EDIT: changes to the question are in **bold**.

Suppose I have a smooth ($C^1$) codimension one foliation $\mathscr{F} \subset P$, the open subset of $R^n$ consisting of points with all positive components. The leaves of $\mathscr{F}$ are connected and proper and each is the graph of a function $f$ defined on **some domain $D$ contained in** $Q$, the open subset of $R^{n-1}$ consisting of points with all positive components **($D$ may depend on the leaf)**. Further, at every point $p \in P$, the normal to the tangent plane of the leaf $L$ passing through $p$, at that point, is non-negative with last coordinate strictly positive. If I now define the *diagonal* **$\Delta \equiv R^n \bigcap \{ x: x_1 = x_2 = \cdots = x_n \} \subset P$** then I know that through every point $t$ of **$\Delta$** passes a leaf $L$ (integral surface) of the foliation and, furthermore, **$\Delta$** is transversal to $L$ (because of the normal condition).

Can I then conclude that **$\Delta$** passes through *every* leaf of $\mathscr{F}$? If not, does there exist some transversal in the positive direction passing through every leaf? I have researched this question; i.e. the existence of a global transversal for a foliation of this type, every way I know. The paper "Foliations and global inversion" by E. Cabral Bereira seems to suggest (Lemma 4.3) that we can always find a such a curve by "standard means", but I may not be applying that result correctly. But Danny Calegari (Section 1.2 of Chapter 4 of notes on his book) states that there only exists a finite collection of transversals which *together* pass through every leaf of a foliation, which of course is not what I need.

I would appreciate any ideas you might have for further research.

**Edit**: After discussion with Sam Nead and some more thought it is also apparent that if we denote by $\delta$ the (orthogonal) projection of $\Delta$ into $Q$ then $\Delta$ does not intersect a leaf $L$ only if the domain of $L$, say $D_L$, does not intersect $\delta$. However, more is necessary as the closure of $D_L$ cannot be contained entirely in $Q \setminus \delta$ as $L$ must separate $P$ into two connected components.

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