Why a vector field on a manifold defines a 1-1 mapping on the manifold?

Question

I am reading paragraph 3.1 (Introduction: how a vector field maps a manifold into itself) of chapter 3 (LIE DERIVATIVES AND LIE GROUPS) of the book Geometrical Methods of Mathematical Physics, by B. Schutz.
"In this paragraph of the book is written:
We have mentioned the idea of a ‘congruence’ in §2.12: a set of curves that fill the manifold, or some part of it, without intersecting. Each point in the region of the manifold M is on one and only one curve. Since each curve is a one dimensional set of points, the set of curves is (n — 1)-dimensional. (With some suitable parameterization, the set of curves is itself a manifold.) The key point from which everything else follows is that the congruence provides a natural mapping of the manifold into itself. If the parameter on the curves is $\lambda$ , then any sufficiently small number $\Delta\lambda$ defines a mapping in which each point is mapped into the one, a parameter distance $\Delta\lambda$ further along the same curve of the congruence (see figure 3.1). This is a 1-1 mapping, at least in any region in which the vector field is sufficiently well-behaved (a $C^\infty$ field will do)."
What I don't understand is why the mapping cited in the last lines is 1-1. That is, if $\alpha$ is a curve of the congruence (I am particularly interested in the case that the congruence is determined by a vector field everywhere different from $0$), why is $\alpha(\lambda_1+\Delta\lambda)\neq\alpha(\lambda_2+\Delta\lambda)$ if $\alpha(\lambda_1)\neq\alpha(\lambda_2)$ as $\alpha$ needs not to be 1-1?