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Let $f : [-1,1] \longrightarrow [-1,1]$ be a continuous function. Then using IVT I have proved that it has a fixed point. Now my question is "Can I prove this result by the technique used in Brouwer's fixed point theorem which I have recently read in the course Algebraic Topology?" That technique also uses method of contradiction and ultimately shows that if it was not the case then $S^1$ would be a retract of $D^2$ for which case $\Bbb Z$ Would become a trivial group which is not true. Hence the result follows. If I use the same technique for one dimensional case I would similarly get $S^0 = \{-1,1 \}$ is a retract of $[-1,1]$. If it happens to be true then the fundamental group of $S^0$ w.r.t. any base point (either $-1$ or $1$) is isomorphic to the fundamental group of $[-1,1]$ which is trivial since $[-1,1]$ is simply connected. But the fundamental group of $S^0$ is also trivial w.r.t. any base point. Hence I think I cannot use the same argument for one dimensional purpose.

Am I correct? Please help me in this regard.

Thank you very much.

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    $\begingroup$ Well you can prove differently that $S^0$ is not a retract of $D^1$: for starters, one is path-connected and the other isn't. "Algebraico-topologically" this can be seen by looking at the fundamental groupoids, or at the $0$th singular homologically, but this is using advanced tools for not much $\endgroup$ – Max Oct 8 '18 at 8:30
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Fundamental groups are not much use for dealing with disconnected spaces since the group cannot see anything outside the component of the base-point. For this case where you want to show there is no surjective $[-1,1] \to \{-1,1\}$ we can use the much cruder topological notion of connectedness. The above map does not exist because the domain is connected and the range isn't.

If you are set on using an algebraic invariant then you can use the $0$th homology group which is just the free group over the number of path components. Or you can use the fundamental groupoid which also contains somehow the number of path components and, unlike the fundamental group, can see all (path) components.

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  • $\begingroup$ I have an argument @Daron similar to you on my own. If $S^0$ is a retract of $[-1,1]$ then there exists a continuous map $r : [-1,1] \longrightarrow \{-1,1 \}$ such that $r(-1)=-1$ and $r(1)=1$. Right? Since $[-1,1]$ is connected so $r([-1,1])$ is a connected subset of $\{-1,1 \}$. Which is only possible if $r([-1,1])$ is singleton which contradicts the fact that $r$ is onto since $r(1)=1$ and $r(-1) = -1$. $\endgroup$ – Dbchatto67 Oct 8 '18 at 9:48
  • $\begingroup$ That's the idea all right! $\endgroup$ – Daron Oct 8 '18 at 14:21

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