regular value of homotopy F is regular value for F_t

Question

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 $t \in I$?

Answer 1

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.