Let
$\require{AMScd}$
\begin{CD}
 P @>{g'}>> B\\
 @Vf'VV @VVfV\\
 A @>>g> C
\end{CD}
be a pullback. $f'$ is iso iff there exists $h:A\to B$ such that 
$\require{AMScd}$
$f\circ h= g$

Proof.
$\Rightarrow$) $f'$ is an iso, so we have $f'^{-1}: A\to P$. We can define $h=g'\circ f'^{-1}$.

$\Leftarrow$) I have problem here. My professor drew this today and said something about some triangles that commute and uniqueness "so $k$ have to be $f'^{-1}$, but I cant' understand how to say this.
That's the diagram:[![enter image description here](https://i.stack.imgur.com/jIDHq.png)](https://i.stack.imgur.com/jIDHq.png)

Thanks to everyone who can help!