# Does there exist a mapping on $S^2$ which is homotopic to indentity has only one fixed point?

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 .

• @MoisheKohan I have edited the post and added my attemptation. Commented Mar 23, 2023 at 4:15
• Do you know any theorems relating to existence of fixed points? Commented Mar 23, 2023 at 4:20
• 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. Commented Mar 23, 2023 at 4:58
• It seems to me that the one point compactification of $(x,y)\rightarrow (x,y+1)$ will do. Commented Mar 23, 2023 at 5:27
• @ConnorMalin do you want to post that as an answer? Commented Mar 23, 2023 at 8:44

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.
• Do you mean a self map of $S^n$? (+1, obviously) Commented Mar 24, 2023 at 4:18