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Let $D^2=\{x\in\mathbb{R}^2:\|x\|\le1\}$, $x\neq a\in D^2$. Find $\pi_1(X\setminus\{x\},a)$ if:

a. $x\in\partial D^2$

b. $x\in \text{int} D^2$

about the first one I think the fundamental group is trivial (means one singleton $\{a\}$) but I can't figure out how can I prove it.

about the second, it seems clear that after removing point from the interior, it's homotopic to $S^1$ (which its fundamental group is known) but I can't find (for example) a retraction which sends the disc with hole in the interior to $S^1$.

So which homomorphism between $\pi_1(D^2\setminus\{x\})\to\mathbb{Z}\setminus\{0\}$ can I pick ? how can I define a deformation retraction between $D^2\setminus\{x\}$ where $x\in\text{int}D^2$ and $S^1$?

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    $\begingroup$ $\mathbb Z$ minus any one of it's elements is not a group $\endgroup$
    – David P
    Commented Dec 9, 2014 at 3:03
  • $\begingroup$ you are obviously right. I fixed this. $\endgroup$
    – user65985
    Commented Dec 9, 2014 at 3:08

1 Answer 1

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(a) Notice that your region is contractible (in fact, starlike).

(b) Note that $S^1$ can be identified with the boundary of $D^2$. So, we can can imagine a homotopy from $D^2 - \{x\}$ to $S^1$ by pulling wider the "hole" at $x$; one way to do precisely is by projecting radially from $x$ onto $S^1$.

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  • $\begingroup$ thanks. (a) Why is it true? You mean pulling $S^1$ to the hole and the $D^2\setminus S^1$ goes by linear lines to the hole? (b) what do you mean "pulling the hole"? to draw radius between the origin and the point and take gradually the line betwene the point and the intersection of the line with $S^1$? $\endgroup$
    – user65985
    Commented Dec 9, 2014 at 3:06
  • $\begingroup$ You're welcome. For (a), do you know the definition of starlike? (Any starlike region, most or less by definition, admits a straight-line deformation retract onto a point.) For (b), yes, that's what I mean by "projecting radially". $\endgroup$ Commented Dec 9, 2014 at 3:15
  • $\begingroup$ For (a): I know the definition but I can't understand how it's applied logically here? You can define a homotopy between $D^2\setminus\{x\}\to\{x\}$ by the proceeding on the line which connects $x,y$ for any y on the disk but somehow it doesn't seem legit. $\endgroup$
    – user65985
    Commented Dec 9, 2014 at 3:22

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