I'm studying for exams and came across this problem
My attempt
I'm going to use van Kampen. Label the bottom horizontal edge $c$ and the bottom right diagonal edge $d$. For our path connected open sets I'm going to choose $U$ to be an open disc inside $Y$, and $V$ to be $Y$ minus a point inside $U$. $U,V, U \cap V$ are all path connected with the latter being an annulus around the point.
Computing the Fundamental Groups:
$\pi_1(U)$ is trivial, as $U$ is an open disc so it's contractible.
$\pi_1(U\cap V) \cong \mathbb{Z} = \langle b| \emptyset \rangle$ as the annulus deformation retracts to $S^1$.
Question:
How do I find the Fundamental Group of $V$? I know that $V$ deformation retracts to just the boundary of this hexagon, so only the vertices and edges, but I can't figure out what the Fundamental Group of this boundary is.
Once I have this the problem should be simple: Using van Kampen
$$\pi_1(Y) \cong \langle \text{generators of $\pi_1(V)$}| \text{relations of $\pi_1(V)$}, i_1(b) = i_2(b) \rangle$$
Where $$i_1:\pi_1(U\cap V) \rightarrow \pi_1(U)$$
$$i_2:\pi_1(U\cap V) \rightarrow \pi_1(V)$$
$i_1$ is the trivial map, so the generator of $\pi_1(U\cap V)$ is sent to 1.
Going off of previous van Kampen problems I think $i_2: r\mapsto cdb^{-1}aab^{-1}$, where the generator of the $\pi_1(U\cap V)$ going once around the circle in the intersection, but goes around the boundary of the hexagon once. This can just be read off of the identification of the boundary. In which case I would get
$$\pi_1(Y) \cong \langle \text{generators of $\pi_1(V)$}| \text{relations of $\pi_1(V)$}, cdb^{-1}aab^{-1} =1 \rangle$$
All that is left is to find the generators and relations of $\pi_1(V)$, but I'm stuck.
Best Answer
You have exactly the right idea! We know the fundamental group of your figure is going to be
$$\langle \text{generators of $\pi_1(V)$} \mid \text{relations of $\pi_1(V)$}, cdb^{-1}aab^{-1} \rangle$$
So what we want to do now is compute $\pi_1(V)$. For this (as with all topology problems), a picture is the most important tool we have. Let's label our hexagon and its vertices:
Now. We know that we're identifying the $b$ edges. So their endpoints must be the same too! This tells us that, actually, $v_3 = v_0$ (since these are both the "back" of $b$) and $v_2 = v_5$ (since these are both the "front"). Analogously, we can identify the fronts and backs of the $a$ edges. This tells us that $v_3 = v_4$ and that $v_4 = v_5$. Putting these equalities together, we find that actually $v_0 = v_2 = v_3 = v_4 = v_5$!
So we can find out what our graph actually looks like by drawing each of these repeated vertices exactly once. This gives:
But now it's easy to read off the fundamental group from this picture. If we take $v_{0,2,3,4,5}$ as our basepoint, we find
$$\pi_1(V) = \langle a,b, cd \rangle$$
is the free group $F_3$ (which is not surprising, since the fundamental group of every graph is free. Do you see why?).
So now we find our final answer (writing $x$ for $cd$):
$$\pi_1(Y) = \langle a,b,x \mid xb^{-1}aab^{-1} \rangle$$
I hope this helps ^_^