Fundamental group of a 3-dimensional figure

Question

Let $X$ be a regular hexagon in $\Bbb C$ with centre in $\mathbf 0$, and name his sides $a$, $b$, $c$, $d$, $e$ and $f$ counterclockwise. Let's consider the equivalence relation $\sim$, that identifies $a$ with $b$, $c$ with $d$ and $e$ with $f$. Now, solutions says that I can use van Kampen with $U=\operatorname B(\mathbf 0,\varepsilon)/\sim\ $and $V=(X\setminus \mathbf 0)/\sim$: clearly $U$ retracts to a point and $U\cap V$ retracts to $S^1$. However, I don't see why $V$ has fundamental group $S^1\vee S^1\vee S^1$: it's true that it retracts to $\partial X/\sim$, but if I glue the consecutive sides of the hexagon I don't obtain a bouquet of circles; I would if I glued the opposite sides (like in the torus, for example). Thank you in advance.

Answer 1

You do obtain a bouquet of circles : $a = b$ so $a$ is actually a circle (its endpoint is the beginning point of $b$ so the beginning point of $a$), same for $c$ and for $e$, and these three circles only touch eachother at their respective endpoints, so they form a bouquet of circles (note that in $\partial X$ you only have the "outer" hexagon, it's not filled in)