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Let $g\colon N\to M$ be an immersion.

Then, I think that $g^{-1}(p)$ is finite set or $0$-dimensional manifolds for all $p\in M$.

Now, let $g_t\colon N\to M$ be an one-parameter family of an immersion.

Let $F\colon N\times [0,1]\to M$ be the corresponding map given by $F(x,t)=g_t(x)$.

Is it true that $F^{-1}(p)$ is an one-dimensional manifold (possibly empty, of course)?

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I think if you were to write up a proof of your 1st claim you'd see an argument for how to answer your question. –  Ryan Budney Apr 7 '11 at 4:22

1 Answer 1

If $M$ and $N$ have the same dimension then $g$ being an immersion implies that $F$ is a submersion, so in this case yes.

Otherwise no. E.g. $N=\mathbb{R}$, $M=\mathbb{R}^2$, $g_t(x)=(x,t)$, $p=(0,1/2)$.

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