In this question and this one, it has been asked how an exact sequence of modules $0 \to K \to A \to F \to 0$ and the pushout of $P \leftarrow K \to A$ give rise to a second exact sequence, namely a commutative diagram with both lines exact : $$\begin{matrix}0&\to&K&\to&A&\to&F&\to&0\\&&\downarrow&&\downarrow&&\downarrow\\0&\to&P&\to&U&\to&F&\to&0\end{matrix}$$
I understand how we get the injectivity of $P\to U$ and the surjectivity of $U \to F$, but both answers are unclear on why $\mathrm{Ker}(U \to F) = \mathrm{Im}(P \to U)$.
If I understand somewhat correctly, we use the pushout property on $A \to F$ and $0 : P \to F$ to get a unique $v : U \to F$ verifying $v \circ (A \to U) = (A \to F)$ and $v \circ (P \to U) = 0$. But I feel like there is no reason for $v$ to be the $U \to F$ morphism of the diagram.
Thanks for the help :)