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Let $F:\mathbb{S}^{2}\times\lbrack0,1]\rightarrow\mathbb{R}$ be a smooth ($C^{\infty}$) function and $f_{t}(x)=F(x,t)$. Suppose that $f_{0}=f_{1\text{ }}$is the projection over $z$-axis, so point $P=(0,0,1)$ is an absolute maximum of both $f_{0}$ and $f_{1}$. Let $A_{t}$ be the critical points set of $f_{t}$ and let $A=\cup_{t}(A_{t},t)$. My question is whether points $(P,0)$ and $(P,1)$ belong to one and the same component of set $A$? If so, this would be some kind of stability result for the critical points set under homotopy.

If this is something well-known or a counter-example exists, any references are welcome. (Of course, the same may be asked in a fairly more general setting, for manifolds etc...)

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  • $\begingroup$ Did you check the projection of a generic sphere eversion? $\endgroup$
    – Anton Petrunin
    Commented Jun 19, 2016 at 16:06
  • $\begingroup$ No, I didn't check sphere eversion, as I don't really understand it. Is it a counter-example? Anyway, if Q=(0,0,-1) and (P,0), (Q,1) belong to one and the same component of set A, it does work to me. $\endgroup$
    – user94090
    Commented Jun 19, 2016 at 16:24

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If $f_t$ is generic then $A$ is formed by a collection of curves in $S^2\times [0,1]$. So, $(P,0)$ is connected to either

  1. $(P,1)$ --- this happens for the constant $f_t$.

  2. $(-P,1)$ --- this happens for the projection of a generic sphere eversion, otherwise the orientation would not change.

  3. $(-P,0)$ --- It can not happen --- this follow since the number of critical points counted with singes ("$+$" for min and max and "$-$" for saddle) has to be $2$ for every $t$.

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  • $\begingroup$ So, it follows that for arbitrary $f_{t}$ some component of $A$ intersects both $\mathbb{S}^{2}\times\{0\}$ and $\mathbb{S}^{2}\times\{1\}$, isn't it? And we may claim such "stability" for any critical point of non zero index? Thanks for the answer. $\endgroup$
    – user94090
    Commented Jun 20, 2016 at 4:36

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