Let $A_0$, $A_1$ and $A_2$ be three finitely presented groups, together with a diagram $A_1 \xleftarrow{j_1} A_0 \xrightarrow{j_2} A_2$. Let $A$ be the pushforward of the diagram, $f_i: A_i \to A$ the characteristic map and suppose first that $j_i$ is a monomorphism for both $i=1,2$ . It seems like an almost trivial thing to answer, but somehow I cannot wrap my head around the following question:
Is it true that each $f_i$ is again a monomorphism ?
I have the following idea, i don't know if it is all sound, and it goes as this:
Let $\{a_k\}_{k=1}^l$ be a (finite) generating set of $A_0$. The characteristic map $f_1: A_1 \to A$ is the composition of the inclusion $g_1: A_1 \to A_1 \ast A_2$ and the projection $\pi: A_1 \ast A_2 \to A$, where we divide out the smallest normal subgroup $N \unlhd A_1 \ast A_2$ generated by words of the form $j_1(a_k) \ast j_2(a_k)^{-1}$ for $k=1,..,l$. The fact that $j_2$ is a monomorphism seems to already imply that $g_1(A_1) \cap N = \{1\}$. That is since $j_2$ is injective, we get $j_2(a_k) \neq 1$ for each $k$, so each non-trivial word in $N$ must have some letters in $A_2$. Is this true ? Since $g_1$ is trivially injective, so must be $f_1$. By symmetry, we therefore have the following implications:
1) $j_2$ injective $\implies$ $f_1$ injective
2) $j_1$ injective $\implies$ $f_2$ injective.
Is this the right idea to go about ? Any help is appreciated