I have an epimorphism of groups $f:G\to H*K$. Is it true that $G=H'*K'$ for the preimages $H'=f^{-1}(H)$ and $K'=f^{-1}(K)$?
UPDATE: (after Tsemo's answer) I have a group $G$ generated by two of its subgroups: $G=\langle H',K'\rangle$, and an epimorphism of $G$ onto a free product of two groups: $f:G\to H*K$. Under this epimorphism $f$, $H'$ and $K'$ map isomorphically onto $H$ and $K$ respectively. Is it true then, that $G=H'*K'$?
No, take the product of a free group $G$ which $H$ which is not free and the projection onto $G$.
Let $g\in Ker f$, $g=h_1k_1h_2k_2...h_nk_n, h_i\in H, k_i\in K$, $f(g)=f(h_1)f(k_1)..f(h_n)f(k_n)=1$ where $f(h_i)\in H, f(k_i)\in K$, since $H*K$ is a free product, $f(h_i)=f(k_i)=1$ and $h_i=k_i=1$ since the restrictions of $f$ to $H'$ and $K'$ are injective.