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.

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Maxime Ramzi
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Dr. Scotti
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1How are $a$ and $b$ identified exactly? I think you should add a picture to your post where you represent $X$ as the hexagon with the sides identifications as in example #5 in this link. Also you wrote "clearly $U$ is homeomorphic to a point" I think you meant "$U$ is homeomorphic to a disk" or maybe "$U$ retracts onto a point". – Adam Chalumeau Jul 27 '19 at 10:28
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You are right, I made an edit – Dr. Scotti Jul 27 '19 at 11:28
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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)
Maxime Ramzi
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I thought I had to glue the whole sides not only the vertices; for example to obtain a torus I glue the opposite sides of a square. Why here I have to glue only the vertices? – Dr. Scotti Jul 27 '19 at 13:26
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You glue the whole sides : that's why I said $a=b$. To get a torus, first of all, you have a full square ( here you only have the boundary of the hexagon) and you glue the whole sides, just like here – Maxime Ramzi Jul 27 '19 at 14:26