{pmatrix}. \end{align*} Therefore there is a unique morphism \begin{align*} c:P\to A\oplus A' \end{align*} such that $mc=\ell$. The pushout relation gives \begin{align*} \pi m = \pi \begin{pmatrix} i&0\\ 0&1_{A'} \end{pmatrix} = i'\begin{pmatrix}v&1_{A'}\end{pmatrix}. \end{align*} Hence, with \begin{align*} d:=\begin{pmatrix}v&1_{A'}\end{pmatrix}c:P\to A', \end{align*} we have \begin{align*} a\rho = \pi\ell = \pi mc = i'd. \end{align*} Because $i'$ is monic, the two morphisms $d\operatorname{pr}_1,d\operatorname{pr}_2:P\times_Y P\to A'$ agree: after composing with $i'$ both become $a\rho\operatorname{pr}_1=a\rho\operatorname{pr}_2$. Thus $d$ descends along the epimorphism $\rho$ to a unique morphism $b:Y\to A'$ satisfying \begin{align*} b\rho=d. \end{align*} Then \begin{align*} i'b\rho=i'd=a\rho, \end{align*} and since $\rho$ is an epimorphism, $i'b=a$. Uniqueness of $b$ follows from monicity of $i'$: if $b_1,b_2:Y\to A'$ satisfy $i'b_1=i'b_2=a$, then $b_1=b_2$. Therefore $i'$ satisfies the universal property of $\ker p'$.[/step]

[guided]We must prove the actual kernel universal property: $p'i'=0_{A',C}$, every morphism killed by $p'$ factors through $i'$, and that factorisation is unique. The equality $p'i'=0_{A',C}$ was built into the definition of $p'$, so the work is to prove existence and uniqueness of the factorisation without assuming that an arbitrary morphism into $E'$ lifts through the quotient map. We use the standard construction of pushouts in an abelian category. Define \begin{align*} j:A\to E\oplus A', \qquad j=\begin{pmatrix} i \\ -v \end{pmatrix}. \end{align*} The morphism $j$ is monic because its first component is $i$, and $i$ is monic as the kernel of $p$: if $t:Y\to A$ satisfies $jt=0$, then $it=0$, hence $t=0$. Let \begin{align*} \pi:E\oplus A'\to E' \end{align*} be the cokernel of $j$. The equality $\pi j=0$ says \begin{align*} \pi\begin{pmatrix}i\\0\end{pmatrix} = \pi\begin{pmatrix}0\\v\end{pmatrix}, \end{align*} so the maps \begin{align*} \tilde v=\pi\begin{pmatrix}1_E\\0\end{pmatrix}, \qquad i'=\pi\begin{pmatrix}0\\1_{A'}\end{pmatrix} \end{align*} satisfy the pushout commutativity relation $\tilde v i=i'v$. Conversely, any compatible pair of morphisms out of $E$ and $A'$ kills $j$, so it factors uniquely through the cokernel $\pi$. This proves that the cokernel construction represents the given pushout. Now define the morphism \begin{align*} \bar p:E\oplus A'\to C, \qquad \bar p=\begin{pmatrix}p&0_{A',C}\end{pmatrix}. \end{align*} Since $pi=0_{A,C}$, we compute \begin{align*} \bar p j = \begin{pmatrix}p&0_{A',C}\end{pmatrix} \begin{pmatrix}i\\-v\end{pmatrix} = pi-0_{A',C}v = 0_{A,C}. \end{align*} Therefore $\bar p$ factors uniquely through the cokernel $\pi$. The induced morphism is the same $p':E'\to C$ constructed earlier, because it satisfies \begin{align*} p'\tilde v=p, \qquad p'i'=0_{A',C}. \end{align*} Equivalently, \begin{align*} p'\pi=\bar p. \end{align*} Before proving the factorisation property, we prove that $i'$ is monic; this avoids a circular uniqueness argument later. Let $b:Y\to A'$ satisfy $i'b=0$. Then \begin{align*} \pi\begin{pmatrix}0\\b\end{pmatrix}=0. \end{align*} Because $j$ is monic and $\pi=\operatorname{coker}j$, abelian categories identify every monomorphism with the kernel of its cokernel, so $j=\ker\pi$. Hence there is a morphism $t:Y\to A$ with \begin{align*} \begin{pmatrix}0\\b\end{pmatrix} = \begin{pmatrix}i\\-v\end{pmatrix}t. \end{align*} Taking components gives $it=0$ and $b=-vt$. Since $i$ is monic, $t=0$, and hence $b=0$. Thus $i'$ is monic. Now take an arbitrary object $Y$ and a morphism $a:Y\to E'$ satisfying $p'a=0_{Y,C}$. We cannot assume that $a$ itself lifts through $\pi:E\oplus A'\to E'$. Instead, we pull back the epimorphism $\pi$ along $a$: \begin{align*} \begin{array}{ccc} P & \xrightarrow{\ell} & E\oplus A' \\ \downarrow \rho & & \downarrow \pi \\ Y & \xrightarrow{a} & E'. \end{array} \end{align*} Cokernels are epimorphisms, so $\pi$ is an epimorphism; epimorphisms are stable under pullback in an abelian category, so $\rho:P\to Y$ is an epimorphism. The pullback relation is \begin{align*} \pi\ell=a\rho. \end{align*} Using $p'a=0_{Y,C}$ and $p'\pi=\bar p$, we get \begin{align*} \bar p\ell = p'\pi\ell = p'a\rho = 0_{P,C}. \end{align*} Thus $\ell$ lands in the kernel of $\bar p$. We identify that kernel explicitly. Since $i=\ker p$, a morphism into $E\oplus A'$ is killed by $\bar p=\begin{pmatrix}p&0_{A',C}\end{pmatrix}$ exactly when its $E$-component is killed by $p$, and therefore exactly when its $E$-component factors uniquely through $i$. Hence \begin{align*} m:A\oplus A'\to E\oplus A', \qquad m= \begin{pmatrix} i&0\\ 0&1_{A'} \end{pmatrix} \end{align*} is $\ker\bar p$. Since $\bar p\ell=0$, there is a unique morphism \begin{align*} c:P\to A\oplus A' \end{align*} with $mc=\ell$. Now we use the pushout relation to see what happens after applying $\pi$. We compute \begin{align*} \pi m = \pi \begin{pmatrix} i&0\\ 0&1_{A'} \end{pmatrix} = i'\begin{pmatrix}v&1_{A'}\end{pmatrix}. \end{align*} Define \begin{align*} d:=\begin{pmatrix}v&1_{A'}\end{pmatrix}c:P\to A'. \end{align*} Then \begin{align*} a\rho = \pi\ell = \pi mc = i'd. \end{align*} This proves that $a$ factors through $i'$ after the epimorphic cover $\rho:P\to Y$. It remains to descend this local factorisation from $P$ to $Y$. Let $\operatorname{pr}_1,\operatorname{pr}_2:P\times_Y P\to P$ be the two projections. Since $\rho\operatorname{pr}_1=\rho\operatorname{pr}_2$, we have \begin{align*} i'd\operatorname{pr}_1 = a\rho\operatorname{pr}_1 = a\rho\operatorname{pr}_2 = i'd\operatorname{pr}_2. \end{align*} Because $i'$ is monic, this implies $d\operatorname{pr}_1=d\operatorname{pr}_2$. The universal property of the coequalizer of the kernel pair of the epimorphism $\rho$ gives a unique morphism $b:Y\to A'$ such that \begin{align*} b\rho=d. \end{align*} Then \begin{align*} i'b\rho=i'd=a\rho, \end{align*} and epimorphy of $\rho$ gives $i'b=a$. If $b_1,b_2:Y\to A'$ both satisfy $i'b_1=i'b_2=a$, then monicity of $i'$ gives $b_1=b_2$. Therefore every morphism killed by $p'$ factors uniquely through $i'$, so $i'=\ker p'$.[/guided]

[step:Conclude that the pushed-out sequence is short exact] We have proved that $i'=\ker p'$ and $p'=\operatorname{coker} i'$. Hence the sequence \begin{align*} 0 \longrightarrow A' \xrightarrow{i'} E' \xrightarrow{p'} C \longrightarrow 0 \end{align*} is exact at $A'$, $E'$, and $C$. Since every kernel morphism is a monomorphism and every cokernel morphism is an epimorphism in an abelian category, the displayed sequence is short exact. This proves the assertion. [/step]