a question about functional analysis conclusion,and I am not sure whether it is true or not?

we have $R^n$,$R^m$ spaces, suppose open set $O_{1}\subset R^n$ and $O_{2}\subset R^m$,

$f:O_{1}->O_{2}$ is k-times differentiable$(1<=k<=\infty)$,then at $x_{0}\in O_{1}$,$rank(f)(x_{0})$ is defined as $rank (f'(x_{0}))$ (linear map from $R^n$ to $R^m$).

Then,can we prove that $rank(f)(x_{0})=p$ can imply that there exits a r>0 such that $rank(f)(x)>=p$ for any $x\in B_{r}(x_{0})$.

Can someone tell me whether it is true or not? I have no idea how to prove it is right or not? Can someone help me?

first reduce to the case where $m=p.$ To do this rotate so that the image of the differential map is the first $p$ coordinates and then project onto the first $p$coordinates. If this map has rank $p$ nearby that is enough.
If $m =p$ then it follows from the inverse function theorem.
If $m>p,$ then rotate the domain to make the differential applied to the first $p$ dimensions be injective and the one applied to the first last $m-p$ zero. Now define a map from $R^m$ to $R^m$ by appending the vector $(x_{p+1},\dots,x_m)$ to the image. Apply the inverse function theorem to this map and you are done.
Suppose $u_1(x),\dots , u_p(x) : O_1 \to O_2$ are continuous and satisfy $|u_j(x)| = 1, x \in O_1, j = 1,\dots , p.$ If $u_1(x_0),\dots , u_p(x_0)$ are linearly independent (LI) for some $x_0\in O_1,$ then $u_1(x),\dots , u_p(x)$ are LI for $x$ in a neighborhood of $x_0.$ Proof: Consider the $p$ functions $$u_j(x) - \sum_{k\ne j}\langle u_j(x), u_k(x) \rangle u_k(x), \,\,\,\,j=1,\dots ,p.$$ None of these is $0$ at $x_0.$ Hence by continuity there is a neighborhood of $x_0$ where all of them are nonzero. In that neighborhood, $u_1(x),\dots , u_p(x)$ are LI.
In the case of non unit-vector maps $v_1,\dots v_p$ that are LI at $x_0,$ we can apply the above to $v_1/|v_1|,\dots, v_p/|v_p|$ to see that $v_1,\dots v_p$ are LI near $x_0.$