Exact sequence involving pushout I think I have a bit of a misunderstanding of pushouts. Let $U, X, Y$ be objects of an abelian category $A$ with morphisms $f: U \rightarrow X$ and $g: U \rightarrow Y.$ Let $Z$ be the pushout of this diagram with natural morphisms $f': X \rightarrow Z$ and $g': Y \rightarrow Z.$ Then apparently we have a short exact sequence $$0 \rightarrow U \rightarrow X \oplus Y \rightarrow Z \rightarrow 0 $$ where the map $U \rightarrow X \oplus Y$ is defined as $u \mapsto (f(u), g(u))$ and the map $X \oplus Y \rightarrow Z$ is denoted by $(f', g').$ I cannot see how this is true. My book states the mapping of the short exact sequence in this way but I think the map $X \oplus Y \rightarrow Z$ should rather be $(-f', g')$ or $(f', -g')$ where $(-f', g')(x, y) = -f'(x) + g'(y).$ Am I right or am I misunderstanding something?

Here is a screenshot of the passage. The arrow notation is a bit different...
 A: 
$\require{AMScd}$In an abelian category, a commutative square
  \begin{CD}
U@>f>> X\\
@VgVV@VVg'V\\
Y@>>f'> Z
\end{CD}
  is a pushout if and only if the sequence
  $$U\xrightarrow{[f,g]}X\oplus Y\xrightarrow{\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]}Z\to 0$$
  is exact.

The exactness of the sequence is equivalent to the fact that $\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]$ is the cokernel of $[f,g]$.
The square is commutative if and only if $fg'=gf'$ (diagrammatic composition order) which is equivalent to
$$[f,g]\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]=fg'-gf'=0$$
On the other hand, if $[f,g]\left[\genfrac{}{}{0pt}{}{p}{q}\right]=fp+gq=0:U\to V$, then $fp=-gq$ hence there exists one and only one morphism $h:Z\to V$ such that $p=g'h$ and $-q=f'h$, hence
\begin{align}
\left[\genfrac{}{}{0pt}{}{p}{q}\right]
&=\left[\genfrac{}{}{0pt}{}{g'h}{-f'h}\right]\\
&=\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]h
\end{align}
thus proving that $\left[\genfrac{}{}{0pt}{}{g'}{-f'}\right]$ is the cokernel of $[f,g]$.
