Given $f:\mathbb{R}^m \supset U \rightarrow \mathbb{R}^n$ differential. Relation of rank of $f'$ and the property of image of $f$?

Question

Let $f:\mathbb{R}^m \supset U \rightarrow \mathbb{R}^n$ a differential map, $U$ open, and $a \in U$. Are the followings true:

1. If $m > n$ and $f(U)$ is open set in $\mathbb{R}^n$ then $\text{rank}f'_a = n$?

2. If $m < n$ and $\pi\big(f(U)\big)$ is open set in $\mathbb{R}^m$ (with $\pi:f(U)\ni x=(x_1,...,x_n)\mapsto (x_1,...,x_m)\in \mathbb{R}^m$), then $\text{rank}f'_a = m$?

I don't know how to get information on Jacobian of $f$ to know about its rank. I tried a proof by contradiction as follows:

In 1's problem, If $\text{rank}f'_a = k < n$ then $f'(a):\mathbb{R}^m\rightarrow \mathbb{R}^k$. We have, when $h\rightarrow 0$, $\mathbb{R}^n \ni f(a + h) - f(a)\rightarrow f'_a(h)\in \mathbb{R}^k$. But, I don't see any reason to forbid $f(a + h) - f(a)\in \mathbb{R}^n$ close to a point $\mathbb{R}^k$. So may be the conclusions are wrong.

If they are wrong, could you give a counterexample? And under which conditions they're right?

Edit: the more general question of the two questions: is the following true:

If $\pi\big(f(U)\big)$ is open in $\mathbb{R}^k$ (with $\pi:f(U)\ni x=(x_1,...,x_n)\mapsto (x_1,...,x_k)\in \mathbb{R}^k$ and $k\le\min\{m,n\}$), then $\text{rank}f'_a = k$?

If it's wrong, give a counterexample. And, under which conditions it's right?

Partly answer: The counterexample of the first claim is provide in the Amirhossein's answer; generalize this counterexample, we get the counterexample of more general question is the map $\mathbb{R}^m\ni (x_1,...,x_m)\mapsto (f_1(x)=x _{1}^{3} +...+ x _{m}^{3},...,f_k(x)=x _{1}^{3} +...+ x _{m}^{3}, f _{k+1}(x)=0,...,f _{n}(x)=0)$ they all has rank $=0$ at $0\in \mathbb{R}^n$.

So, the remain question is: Under which conditions the claims is true?

Answer 1

The answer is no for both of them:

For the first claim:

Consider the the function $f: \Bbb R ^{2} \to \Bbb R, \hspace{0.2cm} f(x,y) = x^{3}.$

Clearly a smooth function with $f(\Bbb R ^{2}) = \Bbb R$ which is open. but $f^{\prime}_{(0,0)} =(0,0)$.

And your second claim can not occur because $f(U)$ is of measure zero in $\Bbb R^{n}$. So it cannot be an open set.