0
$\begingroup$

I want to compute the fundamental group of the set C defined below :

$A_{1}:=[0,1]²,A_2:=[-1,0]\times[0,1],C=\partial A_1 \cup \partial A_2$.

I have to use the Van Kampen theorem and so I know that I must exhibit two open sets that are arc-connected [and cover the space] but I do not see how.

$\endgroup$
2
  • $\begingroup$ There are lots of open sets that are arc-connected. What exactly are you trying to do? $\endgroup$
    – anomaly
    Mar 9, 2015 at 13:58
  • $\begingroup$ @anomaly i am triying to compute the fundamental group of C via Van Kampen theorem. $\endgroup$
    – Far
    Mar 9, 2015 at 14:12

2 Answers 2

1
$\begingroup$

consider $U =(-1/2,1]$x$[0,1] \cap C$ and $V= [-1,1/2)$x$[0,1] \cap C$ then $U$ & $V$ is d.r to a circle...and $U\cap V$ is contractible...so fundametal group will be $\mathbb{Z*Z}$...to see this draw pictures

$\endgroup$
6
  • $\begingroup$ Thank you. Your U and V are they open sets ? I have tried to draw a picture but it seems complicated. $\endgroup$
    – Far
    Mar 9, 2015 at 14:15
  • $\begingroup$ they are open in $C$...why it is complicated $C$ is basically looks like two square are gluing together with respect to an edge...can you see this??? $\endgroup$ Mar 9, 2015 at 14:17
  • $\begingroup$ You are right! They are open in C.So they do not have to be open sets in $\mathbb{R}^2$?. Thank you. $\endgroup$
    – Far
    Mar 9, 2015 at 14:25
  • $\begingroup$ So $U\cap V = ]-1/2,1/2[\times [0,1]\cup C$ and $U\cup V=A_1 \cup A_2$ not equal to $\partial A_1 \cup \partial A_2$.Sorry i just started learning this. So i understand slowly! $\endgroup$
    – Far
    Mar 9, 2015 at 14:50
  • $\begingroup$ sorrry...that was a big TYPO $\endgroup$ Mar 9, 2015 at 17:24
0
$\begingroup$

Ok. I can take $U:= \partial A_1 \cup ([0,-1/2[\times \{0\})\cup ( [0,-1/2[\times \{1\}) $ idem for V : $V:= \partial A_2 \cup ([0,1/2[\times \{0\})\cup ([0,1/2[\times \{1\}) $.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .