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Let $X=\{p,q,r,s\}$ with topology induced by basis $\{p\}$,$\{r\}$,$\{p,q,r\}$, $\{p,r,s\}$. Then $X$ is not simply connected.

My Attempt:

$\{p\}$ is open set and $\{q\}$ is closed set.

Define $\phi:[0,1]\to X$, where

$$\phi(x)=\begin{cases} q & x=0,1,\\ p & x\in (0,1). \end{cases}.$$

$\phi$ is a continuous loop. But how to show it is not homotopic to constant?

Any Help will be appreciated.

Arctic Char
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Curious student
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  • Alternatively, can you find the fundamental group of $X$? – Math1000 Nov 30 '19 at 05:05
  • You should try with a slight different curve. There are only two points you can go continuously to, from p, but you can't go from either of them to the other. – deb Nov 30 '19 at 05:15
  • @Math1000 How to find fundamental group using just basis ? I tried to use Van Kampen theorem But unable to do? Please can you give me some hint – Curious student Nov 30 '19 at 05:25
  • If you know the groupoid version of van Kampen then you can use it. If not then you probably have to use covering spaces. – Eric Wofsey Nov 30 '19 at 06:15
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    This question is probably a duplicate though, let me see if I can find a link. – Eric Wofsey Nov 30 '19 at 06:16
  • The question I linked is different, but my answer there answers your question just as well (by constructing a nontrivial covering space of $X$, or alternatively by constructing a weak homotopy equivalence $S^1\to X$). – Eric Wofsey Nov 30 '19 at 06:29
  • See also https://math.stackexchange.com/questions/161580/an-interesting-topological-space-with-4-elements for related discussion and references. – Eric Wofsey Nov 30 '19 at 06:32

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