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I need to prove $A$ is path-connected and its fundamental group is trivial. Path-connectedness is obvious. I know we already have some posts with proofs for this statement. But when we say we can construct a straight line homotopy between the constant map and any other loops, do we use the fact that $A$ is star convex or $A$ is star convex subset of $\Bbb R^n$?

My proof for trivial fundamental group is: let $[f], [g]$ be two homotopy classes of loops based at $x_0\in A$. Then pick $f_0\in [f]$, $g_0\in [g]$ and construct the straight-line homotopy $H(t,x) = (1-t)f_0(x)+tg_0(x)$. Hence, $f_0$ is homotopic to $g_0$.

In my proof of trivial fundamental group, I don't need any property of $A$? Do we use some property of $A$ to show that $H$ is a homotopy between $f_0$ and $g_0$?

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2 Answers 2

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Star convex is a property that makes sense only in vector spaces, so one may assume that $A$ is a subset of $\mathbb{R}^n$.

To be a homotopy, the image of $H$ must be in $A$! This is where your map failed to be a homotopy and why the star convex property of $A$ will come into place.

Let $x_0$ be a point with respect to $A$ is star convex, then: $$H(t,x):=(1-t)f_0(x)+tx_0,$$ is a homotopy between $f_0$ and the constant loop based at $x_0$ and $A$ is simply-connected.

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  • $\begingroup$ I think my map $H$ has some problem, but I don't understand why the image of my $H$ is not completely in $A$. Could you explain? $\endgroup$
    – user546106
    Commented May 1, 2018 at 0:46
  • $\begingroup$ Sure, assume $A$ is a five-pointed star in $\mathbb{R}^2$, $f_0$ and $g_0$ are two constant loops based at two different points of the star, then the segment between those points is not fully contained in $A$ and is the image of $H$. Your map will be a homotopy when $A$ is convex. $\endgroup$
    – C. Falcon
    Commented May 1, 2018 at 0:57
  • $\begingroup$ But my $f_0\in [f]$ and $g_0\in [g]$, where $[f]$ and $[g]$ are different homotopy classes of loops based at $x_0\in A$. $\endgroup$
    – user546106
    Commented May 1, 2018 at 1:05
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    $\begingroup$ Sure, fair point, but just base $g_0$ at the same point than $f_0$ and choose it so that it visits another vertex of the star, then $H$ will fail to be a homotopy for the exact same reason. Examine $H$ at the time $g_0$ is at the vertex. $\endgroup$
    – C. Falcon
    Commented May 1, 2018 at 1:08
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Hint: the straight-line homotopy won't stay in $A$ in general. Instead pick a point $c \in A$ that can "see" all of $A$ and take $c$ as the base point. Now construct a homotopy that spends half its time collapsing $f_0$ to the constant function $t \mapsto c$ and then spends the rest of its time expanding $t \mapsto c$ up to $g_0$.

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