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We define a set $S \subset \Bbb{R}^n$ to be star convex if there exists $a \in S$, such that the line segment connecting $a$ and any other point in $S$ lies entirely in $S$. I would like to show that it's simply connected. Can someone verify my proof?

The set $S$ is certainly path connected since given $x,y \in S$, we can construct a path from $x$ to $a$ and $a$ to $y$ , and so adjoining the paths yields a path from $x$ to $y$. Also given any loop $p(r)$, $r \in [0, 1]$, we have a straight-line homotopy

$$H(r, t) = ta + (1-t)p(r)$$

with $H((r, 0) = p(r)$ and $H(r, 1) = a$, so $p$ is homotopic to a point, meaning $S$ is simply connected.

Libertron
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Student1
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    Looks good to me. – Rankeya Feb 27 '13 at 05:42
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    Agreed, this looks fine to me. – Alex Youcis Feb 27 '13 at 06:24
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    This looks quite correct, the only small quibble I see is that the wording “we can construct a path from $x$ to $a$” is more indefinite than necessary. Why not just say there is a straight line path from $x$ to $a$? When writing proofs (especially if this is for homework) it is good to add justification, however brief, to handwaving phrases like “we can construct”. – Erick Wong Jan 12 '18 at 02:57
  • You also need to specify that the loop $p(r)$ is based at point $a$ so that the straight-line homotopy works. Then, since the set is path connected, you can infer that the fundamental group at every other point is the same and the rest works like you have shown. – Aritra Das Apr 21 '20 at 06:16

3 Answers3

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Yes, your proposed proof is correct.

davidlowryduda
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Not quite correct, you need to check that $H(x,t)$ always lies in the set $S$ (because star convex set need not be convex).

Apurv Anand
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  • This is correct since the straight line from the star-center $a$ to any other point is always contained in $S$--you are thinking of the straight line homotopy between any two points $f(x)$ and $g(x)$, which indeed is not necessarily in $S$ since $S$ is not necessarily convex. – Nap D. Lover Dec 06 '23 at 19:59
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Doesn't look right because to show something is Simply Connected, you need to show that the fundamental group is trivial alongside X being path connected. This means that you would want to create a path homotopy and not just a homotopy.

Sahir Gill
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