Hello!
Let $K = G\ast\_U H$ be the free product of the groups $G$ and $H$ amalgamated along the common subgroup $U$. I have a homomorphism of groups $f:K\to K'$ defined using the universal property of free products (i.e., it is the unique map determined by two homomorphisms $j\_1 : G\to K'$ and $j\_2 : H\to K'$ that agree on $U$). I know that both $j\_1$ and $j\_2$ are injective and that the intersection of their images in $K'$ is just the image of $U$. 


(1) Is it sufficient to conclude that the universal morphism $f$ is injective on $K$? 


(2) If, as I suspect, the answer is no, is there some criterion to derive the injectivity of $f$? For example, let $A$ and $B$ denote the images of $j\_1$ and $j\_2$.
Is it sufficient to prove that for each $h\in H\setminus U$ the double coset $A j\_2(h) A$ is disjoint from $B$, plus the similar statement exchanging the role of the two subgroups?



Injectivity of a group homomorphism with domain a free product
If you have a homomorphism to $K'$ where $K'$ acts on a set $X$, then you could use ping-pong arguments to verify injectivity. In your setting, this is called 1-st Maskit combination theorem, you can find some examples in Maskit's book "Kleinian groups". If $K=G\_1\star G\_2$, then ping-poing setting is a pair of subsets $A\_1, A\_2\subset X$ so that $A\_1\cap A\_2\ne \emptyset$, and $\forall g\in G\_i\setminus 1$, $gA\_i\subset A\_i^c\cap A\_{i+1}$, where $A\_i^c=X\setminus A\_i$, $i=1,2$ mod 2. Under this condition, you do get a faithful action of $K$ on $X$. 


Similar thing works for more general amalgams $K=G\_1 \star\_H G\_2$, but the ping-pong condition becomes more technical. 


*Edit:* Here is the ping-pong setting for general amalgams (Maskit's 1st combination theorem). You have an action of a group $G=G\_1\star\_H G\_2$ on a set $X$, where each $G\_i$ acts faithfully and you are trying to determine if the action of $G$ is faithful. A subset $X'\subset X$ (which need not be $G$-invariant) is partitioned in three (disjoint, nonempty) subsets $U\_1, U\_2, S$. You should think of two complementary components ("open half-spaces" or "halfs of the ping-pong table") to a hypersurface $S$ ("the net on the ping-pong table") in a manifold $X$. If $X$ is a complex manifold, you are in trouble since such $X$ does not come equipped with natural real hypersurfaces. One frequently uses a bisector for a suitably chosen metric on $X$ to define such $S$. Define $V\_i=U\_i\cup S$, $i=1,2$ ("closed half-spaces"). 


The ping-pong assumption then is that each $V\_i$ ($i=1,2$) is *strictly invariant* under the subgroup $H\le G\_i$, which means that $h(V\_i)=V\_i$ for every $h\in H$, and $g(V\_i)\subset U\_{3-i}, \forall g\in G\_i\setminus H$. (The terminology *strictly invariant* is Maskit's. Notice that $U$'s and $V$'s both appear in the definition.) Examples could be found in Maskit's book. 


If you think of $g\in G\_i\setminus H$ as a player on the $i$-th team, then it sends the ball 
(point in $V\_i$) from its side of the table (including the net) to the opponent's side $U\_{3-i}$. 


If such partition of $X$ exists, then the action of $G$ on $X$ is faithful. The proof is a direct application of the normal form for the elements of $G$. Namely, write $g\in G$ as the product, say, $g=h\_0 g\_1 h\_1 g\_2 h\_2 ... g\_{n}$ where each $h\_j\in H$, $g\_{2m+1}\in G\_1\setminus H, g\_{2m}\in G\_2\setminus H$. Now, pick a point $x\in S$ and inductively apply each letter in this product to $x$ and check where the image lands: You will see that it never returns to $S$, unless $g=h\_0\in H$. 


There is a similar combination theorem for HNN extensions ("2nd combination theorem"). 


