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I am trying to solve the following exercise I've found in a qualifying exam in Differential Topology:

Show that there is no $f:\mathbb{S}^1 \times \mathbb{S}^1 \to \mathbb{S}^1$ satisfying $f(x,x)=x$ and $f(x,y)=f(y,x)$.

I think the word "smooth" is implict. I tried to think of some argument using Sard's Theorem, but no succes. I would appreciate some hints.

Thanks in advance.

Arctic Char
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User
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    What kinds of technique are available to you? – Arctic Char Jul 03 '22 at 18:40
  • @ArcticChar Only the initial topics in Differential Topology: for example, vector fields, lie groups, forms, integrals, orientation, transversality, ... – User Jul 04 '22 at 10:29

1 Answers1

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This is straightforward if we are permitted to use homology or fundamental groups and if $f$ is assumed continuous. I will proceed using the fundamental group.

Suppose we had such an $f$. Then consider the induced map $\pi(f) : \mathbb{Z}^2 \to \mathbb{Z}$. I am abusing notation a bit by writing the fundamental groups as $\mathbb{Z}^2$ and $\mathbb{Z}$, but it’s not a big deal.

The constraint $f(x, x) = x$ is tantamount to saying $f \circ \Delta_ \mathbb{S} = 1_\mathbb{S}$, where $\Delta_\mathbb{S} : \mathbb{S} \to \mathbb{S} \times \mathbb{S}$ is the diagonal map. Therefore, we have $\pi(f) \circ \pi_{\Delta_\mathbb{S}} = 1_\mathbb{Z}$. Now since $\pi$ preserves products, it preserves the diagonal map. So we see that $\pi(f) \circ \Delta_\mathbb{Z} = 1_\mathbb{Z}$. That is, we have $\pi(f)(1, 1) = 1$.

The switching clause $f(x, y) = f(y, x)$ is equivalent to saying $f \circ s_\mathbb{S} = f$, where $s_\mathbb{S} : \mathbb{S} \times \mathbb{S} \to \mathbb{S} \times \mathbb{S}$ is the swapping map. Therefore, we have $\pi(f) \circ \pi(s_\mathbb{S}) = \pi(f)$. Since $\pi$ preserves products, it preserves the swapping map. So we have $\pi(f) \circ s_\mathbb{Z} = \pi_f$. That is, $\pi(f)(1, 0) = \pi(f)(0, 1)$.

Now we see that $\pi(f)(1, 1) = \pi(f)(1, 0) + \pi(f)(0, 1) = 2 \pi(f)(1, 0) = 1$. But $1$ is not divisible by $2$. This is a contradiction. $\square$

If we drop the continuity assumption, we can find such an $f$. Pick some point $p \in \mathbb{S}$, and consider the function

$f(x, y) = \begin{cases} x & x = y \\ p & otherwise \end{cases}$

Mark Saving
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    Thank you for your answer! I didn't study homology or fundamental group in this discipline, then I think a solution without this machinery would be more suitable. But thank you! – User Jul 04 '22 at 10:33
  • what if, assuming f is smooth, we use the differential $df_p$ at some point $p$ and try to use the linearity of $df_p$ in the same way you used $\pi$? – User Jul 04 '22 at 11:32
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    @User There does seem to be a way to rephrase the above proof using a certain method of computing winding numbers by integration. You can put a certain exact form $\omega$ on the circle such that $\int_{\gamma} \omega$ is the winding number of $\gamma$. I will type up the details when I have time. – Mark Saving Jul 05 '22 at 04:30