We are given a finite path-connected simplicial complex $X = A \cup B$, where $A$, $B$ and $A \cap B$ are sub-simplical complexes.
If $A \cap B$ is homeomorphic to a circle and bounds discs in each of $A$ and $B$, what can be said about the fundamental group $\pi_1(X)$ in terms of the fundamental groups of the subspaces.
My thoughts:
I tried to find inspiration from the sphere, but $\pi_1(A) = \pi_1(B) = \{e\}$ in this case so this example doesn't help too much. Anyway, we have $\pi_1(A \cap B) = <c> \cong \mathbb{Z}$, where $<c>$ denotes the free group on one letter, I've chosen $c$. Suppose that $\pi_1(A) = <a_1,...,a_n | r_1,..r_p>$ and $\pi_1(B) = <b_1,...,b_m | s_1,..s_q>$.
Next, we need to work out homomorphisms $i_1: \pi_1(A \cap B) \to \pi_1(A) $ and $i_2: \pi_1(A \cap B) \to \pi_1(B) $ to get relation $i_1(c) = i_2(c)$.
I suppose this is where the fact that $A \cap B$ bounds a disc comes in, I think the idea here is that the homomorphisms give us the relation $a_i = b_j$ for some $i,j$ (honestly this is just a stab in the dark.)
So $\pi_1(X) = <a_1,.a_{i-1},a_{i+1},..,a_n,b_1,...,b_m | r_1,..r_p,s_1,..s_q>$
My main problem here is understanding what the significance of $A \cap B$ bounding discs in $A$ and $B$ is?
Thanks!