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Let $F:M\to N$ be a smooth map between smooth manifolds $M$ and $N$. If $F$ has full rank at a point $p \in M$, there is a neighborhood of $p$ in which $F$ has full rank. What if $F$ does not have the full rank? Let $r$ be the rank of $F$ at $p$. Can I still assume that there exists a neighborhood of $p$ in which $F$ has the rank $r$?

A related quesion is here. If $F$ has constant rank at a level set $F^{-1}(c)$, is the level set an embedded submanifold? If the rank equals the dimension of $N$, the answer is yes. It is so called the regular level set theorem. What if $F$ has constant rank at the level set, but it is not full rank? I think that the level set is an embedded submanifold if $F$ has constant rank in an open set containing the level set. However, if I don't know the existence of the open set, can I still believe that the level set is an embedded submanifold?

I found a counterexample for the first question. The map $(x, y, z) \mapsto (x, y, y + z^2)$ has rank 2 at $(0, 0, 0)$ but it has rank 3 for $(0, 0, \epsilon \ne 0)$. However, no progress on the second question.

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    For the first question: I'm sure you're able to find counterexamples for the case $F: \mathbb{R}^n \to \mathbb{R}^n$. Do you have any ideas? – 2017-02-26
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    @menag I edited my post. It was not hard to find. It was my fault not to think about it more seriously. – 2017-02-26

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No. It is well known that any closed subset $A \subseteq \mathbb{R}^n$ can be realized as the zero set $F^{-1}(0)$ of a smooth function. If $\overline{A^{\circ}} = A$ then $F|_{A^{\circ}} = 0$ implies that $dF|_{A^{\circ}} = 0$ and then by continuity we also have $dF|_{A} = 0$. Hence, $F$ has constant rank $0$ on $A$ but $A$ need not be an embedded manifold.

For example, you can take $A = [0,1]^2 \subseteq \mathbb{R}^2$ or if you want an example which is not even a manifold with corners, just take the interior region of a sufficiently ugly simple closed loop in $\mathbb{R}^2$.