0
$\begingroup$

I'm taking an introductory course in topology and we have a homework exercise to compute the fundamental group of X = $\mathbb{R} \big / \{-1, 1\}$ under the group action $\mathbb{Z}_2\times\mathbb{R} \rightarrow \mathbb{R}$ taking $(n, r)$ to $nr$.

The way I want to approach this is using the Van Kampen theorem(not explicitly stated). Is it correct to think of the equivalence classes of the quotient space as the set, $U$, of positive real numbers and the set, $V$, of negative real numbers (since we multiply by either 1 or -1)?

If so, since the union of $U$ and $V$ is X and their intersection is just $\{ 0 \}$, can we apply the Van Kampen theorem to said sets to conclude that $\pi_1(X) = \pi_1(U)*\pi_1(V) = 0$ in the sense that both $U$ and $V$ are both contractible spaces(or simply connected) and have trivial fundamental group.

An alternative approach would be to use the covering space theory, but I don't think the action is properly discontinuous since $\{0\} \cap n*\{0\} \neq \emptyset$, or any interval containing 0 for that matter. So I don't think that is a better approach. Any hint or clarification to how to use the Van Kampen theorem in this example would be appreciated!

$\endgroup$
  • 1
    $\begingroup$ The easiest way is to find that $X$ is homeomorphic to $[0,+\infty)$, and directly compute the fundamental group of that. Why do you want to use the Seifert/van Kampen theorem here? $\endgroup$ – Daniel Fischer Apr 23 '17 at 12:04
  • $\begingroup$ Thanks! Coming from some introductory homological algebra, a good way to proceed in computing homology groups was to use tools such as a mayer-vietoris to avoid doing explicit calculations on the original space. It was simply my first line of though to use other tools than direct computation. Do you recommend to generally work directly with homeomorphisms for computations of the fundamental group? $\endgroup$ – Vincent Haugdahl Apr 23 '17 at 12:44
  • 1
    $\begingroup$ No, when you have an obvious decomposition into parts whose fundamental group is easier to determine, Seifert/van Kampen is natural (there may for some spaces still be easier ways). One should choose one's strategy depending on the situation. If you have an obviously contractible space, you prove that it is contractible. If you have an obvious universal covering, you use that. If you have an obvious decomposition, you use Seifert/van Kampen. Unless you have a reason not to, such as wanting to get practice with a specific method (although even then it's probably better to use examples where… $\endgroup$ – Daniel Fischer Apr 23 '17 at 12:56
  • 1
    $\begingroup$ a different method isn't obviously better suited). In this case, it seems you just didn't see what $X$ is, and so missed the obvious. Happens to all of us more often than we wish. I just wondered whether you had a specific reason to want to use SvK. $\endgroup$ – Daniel Fischer Apr 23 '17 at 12:56
  • $\begingroup$ This particular example can be doone in various ways, as the comments show. But readers should be aware that if a discrete group $G$ acts on a topological space $X$ then it also acts on the fundamental groupoid $\pi_1(X)$ and also on the groupoid $\pi_1(X,A)$ where $A$ is a set of base points,provided $A$ is a union of orbits of the action. The algebra and topological applications of groups acting on groupoids is discussed in Chapter 11 of the book "Topology and Groupoids", with applications to orbit spaces. The algebraic results are due to Higgins and Taylor. $\endgroup$ – Ronnie Brown Apr 23 '17 at 20:28
0
$\begingroup$

Van Kampen doesn't seem like the way to go here to be honest. First of all, their intersection is not just $0$ under the equivalence relation, they are the same set, so this doesn't really work.

I think a better way to approach this is to find a space homeomorphic to the quotient space. Try $\Bbb R_{\geq 0}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.