Direction field non rectifiable Why the direction field induced by the following integral curves and defined on the entire plane is non-rectifiable ?

Edit: A direction field is (locally) rectifiable if there exists, 
in a neighbourhood of each point, a diffeomorphism mapping it into a field of parallel directions.
 A: This vector field is locally rectifiable but not globally rectifiable.
The proof that it is locally rectifiable would be to write down formulas for the required diffeomorphisms in a neighborhood of each point, although to do that one would need to be given formulas for the vector field itself; this should not be too difficult.
For the direction field to be globally rectifiable, there would have to be a diffeomorphism $f : \mathbb{R}^2 \to \mathbb{R}^2$ such that $Df$ takes your vector field to the horizontal vector field $\frac{d}{dx}$, hence takes each of the integral curves in your drawing to a horizontal line. To see why this is impossible, you can use orientations of integral curves. 
In your picture, there are exactly two ways to continuously orient the integral curves. Any orientation of any single curve determines the entire orientation. Your diagram shows the $x_1$-axis (the line $x_2=0$) oriented towards the right, which is then extended by continuity to all integral curves. The other choice would be to orient the $x_1$-axis to the left, and then extend continuously. No matter which of those two choices you make, the horizontal integral curves in your diagram will be alternately oriented ...-left-right-lift-right-... as you move upwards from each horizontal integral curve to the next. So, for instance, in your diagram the horizontal axis $x_2=0$ is oriented to the right, and the next horizontal integral curve above it, which is probably $x_2 = 1$, is oriented to the left.
In the horizontal vector field $\frac{d}{dx}$, there are also exactly two continuous ways to orient the integral curves: all oriented to the right; or all oriented to the left. So, for instance, both of the lines $x_2=0$ and $x_2=1$ will be oriented to the right, \emph{or} both will be oriented to the left.
If there were a diffeomorphism between your vector field and the horizontal vector field, then there would exist a diffeomorphism $f : \mathbb{R} \to [0,1]$ which preserves the orientation of the lower boundary line $\mathbb{R} \times \{0\}$ but reverses the orientation of the upper boundary line $\mathbb{R} \times \{1\}$. But no such diffeomorphism exists: any orientation preserving diffeomorphism from an oriented manifold to itself must preserve the induced boundary orientations (this is a standard theorem in differential topology).
