TITLE: Generalization of Darboux's Theorem
QUESTION [15 upvotes]: Darboux's Theorem. If $f:[a,b]\to\mathbb R$ is differentiable and $f'(a)<\xi<f'(b)$, then there exists a $c\in (a,b)$, such that $\,f'(c)=\xi$.
Does any of the following generalizations

Let $U\subset\mathbb R^n$ connected and $f: U\to \mathbb R$ differentiable. Then $\nabla f[U]$ is connected,
Let $U\subset\mathbb R^n$ convex and $f: U\to \mathbb R$ differentiable. Then $\nabla f[U]$ is convex,
$H_k\big(\nabla f[U],\mathbb{Z}\big) \hookrightarrow H_k(U,\mathbb{Z})$, for all $k$,

hold?

REPLY [13 votes]: It turns out than none of the three potential generalisations holds.
Counterexamples for the last two questions are presented in the answer of Ali Taghavi, and in particular by function $f(x,y)=(\mathrm{e}^x\cos y,\mathrm{e}^x\sin y)$, as $f[\mathbb R^2]=\mathbb R^2\smallsetminus\{(0,0)\}$.
For the first question, a counterexample appears in:
Solution to the gradient problem of C.E. Weil, by Zoltán Buczolich
where the author gives a complete answer to the famous gradient problem of C. E. Weil. On an open set $G\subset \mathbb{R}^{2}$ he constructs a differentiable function $f:G\to\mathbb{R}$, for which there exists an open set $\Omega_{1}\subset\mathbb{R}^{2}$ such that $\nabla f({p})\in \Omega_{1}$ for a ${p}\in G$ but $\nabla f({q})\not\in\Omega_{1}$ for almost every ${q}\in G$. This also shows that the Denjoy-Clarkson property does not hold in higher dimensions.