Suppose we have a free product with amalgamation $G = A \ast_C B$ with $A \neq C$ and $B \neq C$.
I want to show that if we have a surjective homomorphism $f : H \to G$, where $H$ is some group (no conditions on it), then $H = A' \ast_{C'} B'$ with $A' \neq C' \neq B'$.
My ideas so far:
The map is surjective so $\frac{H}{\ker f} \cong G$. If your quotient is an amalgamated free product, are you also one? Probably not.
Maybe from a geometric point of view: $G$ acts on a tree $T$ with fundamental domain $e = [P,Q]$ (a single edge with two vertices), with $\text{stab}(P) = A$, $\text{stab}(Q) = B$ and $\text{stab}(e) = C$. $H$ acts on this tree too, via $f$. I think maybe the fundamental domain $T/H$ is also just one edge so $H$ is an amalgamated free product. But I can't really see why this is true.
The map $f\colon H \to G$ gives an action of $H$ on a tree with identical quotient but different stabilizer information. In this case, write $K = \ker f$. Then $\operatorname{stab}(P) = f^{-1}(A)$, $\operatorname{stab}(Q) = f^{-1}(B)$, and $\operatorname{stab}(e) = f^{-1}(C)$. One way to see this is that under this action, the stabilizer in $H$ with this action should be the full preimage of the stabilizer in $G$, which is exactly what we've written.