[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 $\ker p'$.
[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]
[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]
