Consider the system of differential equation
$$\dot x = f(x) $$ with $x\in{M}$ where $f$ is a function of the only $x$, so the system is autonomus. Now let $\phi^t(x)$ be the flow of the system, that is the solution at time $t$ with initial condition $x(0)=x$.
Fixing a time $\tau$ one can consider the map
$$\phi^\tau: x\in M\rightarrow \phi^\tau(x) \in M$$.
So one can consider the sequence ${x_k}$ defined by $x_k= \phi^{k\tau}(x_0)$ given $x_0\in M$. This is called orbit.
An invariant set $D$ is a set such that $\phi^\tau(D) \subseteq D$.
My question is: given two invariant curves $\gamma_1,\gamma_2$ for the map above, can these $\gamma_1$ intersect $\gamma_2$? I know that this is true in general for a map, but what happen if the map coming from a time independent vector field, as in this case? I think that this is not true ( there's no intersection ) in view the property of the phase space of autonomus ODE: for any $x_0$ in $M$ there exists only one solution passing through this point.
In the notes of my teacher i read this sentence: " intersection between two invariant curve cannot happen for that map which comes from the discretization of flow of vector field independent with respect to time". So, is this sentence wrong?
It can happen: as a trivial counterexample imagine zero vector field, ie. nothing moves at all. Then the time-$\tau$ map $\phi^\tau$ is the identity for any $\tau$ so any curve is an invariant curve for the map and you are free to choose whatever you want as your two invariant curves.
But it is a very common intutiton to think the opposite. The problem is that one tends to imagine considered invariant curves as orbits of the flow $\phi$ and for these the statement obviously holds, more precisely, in general, two orbits of a flow are either identical or disjoint. This is still true for the orbits of any time-$\tau$ map because it is a homeomorphism but an invariant set for this map consists, in principle, of many orbits of the map.
It is not that difficult to construct also nontrivial examples: think, for example, about a flow on an infinite horizontal tube (cylinder) with constant unit speed in horizontal direction. Then take as the first invariant curve any horizontal line on the tube. Now fix any $\tau$ (eg. $\tau = 1$) and construct the second invariant curve in such a way that it wraps around the tube and does one revolution in exactly time $\tau$.