Let $M,N$ be smooth manifolds of the same dimension, let $I=[0,1]$ and say we have a smooth map $F:M \times I \to N$ with regular value $y \in N$. Is $y$ then also a regular value of $F_t := F(\cdot,t)$ for all $t \in I$?
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No. Consider $M = N = S^1$, parameterized by an angle $\theta$. Define
$$ F(\theta, t) = t $$ Then the derivative of $F$ at $(\theta, t)$ is $$ \pmatrix{0\\1} $$ which is maximal rank, hence $F$ is regular at $(\theta, t)$.
On the other hand, $F_0(\theta, t) = 0$ for all $\theta$, hence its derivative is the $0$-matrix, which does not have maximal rank.
John Hughes
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I guess you meant to say $N=R$? Anyway I see the point. Thanks! – gogoog Mar 18 '21 at 12:33
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No, I meant $N = S^1$. Basically, at time $t$ the entire circle $M$ is sent to the single point of $N$ at angle $t$. But you could also use $N = \Bbb R$ and it works equally well. – John Hughes Mar 18 '21 at 12:37