A pushout exists for a diagram on three groups? From Rotman's Algebraic Topology, I have a question about a particular part of this proof.

If $f_1: B \rightarrow A_1$ and $f_2:B \rightarrow A_2$ is a diagram of groups then a pushout exists.

Proof:  Let $C$ be the pushout defined as $C=(A_1 * A_2)/N$, where $A_1 * A_2$ is the free product and $N$ is the normal subgroup generated by $\{f_1(b)f_2(b^{-1}): b \in B\}$.
Define $g_i(a_i) = a_iN$ as a map between $A_i \rightarrow C$, then it should follow that the diagram commutes as $C$ is a solution.
I see that I would have to show that $g_1f_1 = g_2f_2$, but I'm not sure how I'd show that $f_1(b)N = f_2(b)N$.
I can see that $N=\langle \prod_{b \in B} f_1(b)f_2(b^{-1})\rangle$, but I'm not sure how to proceed.
Anyone have any ideas?
 A: You want $f_1(b)N=f_2(b)N$, and this is equivalent to $N=f_1(b)^{-1}f_2(b)N$, i.e., $f_1(b)^{-1}f_2(b)\in N$. This isn't quite what you have, but $f_1(b)^{-1}=f_1(b^{-1})$ since $f_1$ is a homomorphism, and if $b\in B$ then so is $b^{-1}$, so we may replace $b$ by $b^{-1}$ to see that $f_1(b)^{-1}f_2(b)\in N$ if and only if $f_1(b)f_2(b^{-1})\in N$.
I don't know if this is what your problem with it is, or what you particularly want, but if you let me know perhaps I can either explain what's going on or give you a source that also proves it.
A: Part of what's giving you trouble is that the equation $g_i(a_i)=a_i N$ does not make sense: $a_i$ is an element of $A_i$, whereas $N$ is a subgroup of $A_1 * A_2$, therefore $a_i N$ is not defined.
What you can instead do is to define the homomorphism
$$g_i : A_i \xrightarrow{j_i} A_1 * A_2 \xrightarrow{q} A_1 * A_2 / N = C
$$
where 


*

*$j_i : A_i \to A_1 * A_2$ is the canonical injection (defined for each $i=1,2$),

*$q : A_1 * A_2 \to A_1 * A_2 / N$ is the canonical surjection defined by $q(x)=xN$ for all $x \in A_1 * A_2$.


With this new definition of $g_1$ and $g_2$, probably you can finish the proof with ease.
