Pull-backs of diagrams of groups with free product.

Question

Until recently I calculated only pull-back of diagrams of finite groups. Now I am trying to calculate the pull-back of diagram of groups when the groups are free products of other groups. It seems very hard to prove that a group $G$ is a pull-back of such diagram by definition. Is there an easier way to either calculate the pull-back or proving that a group is the pull-back of such diagram. My easiest example I am interested is the following: $G_1=F_1*C_3$, $G_2=F_2*C_2$ with natural epimorphisms to the group $C_6$ (the generator of $F_1$ goes to an element of order 2 in $C_6$ and generators of $F_2$ goes to elements of order 3 in $C_6$). Here $F_n$ is the free group with $n$ generators and $*$ is the free product.

Answer 1

I'm assuming that you've been given the definition of the pullback in terms of the universal property, which can be difficult to work with in concrete calculations (but is usually better for proving general theorems). I'll give the equivalent definition in terms of elements of the groups, which will hopefully make things easier for you.

Definition If $f\colon H\to G$ and $p\colon H'\to G$ are group homomorphisms and $p$ is further surjective, then the pullback $f^{\ast}H'$ is the group of pairs $(h,h')$ such that $f(h)=p(h')$ with component-wise multiplication.

You can also think of this group equivalently as the preimage of the diagonal subgroup $\Delta G=\{(g,g)\mid g\in G\}$ of the group $G\times G$ under the product homomorphism $f\times p\colon H\times H'\to G\times G$ and so $$f^{\ast}H'=(f\times p)^{-1}(\Delta G).$$

As $f^{\ast}H'$ is a subgroup of $H\times H'$, it also comes equipped with a pair of projection maps onto $H$ and $H'$ which are just the restrictions of the usual projections to the subgroup $f^{\ast}H'$ and it's easy to verify that these maps complete the pullback diagram.