Suppose that $F: M \to N$ is a smooth map such that $\dim(M) \ge \dim(N)$ and rank $dF(m)=\dim N$ for some point $m \in M$. Prove that there exists an open neighborhood $V$ of the point $F(m)$ in $N$ that belongs to the image of $F$.
Now $dF_m:T_mM \to T_{F(m)}N$. Since $dF(m)=\dim N $ and $\dim(M) \le \dim(N)$, $F$ has a full rank at $m$. I think that I have to use Inverse Function Theorem to make a conclusion. But I am unable to.
Any hint will be appreciated. Thanks for the help!!