A taut foliation of a compact connected manifold has a single closed transversal

A nonempty manifold with a regular codimension-one foliation has dimension at least one. If $\dim M=1$, its leaves are points. The coorientation and finitely many smooth chart bumps give a smooth nowhere-zero positive field $X$, whose flow is complete by [F4]. Each orbit is open because its orbit map has nonzero derivative, so connectedness makes $M$ one orbit. The orbit map $\mathbb{R}\to M$ is a surjective local diffeomorphism. If it were injective, it would be a homeomorphism, contradicting compactness of $M$. Thus its period subgroup is nontrivial; it is closed and misses a neighbourhood of zero by local injectivity, so it has a least positive element $T$. The induced map $\mathbb{R}\mathrm{/}T\mathbb{Z}\to M$ is an embedded positive circle onto $M$, meeting every point leaf. This proves the claim in dimension one. In the remaining steps assume $\dim M\ge 2$.

For a closed transversal $\gamma$ let $N_{\gamma }$ be the union of the leaves it meets; transversality is open and in a product chart a transversal meets all nearby plaques, so $N_{\gamma }$ is open and saturated, and the tautness hypothesis [F1] says that the family of all $N_{\gamma }$ covers $M$.

By compactness choose finitely many positive closed transversals ${\gamma }_1,\dots ,{\gamma }_k$ whose saturations cover $M$; reverse their orientations if necessary. This covering property survives sufficiently small $C^1$ perturbations of the finite curves. Indeed, for each $x\in M$ a finite plaque chain joins $x$ to a point of one curve. Restrict a positive parameter arc of that curve to a foliation box, and choose a smaller closed transverse-height interval strictly inside its height range. Transport this smaller interval along the chain using [F5]. Its saturation contains a neighbourhood of $x$. A sufficiently close perturbed arc still crosses every height of that smaller interval, by its positive derivative and the strict endpoint inequalities. Choose finitely many such neighbourhoods covering compact $M$ and take the minimum of their finitely many perturbation margins. Every perturbed family within that margin therefore still meets every leaf. The same argument applies to a finite family of positive immersed curves, because only regular parameter arcs were used.

We use the following finite approximation construction. A compact piecewise $C^2$ positive curve with positive one-sided derivatives can be rounded at its finitely many seams in foliation boxes: mollify the continuous chart curve; positivity of the transverse derivative persists because the mollifier is nonnegative, and a cutoff restores the old curve off the seam with derivative error tending to zero. A $C^2$ curve can then be approximated in $C^1$ by a smooth ambient curve: cover its parameter circle by finitely many smaller intervals mapping into smooth ambient charts, mollify the coordinate functions on each interval, blend back by a fixed smooth cutoff, and carry out these finitely many replacements. On the smaller intervals the replacement is smooth; subsequent replacements preserve smoothness where already obtained, and errors can be made smaller than any prescribed total $C^1$ margin. For a positive piecewise curve, the same covering margin is available even at a seam: its continuous transverse height is strictly increasing through the seam, so choose a smaller closed height interval strictly between the heights at the ends of an arc crossing it. Rounding and approximation preserve these strict endpoint inequalities. Thus positivity and the leaf-covering property of step 2.1 persist. For an initially embedded finite disjoint family, small $C^1$ errors also preserve that property: local injectivity follows from a nonzero coordinate derivative on finitely many parameter intervals, and pairs of parameters outside these intervals have images a positive distance apart in finitely many compact coordinate neighbourhoods.

Assume first $\dim M\ge 3$. The finite intersection graph of the open saturated sets $N_{{\gamma }_i}$ is connected: otherwise the unions belonging to two graph components would separate $M$. Two positive curves whose saturations meet have points in a common leaf, joined by a finite leafwise path. A finite plaque chain along that path supplies a foliated strip $H:[0,1]\times (-a,a)\to M$: for each $t$ the path $s\mapsto H(s,t)$ stays in one leaf, and the transverse derivative in $t$ is positive; its end transversals are short arcs of the two curves. To construct the strip, subdivide the path into finitely many convex plaque charts, interpolate its leafwise coordinate in each chart, transport the transverse coordinate by [F5], and smooth the leafwise seams inside their common plaques. This uses only finitely many compatible chart pieces; the strip need not be an embedding.

The general-position perturbations needed here follow from [F6] with finite parameters. For a smooth immersed circle, a fixed neighbourhood of the parameter diagonal contains no distinct coincident image pair, by the local injectivity and finite compact cover just used. On the remaining compact set of possible coincident pairs, finitely many smooth bumps supported in disjoint parameter intervals move either image independently in all ambient coordinate directions. They can be realized by local ambient coordinate translations on the curve with cutoff in its parameter. The resulting two-point evaluation is a submersion near its inverse image of the target diagonal; shrink the parameter ball so this remains true and no other coincidence enters. Apply [F6] there. The diagonal has codimension $\dim M$, so when $\dim M\ge 3$ a good slice has empty two-point coincidence set and is an embedding. For a finite family of embedded circles in a surface, the same independent parameters for distinct circles make all pair evaluations transverse to the diagonal and all three-point evaluations transverse to the small diagonal. Their respective source dimensions and target codimensions are $2,2$ and $3,4$. Thus pair intersections are isolated and, by compactness, finite, while triple intersections are absent. Each individual circle stays embedded. A finite union of null bad-parameter sets cannot fill any parameter ball, so these perturbations can be arbitrarily small and preserve the margins of step 2.1.

Delete from the two curves the short arcs parametrized at the ends of that strip by $-ϵ\le t\le ϵ$, where $0<ϵ<a$. Join the first lower endpoint to the second upper endpoint by $H(s,t(s))$, and the second lower endpoint to the first upper endpoint by $H(1-s,t(s))$, with $t$ strictly increasing from $-ϵ$ to $ϵ$. These joins are positive. Following the undeleted part of each circle and these two joins gives one piecewise positive immersed circle. Every removed plaque is still met, because each join crosses the whole transverse interval and its height-$t$ point lies in the leaf of each end plaque of height $t$. Consequently its saturation contains the saturations of both old curves. Before rounding, the merged circle and the unchanged remaining curves still cover all leaves. Round with the finite covering margin of steps 2.1 and 3.1; the entire remaining family therefore still covers all leaves. Its intersection graph of open saturations is again connected by the argument of step 3.2. Choose an overlapping pair again and repeat. Each merge reduces the finite family size by one, so eventually a single positive immersed circle meets every leaf. Steps 2.1, 3.1 and 4.1 smooth it and perturb it to an embedded positive circle while retaining this covering property. This proves the result in dimensions at least three.

Now let $\dim M=2$. Apply the surface perturbation of step 4.1 to the finite family from step 2.1. Around each of its finitely many crossings choose pairwise disjoint foliation rectangles containing exactly the two crossing arcs. With plaques $y=\text{constant}$, these arcs are graphs $x_1(y),x_2(y)$, since both are positive. They cross once transversely. Replace them inside the rectangle by the ordered graphs $x_{\pm }(y)=\frac{x_1(y)+x_2(y)}{2}\pm \frac{1}{2}\sqrt{(x_1(y)-x_2(y){)}^2+ϵ(y{)}^2},$ where $ϵ$ is positive near the crossing, zero on endpoint collars, and sufficiently small to keep both graphs in the rectangle. Away from the crossing its zero set lies where $x_1-x_2\mathrm{\ne }0$, so both replacements are $C^2$ and agree with the old pair on the endpoint collars. They are disjoint positive arcs, and every plaque meeting a removed arc still meets the replacement pair. Performing all these resolutions yields finitely many disjoint embedded $C^2$ positive circles whose union still meets every leaf: the remaining finite arc graph has degree two everywhere and has no crossings. Smooth this disjoint family by step 3.1, using step 2.1 for the covering margin. Denote its union by $C$.

Choose a connected component $Q$ of the orientation cover from [F7]. It is compact, oriented, and surjects onto $M$. Pull back $F$ and $C$; write $\tilde{C}$ for the finite disjoint union of embedded circles over $C$. The ambient orientation and normal coorientation orient the tangent line field of the lifted foliation. Its positive local $C^1$ sections, combined with finitely many smooth chart bumps, give a nowhere-zero $C^1$ tangent field $X$ on $Q$. Its flow is complete by [F8] and compactness. Each orbit is open in its one-dimensional leaf by a flow box, so a connected leaf is exactly one complete orbit. Every such lifted leaf meets $\tilde{C}$: its projection is a whole base leaf, because any leafwise path lifts through the covering, and the base leaf meets $C$. In particular no complete $X$-orbit avoids $\tilde{C}$.

Cut $Q$ along $\tilde{C}$. This construction needs no surface classification. Each oriented embedded circle has a two-sided smooth collar: choose along it a smooth transverse field in the side cone selected by $X$, patch with finitely many bumps, and use its short smooth flow and the local inverse theorem. Compactness makes the collar injective after a common shrink, by local injectivity and separation of distant circle parameters. Replace each collar coordinate $r\in (-a,a)$ by its two labelled halves, retaining two copies of $r=0$. Glue to the unchanged complement. The resulting $P$ is a compact surface with boundary, and its natural map to $Q$ is a local diffeomorphism on each half-chart. Its boundary circles are the two copies of each component of $\tilde{C}$, and the pulled-back $X$ is everywhere transverse to them. There are finitely many connected components of $P$ by a finite cover by connected disk and half-disk charts. Each component has boundary: a boundaryless component would map to a nonempty open-and-closed subset of connected $Q$ disjoint from $\tilde{C}$.

In each component $P_0$ of $P$, every maximal interior trajectory of $X$ reaches its boundary in finite positive and negative time. For otherwise a forward trajectory remaining in compact $P_0$ for all $t\ge 0$ has a nonempty compact limit set $K$, the intersection of the closures of its tails. Local flow continuity shows that $K$ is invariant in both time directions: translate a sequence of times tending to infinity by any fixed sufficiently small positive or negative time, and then iterate. It cannot meet $\mathrm{\partial }{P}_0$, since a field transverse to that boundary has, in one time direction, points outside the half-chart; invariance would put such points back in $P_0$. Thus $K\subset \mathrm{Int}{P}_0$, and any point of $K$ has a complete orbit in $K$. Its image in $Q$ is a complete orbit avoiding $\tilde{C}$, contradicting step 6.1. The negative-time argument is identical. Nonemptiness of the limit set uses the finite-intersection property of compact sets, not a choice of successive times. Sequence arguments, when used in the local metrizable charts, require at most the standing ${\mathrm{A}\mathrm{C}}_{\omega }$.

Let $B_{-}$ be the incoming boundary of $P_0$. The sign of the boundary crossing is constant on each boundary circle. Step 8.1 gives, for each $p\in B_{-}$, a finite first exit time $T(p)>0$ at an outgoing boundary. This is a $C^1$ function: the exit boundary is a regular scalar equation along the flow, so [F8] gives the local hitting time; the compact segment between its incoming and outgoing collars stays in the interior, excluding any earlier exit for nearby initial points. The map $(p,s)\mapsto {\phi }_{sT(p)}(p)$ is a $C^1$ product identification $B_{-}\times [0,1]\to P_0$. It is onto by tracing each point backwards to its first boundary hit, injective by uniqueness and absence of intermediate boundary hits, and has a local $C^1$ inverse by flow boxes and transversality at the two ends. Connectedness of $P_0$ implies that $B_{-}$ is one circle; its outgoing boundary is also one circle. Thus the two boundary circles of each cut component lie on exactly the same lifted leaves.

Form the finite graph with vertices the components of $\tilde{C}$ and one edge for each cut component, incident to the circles which are the images of its two boundary circles; loops are allowed. This graph is connected, since otherwise the unions of the cut components and circle collars belonging to its distinct graph components would separate $Q$. Step 9.1 says that the saturations of the vertices at the two ends of every edge are equal. Therefore all these circle saturations are equal; their union is $Q$ by step 6.1, so any one component of $\tilde{C}$ meets every lifted leaf. Its image in $M$ is one of the embedded positive circles of $C$. Every base leaf has a lifted leaf in $Q$, since $Q\to M$ is onto and leafwise paths lift. The selected base circle consequently meets every base leaf. Together with step 1.1 and step 5.1 this proves the statement in every dimension, using only finite constructions and the standing countable choice.