# Free amalgated product and injectivity (using universal property)

Consider the following commutative diagram of group homomorphisms ( I know it looks ugly): $$\begin{array}{ccccccccc} & & G_1 & \\ & \nearrow{i_1} & & \searrow{f_1} & \\ A & & & & G \\ & \searrow{i_2} & & \nearrow{f_2} & \\ & & G_2 & \end{array}$$

here $i_1,i_2$ are injective. We say that $G$ is the amlagated product $G_1*_A G_2$ if for any pair of homomorphisms $\phi_i:G_i\to H$ with $\phi_1 \circ i_1=\phi_2 \circ i_2$ there exists a unique homomorphism $\phi:G\to H$ such that $\phi \circ f_i=\phi_i$.

I want to show that then the composition $f_1 \circ i_1$ is injecetive, using the universal property above.

I already know how to show it using the explicit construction of $G$ as a quotient of $G_1 *G_2$, but I was not able to mimic the proof.

• math.meta.stackexchange.com/questions/2324/… You can't do diagonal arrows with AMScd, but that should be fine enough for this example (just rotate it 45 degrees clockwise). – Derek Elkins Jul 8 '18 at 15:20
• Are you sure it is $\phi\circ i_i=\phi_i$ and not $f_i$ instead of $i_i$?. – Javi Jul 8 '18 at 15:33
• In general, one can ask if there is a conceptual way to prove that colimits are 'large' (that they are small, i.e. generated by the images of the arrows to it, follows from the universal property). I don't know if there is. It seems that we always need to rely on the explicit construction. (E.g.: why is the dimension of the tensor product of vector spaces at least the product of the dimensions? Why is the polynomial ring $R[X]$ strictly larger than $R$?) – punctured dusk Jul 8 '18 at 16:28
• @Javi, you are right, I will fix it – klirk Jul 8 '18 at 16:30
• @barto: what do you mean with "large" – klirk Jul 8 '18 at 17:43