Global transversals of a codimension one foliation

Question

EDIT: changes to the question are in bold.

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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.

Answer 1

Trying to answer the new version of the question.

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We claim that, yes, the diagonal $\Delta$ meets all leaves.

Let $R$ be the plane spanned by $(0, 0, 0, \ldots, 0, 1)$ and $(1, 1, 1, \ldots, 1, 1)$. Note that $R$ contains the ray $\Delta$. Let $\Gamma = R \cap Q$ - that is, $\Gamma$ is the orthogonal projection of $\Delta$ "down" to $Q$.

Suppose that $L$ is a leaf of the given foliation $\mathcal{F} \subset P$. Let $D$ be the image of the orthogonal projection of $L$ into the hyperplane $Q$. We are told that the "inverse" of this projection is a continuous map $f \colon D \to L$. Let $x$ be any point of $D$. Define $R_x$ to be the two-dimensional plane which (1) contains $x$ and (2) is parallel to $R$. Define $D_x = D \cap R_x$ and $L_x = L \cap R_x$. Finally define $f_x = f|D_x \colon D_x \to L_x$. It is an exercise to show that $f_x$ is non-strictly monotonically decreasing. So we have shown the following.

Lemma: If $D$ meets $\Gamma$, then $L$ meets $\Delta$.

[Now things become a bit vague.]

If $D$ has no boundary, then $D = Q$ and we are done. Suppose that the boundary $\partial D$ is non-empty and "nice" (a smooth codimension-one submanifold of $Q$, properly embedded in $Q$). Let $y$ be a point of $\partial D$ and let $n_y$ be the unit vector at $y$ which (1) is normal to $\partial D$, (2) is tangent to $Q$, and (3) points into $D$. Note that the last coordinate of $n_y$ is zero.

Note that $f$ either blows up, or vanishes, at $y$. I claim that if $f$ blows up at $y$ then all coordinates of $n_y$ (except the last) are positive. On the other hand, if $f$ vanishes at $y$ then all coordinates of $n_y$ (except the last) are negative. Thus $\partial D$ has a "convexity property", and so meets $\Gamma$, and we win.

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Below we answer a previous version of the question - see the edit history.

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Yes, $\Delta$ meets all leaves. Let $R$ be the plane spanned by $(0, 0, 0, \ldots, 0, 1)$ and $(1, 1, 1, \ldots, 1, 1)$. Note that $R$ contains $\Delta$. Every leaf meets $R$ and its intersection with $R$ is a graph over $Q \cap R$. By your hypotheses these graphs are (non-strictly) monotonically decreasing. So by the intermediate value theorem they meet $\Delta$.