Can the graph of a $C^{1}([a,b])$ function does not split an open exactly into two connected components? I was reading Elementary Theory of Analytic Functions by Henri Cartan, (In particular chapter 2 paragraph 9 : "Complements : oriented boundary of a compact set") to understand what orientation of boundary of a compact set means.
In doing so, at page 64 there's the following lemma : 

Lemma : If a path $\gamma$ is continuosly differentiable and if its derivative $\gamma'$ is everywhere $\ne 0$, then, in a neighbourhood of each value of the parameter $t$, the mapping $t \to \gamma(t)$ is injective and its image cuts the plane (locally) into two regions.

The proof uses the Implicit function theorem. 
Since I was unable to understand to what function apply the implicit function theorem, I reduced the problem to the following :  
Let's be $f \in C^{1}([a,b])$ and let's take $t_{0} \in (a,b)$. 
It exists or not an open set $U \subseteq \mathbb{R}^{2}$ such that $(t_{0},f(t_{0})) \in U$ and such that $U - Graph(f)$ has exactly two connected components ? 
Where $Graph(f) = \left\lbrace (x,y) \in \mathbb{R}\times \mathbb{R} : x \in [a,b],y = f(x) \right\rbrace$
I'm following this way because I don't want to use the Jordan curve theorem since I don't have the tools to understand it properly, but I'd like to understand, at least locally, what's the behaviour of a $C^{1}$ curve.
 A: The continuous image of a connected set is connected.  So not if the open set is connected.
A: Let us prove the following more general theorem:

Let $f : X \to \mathbb R$ be a continuous function defined on a topological space $X$,  let $\xi_0 = (x_0,f(x_0)) \in G(f) = \{ (x,f(x)) \mid x \in X \}$ and let $U \subset X \times \mathbb R$ be a connected open neighborhood of $\xi_0$. Then $U \setminus G(f)$ is not connected.

Define $\phi : X \times \mathbb R \to X \times \mathbb R, \phi(x,t) = (x,t - f(x))$. This is a homeomorphism with inverse $\phi^{-1}(x,t) = (x,t + f(x))$.  It therefore suffices to show that $\phi(U \setminus G(f)) = \phi(U) \setminus \phi(G(f))$ is not connected.
$\phi(U)$ is a connected open neighborhood of $\eta_0 = \phi(x_0,f(x_0)) = (x_0,0)$ and $\phi(G(f)) = X \times \{0\}$. Obviously $V = \phi(U) \setminus X \times \{0\}$ is not connected because $V_+ = \phi(U) \setminus X \times [0,\infty)$ and $V_- = \phi(U) \setminus X \times (-\infty,0]$ are nonempty disjoint open subsets of $V$ whose union is $V$.
What can be said about the number of components of $U \setminus G(f)$ ? Unfortunately  nothing, the number may be bigger than two. As an example take $f : [-2,2] \to \mathbb R,f(x) = 0$. Let $U = \{ (x,y) \mid  1 < x^2 + y^2 < 2, y < 1\}$, $x_0 = 3/2$. Then $U$ is a connected open neighborhood of $\xi_0 = (x_0,f(x_0))$, but $U \setminus G(f) = U \setminus [-2,2] \times \{0\}$ has three components.
