[proofplan]
We construct $i'$ from the [pullback](/page/Pullback) universal property using the compatible pair consisting of $i:A\to E$ and the zero morphism $A\to C'$. We then prove that $i'$ is a [kernel](/page/Kernel) of $p'$ by testing an arbitrary morphism killed by $p'$. Finally, we prove that $p'$ is a [cokernel](/page/Cokernel) of $i'$ by invoking the standard stability theorem that cokernels in an [abelian category](/page/Abelian%20Category) are preserved by pullback, after verifying that $p$ is the cokernel in the original [short exact sequence](/page/Short%20Exact%20Sequence) and that the displayed square is the required pullback.
[/proofplan]
[step:Construct the induced morphism $i':A\to E'$ from the pullback property]
Let $0_{A,C'}:A\to C'$ denote the zero morphism. Since $i$ is a kernel of $p$, we have
\begin{align*}
p\circ i=0.
\end{align*}
Also
\begin{align*}
u\circ 0_{A,C'}=0.
\end{align*}
Thus
\begin{align*}
p\circ i=u\circ 0_{A,C'}.
\end{align*}
By the universal property of the [pullback](/page/Pullback) square, there exists a unique morphism
\begin{align*}
i':A\to E'
\end{align*}
such that
\begin{align*}
\tilde u\circ i'=i,
\qquad
p'\circ i'=0_{A,C'}.
\end{align*}
[/step]
[step:Show that $i'$ is a kernel of $p'$]
We prove the universal property of the [kernel](/page/Kernel). Since $p'\circ i'=0$, the morphism $i'$ is killed by $p'$.
Let $X$ be an object of $\mathcal A$, and let $f:X\to E'$ be a morphism satisfying
\begin{align*}
p'\circ f=0.
\end{align*}
Then
\begin{align*}
p\circ \tilde u\circ f
=
u\circ p'\circ f
=
0,
\end{align*}
where the first equality is the commutativity of the pullback square. Since $i:A\to E$ is a kernel of $p:E\to C$, there exists a unique morphism
\begin{align*}
a:X\to A
\end{align*}
such that
\begin{align*}
i\circ a=\tilde u\circ f.
\end{align*}
We claim that $i'\circ a=f$. Indeed,
\begin{align*}
\tilde u\circ i'\circ a=i\circ a=\tilde u\circ f
\end{align*}
and
\begin{align*}
p'\circ i'\circ a=0=p'\circ f.
\end{align*}
The pullback universal property gives uniqueness of a morphism $X\to E'$ with these two composites, hence $i'\circ a=f$.
If $b:X\to A$ also satisfies $i'\circ b=f$, then
\begin{align*}
i\circ b=\tilde u\circ i'\circ b=\tilde u\circ f=i\circ a.
\end{align*}
Since [kernel](/page/Kernel)s are [monomorphism](/page/Monomorphism)s, $i$ is monic, so $b=a$. Therefore $i'$ is a kernel of $p'$.
[guided]
We must show that $i':A\to E'$ has exactly the universal property that defines $\ker p'$. We already know
\begin{align*}
p'\circ i'=0
\end{align*}
from the construction of $i'$.
Now take any object $X$ of $\mathcal A$ and any morphism $f:X\to E'$ such that
\begin{align*}
p'\circ f=0.
\end{align*}
The goal is to force $f$ to factor uniquely through $i'$. Composing $f$ with the other pullback projection gives $\tilde u\circ f:X\to E$. Because the square commutes, we have
\begin{align*}
p\circ \tilde u\circ f
=
u\circ p'\circ f
=
0.
\end{align*}
Thus $\tilde u\circ f$ is killed by $p$. Since $i:A\to E$ is a kernel of $p$, there is a unique morphism $a:X\to A$ such that
\begin{align*}
i\circ a=\tilde u\circ f.
\end{align*}
It remains to check that this $a$ really gives the desired factorization through $E'$. The two morphisms $i'\circ a:X\to E'$ and $f:X\to E'$ have the same composites with both pullback projections:
\begin{align*}
\tilde u\circ i'\circ a=i\circ a=\tilde u\circ f
\end{align*}
and
\begin{align*}
p'\circ i'\circ a=0=p'\circ f.
\end{align*}
The pullback universal property says that a morphism into $E'$ is uniquely determined by these two compatible composites, so
\begin{align*}
i'\circ a=f.
\end{align*}
For uniqueness, suppose $b:X\to A$ also satisfies $i'\circ b=f$. Composing with $\tilde u$ gives
\begin{align*}
i\circ b=\tilde u\circ i'\circ b=\tilde u\circ f=i\circ a.
\end{align*}
A [kernel](/page/Kernel) is a [monomorphism](/page/Monomorphism), so $i$ is monic, and therefore $b=a$. This proves that $i'$ is a kernel of $p'$.
[/guided]
[/step]
[step:Show that $p'$ is a cokernel of $i'$]
We use the Pullbacks Preserve Cokernels in Abelian Categories theorem. Its hypotheses are: an abelian category, a cokernel $p:E\to C$, a morphism $u:C'\to C$, and a pullback square of $p$ along $u$. These hypotheses hold here because $\mathcal A$ is abelian, the original sequence is short exact so $p$ is a cokernel of $i$, $u:C'\to C$ is the given morphism, and the displayed square defines $E'$ as the pullback. The theorem therefore implies that the pullback projection $p':E'\to C'$ is a cokernel of the induced morphism $i':A\to E'$.
Equivalently, for every object $X$ of $\mathcal A$ and every morphism $f:E'\to X$ satisfying
\begin{align*}
f\circ i'=0,
\end{align*}
there exists a unique morphism
\begin{align*}
g:C'\to X
\end{align*}
such that
\begin{align*}
g\circ p'=f.
\end{align*}
Thus $p'$ is a cokernel of $i'$.
[guided]
The remaining exactness condition is the [cokernel](/page/Cokernel) condition. We already know
\begin{align*}
p'\circ i'=0,
\end{align*}
so $p'$ annihilates $i'$. What remains is the universal factorization property: every morphism out of $E'$ that annihilates $i'$ must factor uniquely through $p'$.
We now apply the Pullbacks Preserve Cokernels in Abelian Categories theorem. The theorem says that if $\mathcal A$ is an abelian category, $p:E\to C$ is a cokernel, $u:C'\to C$ is any morphism, and
\begin{align*}
\begin{array}{ccc}
E' & \xrightarrow{\tilde u} & E \\
\downarrow p' & & \downarrow p \\
C' & \xrightarrow{u} & C
\end{array}
\end{align*}
is a pullback square, then $p':E'\to C'$ is a cokernel of the morphism induced from the kernel of $p$.
We verify the hypotheses one by one. First, $\mathcal A$ is abelian by hypothesis. Second, the original sequence
\begin{align*}
0 \longrightarrow A \xrightarrow{i} E \xrightarrow{p} C \longrightarrow 0
\end{align*}
is short exact, so $p:E\to C$ is a cokernel of $i:A\to E$. Third, $u:C'\to C$ is the morphism specified in the theorem statement. Fourth, the square defining $E'$ is explicitly assumed to be a pullback square. Finally, the morphism induced from the kernel of $p$ is precisely the morphism $i':A\to E'$ constructed earlier: it is characterized by
\begin{align*}
\tilde u\circ i'=i,
\qquad
p'\circ i'=0_{A,C'}.
\end{align*}
Therefore the theorem applies and gives that $p':E'\to C'$ is a cokernel of $i':A\to E'$.
Unpacking this cokernel conclusion gives the desired factorization statement. Let $X$ be an object of $\mathcal A$, and let $f:E'\to X$ be a morphism satisfying
\begin{align*}
f\circ i'=0.
\end{align*}
Since $p'$ is a cokernel of $i'$, there exists a unique morphism
\begin{align*}
g:C'\to X
\end{align*}
such that
\begin{align*}
g\circ p'=f.
\end{align*}
This is exactly the universal property required to prove that $p'$ is a cokernel of $i'$.
[/guided]
[/step]
[step:Conclude short exactness of the pulled back sequence]
The morphism $i'$ is a [kernel](/page/Kernel) of $p'$, and $p'$ is a [cokernel](/page/Cokernel) of $i'$. Hence
\begin{align*}
0 \longrightarrow A \xrightarrow{i'} E' \xrightarrow{p'} C' \longrightarrow 0
\end{align*}
is [short exact](/page/Short%20Exact%20Sequence) in $\mathcal A$.
[/step]
