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It is well-known that if a mapping $f:S^2\to S^2$ is homotopic to identity, then it must have a fixed point. I am wondering that if there exists a map on $S^2$ which is homotopic to identity and has exactly one fixed point.

My attemptation:

I try to construct a vector field over $S^2$ which has a single zero. The flow $f_t$ generated by this vector is homotopic to $id$. However, I can't ensure that there exists some $t$, such that $f_t$ has only one fixed point and is exactly where the corresponding vector field vanishes .

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  • $\begingroup$ @MoisheKohan I have edited the post and added my attemptation. $\endgroup$
    – Y.Wayne
    Mar 23, 2023 at 4:15
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    $\begingroup$ Do you know any theorems relating to existence of fixed points? $\endgroup$ Mar 23, 2023 at 4:20
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    $\begingroup$ Well, actually, that’s not true. If you know some differential topology and not just algebraic topology, the Lefschetz number of a smooth map is the sum of local Lefschetz numbers at the fixed points. See Guillemin & Pollack or Bott & Tu. $\endgroup$ Mar 23, 2023 at 4:58
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    $\begingroup$ It seems to me that the one point compactification of $(x,y)\rightarrow (x,y+1)$ will do. $\endgroup$ Mar 23, 2023 at 5:27
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    $\begingroup$ @ConnorMalin do you want to post that as an answer? $\endgroup$
    – ronno
    Mar 23, 2023 at 8:44

1 Answer 1

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For any $n$ there exists a self map of $S^n$ which is homotopic to the identity and has exactly one fixed point. The map is obtained by starting with the homeomorphism $\mathbb{R}^n \rightarrow \mathbb{R}^n$ given by $(x_1,\dots,x_n) \rightarrow (x_1+1,\dots,x_n)$ and applying one point compactification. This evidently has a single fixed point, and it is homotopic to the identity because the original map $\mathbb{R}^n \rightarrow \mathbb{R}^n$ is homotopic to the identity through homeomorphisms.

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  • $\begingroup$ Do you mean a self map of $S^n$? (+1, obviously) $\endgroup$ Mar 24, 2023 at 4:18
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    $\begingroup$ @HallaSurvivor Thanks $\endgroup$ Mar 24, 2023 at 4:56

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