Every closed oriented smooth three-manifold bounds a compact oriented smooth four-manifold

Smooth disk isotopy: vector fields. Put $Q=[0,1{]}^2$. Let $\mathcal{F}$ be its orientation-preserving diffeomorphisms equal to identity near the boundary. Enlarge this to $\mathcal{E}$ consisting of diffeomorphisms equal to identity near the left, top and bottom edges and of the form $(x,g(y))$ near the right edge. For $f\in \mathcal{E}$, the nonzero plane field $v=f_{\ast }{e}_1$ is $e_1=(1,0)$ near every edge. Lift $v:Q\to {\mathbb{C}}^{\times }$ through the exponential to a smooth logarithm $\mathrm{\ell }$ with value zero at the lower left corner. Existence follows by continuation of local logarithms on the simply connected square; uniqueness makes them agree. The boundary collar is connected, so $\mathrm{\ell }=0$ there. Set $v_t=\exp ((1-t)\mathrm{\ell })$. These fields are nonzero, equal $e_1$ on a common collar, and interpolate from $f_{\ast }{e}_1$ to $e_1$.

Surface cutting and twists. All surfaces below are compact, connected and oriented; original boundary collars remain fixed. A twist on an annulus is $(\theta ,u)\mapsto (\theta +2\pi b(u),u)$, where $b$ equals zero and one at opposite ends. The negative sign gives its inverse. A smooth disk-and-bands model follows from [F1] in dimension two: cancel a maximal tree of zero-handles and joining bands. Sliding band ends in boundary collars puts intersecting pairs into one-holed-torus blocks; remove such blocks successively. With no intersecting pair the remaining model is a disk with holes. These are smooth operations: each band slide moves adjacent attaching intervals along an embedded arc in a disk by the supported velocity flow [F3, F4]. In particular cutting a nonseparating curve lowers genus by one. For two disjoint nonseparating curves, the complement is connected or has two components each incident to both curves: a component incident only to one would make that curve separating. No conversion of an arbitrary topological homeomorphism into a diffeomorphism is used.

No trapped trajectory. Extend each $v_t$ by $e_1$ beyond the square. A forward trajectory entering the left edge cannot cross the horizontal boundary collars and can exit only through the right. If it never exits, its closure is compact in a slightly larger rectangle; there is no equilibrium. By [F2] its limit set is a periodic orbit. Uniqueness makes such an orbit a smooth simple closed curve, whose tangent winding is $\pm 1$ by [F2]. But $\mathrm{\ell }$ extends by zero, so its restriction to that orbit gives a single-valued tangent-angle lift and winding zero, a contradiction. The same argument backwards shows every interior trajectory reaches the left edge in finite negative time. Thus trajectories from the left parametrize the whole square without overlap.

Intersection reduction. For nonseparating smooth curves $\alpha ,\beta$, remove innermost bigons until they are in transverse minimal position. Each removal is a smooth disk-supported isotopy lowering the finite intersection count by two. If $N=\mathrm{\mid }\alpha \cap \beta \mathrm{\mid }\ge 2$, take an arc $\delta$ of $\beta$ between consecutive intersections. If its endpoints lie on opposite banks of $\alpha$, join $\delta$ to a subarc of $\alpha$, round corners and shift one endpoint across $\alpha$. The resulting simple curve $\gamma$ crosses $\alpha$ once and $\beta$ at most $N-1$ times: outside the small rectangle about $\delta$ it follows a parallel copy of that subarc. Its single crossing makes it nonseparating. If the endpoints lie on the same bank, band surgery gives two curves ${\gamma }_1,{\gamma }_2$ disjoint from $\alpha$, with oriented classes $[\alpha ]=[{\gamma }_1]+[{\gamma }_2]$. The class of $\alpha$ in the homology of the capped surface is nonzero; hence one ${\gamma }_i$ has nonzero capped class and is nonseparating. Each uses one complementary subarc of $\alpha$ and two parallels of $\delta$; pushing these off $\beta$ gives at most $N-2$ intersections with $\beta$. Induction on $N$, with bases zero and one, gives a path of nonseparating curves with successive intersections zero or one. All moves avoid protected collars.

Smooth time normalization. Let ${\mathrm{\Gamma }}_t(s,y)$ be the flow from $(0,y)$ and $r_t(y)>0$ its first right-edge exit time. At exit ${\mathrm{\partial }}_s{\mathrm{\Gamma }}_t^1=1$, so the implicit function theorem and [F3] give smooth exit times. Compactness gives $0<m\le r_t(y)\le M$. Choose a smooth function $\rho \ge 0$ zero on endpoint collars of total length $c<m$ and positive elsewhere. The function $a\mapsto {\int }_0^1\exp (a\rho (u))du$ is smooth strictly increasing with image $(c,\mathrm{\infty })$. Its smooth inverse defines $a(r)$ for $m\le r\le M$. Put $\chi (x,r)={\int }_0^x\exp (a(r)\rho (u))du$. Then $\chi (0,r)=0$, $\chi (1,r)=r$, ${\mathrm{\partial }}_x\chi >0$, ${\mathrm{\partial }}_x\chi =1$ near the endpoints, and $\chi (x,1)=x$. Therefore $E_t(f)(x,y)={\mathrm{\Gamma }}_t(\chi (x,r_t(y)),y)$ is a smooth diffeomorphism in $\mathcal{E}$. Its inverse follows the trajectory to the left and inverts $\chi$. At $t=0$ it is $f$, and at $t=1$ it is identity.

Replace disjointness by one crossing. For disjoint nonisotopic nonseparating curves $a,b$, cut along both. In a connected complement join one bank of $a$ to one bank of $b$; cutting along this arc leaves the surface connected because it joins distinct boundary components. Join the other two banks by a second disjoint arc. If the complement has two components, join its $a$-bank to its $b$-bank in each component. Regluing and rounding gives a curve $d$ crossing each once; it is nonseparating since it crosses $a$ once. Thus every zero-intersection step can be replaced by two one-intersection steps. The intersection-one graph is connected for every positive genus, including genus one with arbitrary boundary.

Fix the right edge. For $h\in \mathcal{E}$, write $h(x,y)=(x,g_h(y))$ near the right edge. For an increasing interval diffeomorphism define $G_s(g)(y)=(1-s)g(y)+sy$, still increasing. Choose $\beta (x)=1$ near the left and zero near the right, and put ${\mathrm{\Psi }}_h(x,y)=(x,G_{\beta (x)}(g_h^{-1})(y))$. This is identity near the left, top and bottom and equals $(x,g_h^{-1}(y))$ near the right. Thus $p(h)={\mathrm{\Psi }}_h\circ h$ lies in $\mathcal{F}$, with $p{\mathrm{\mid }}_{\mathcal{F}}=\mathrm{i}\mathrm{d}$. For $f\in \mathcal{F}$, the family $p(E_t(f))$ is the required smooth isotopy relative to a common boundary collar. Rounding the square corners away from its support gives the disk assertion. A smooth time reparametrization makes this and subsequent isotopies stationary at their endpoints for concatenation.

Transport by twists. A neighborhood of two curves $a,b$ with one crossing is a one-holed torus. Cutting it to a square, the twists act on the coordinate arcs by the two elementary shears. The three shears $T_a{T}_b{T}_a$ exchange the two coordinate directions, reversing one. Their square acts as $-I$, so it reverses the orientation of $a$. These representatives are supported in the one-holed torus and fix its boundary collar. By step 3.2, a finite twist word takes any nonseparating curve to any other; the squared three-twist word corrects any unwanted reversal of its orientation.

Planar induction and its kernel. Every diffeomorphism of a disk with $n$ holes, fixed on boundary collars, is smoothly isotopic to a twist word. For $n=0$ use step 4.1. Cap one inner hole by a parametrized disk $D$. By induction subtract a twist word for the extended map, choosing its curves to miss $D$ by isotopy. The remaining extended map has a smooth isotopy to identity. Its restriction to $D$ is a loop of disk embeddings. The kernel is generated by moving the center of this disk along loops and rotating its frame; the disk-family and lifting constructions establish this assertion, including the compact-family and homotopy-lifting details.

Genus reduction and the whole first jet. For positive genus choose an oriented nonseparating curve $a$. Step 4.2 gives a twist word $h$ such that $hf$ preserves its class and orientation. An ambient isotopy makes the image curve exactly $a$. Correct its restriction to $a$ in an annular collar: the convex interpolation of an increasing circle lift with the identity is increasing. Now the resulting map $k$ fixes $a$ pointwise and, in collar coordinates $(\theta ,u)$ with $a=\{u=0\}$, has derivative $A(\theta )=\left(\begin{array}{cc}1& a(\theta )\\ 0& b(\theta )\end{array}\right)$ with $b>0$. Define $J_t=\left(\begin{array}{cc}1& -ta\mathrm{/}b\\ 0& 1-t+t\mathrm{/}b\end{array}\right)$, so $J_0=I$ and $J_1=A^{-1}$. Write $J_t^{\mathrm{\prime }}{J}_t^{-1}=\left(\begin{array}{cc}0& c_t\\ 0& d_t\end{array}\right)$. The supported field $\rho (u)(c_t(\theta )u,d_t(\theta )u)$ has a flow ${\psi }_t$ fixing the curve and with derivative $J_t$ there, by the variational ODE. Consequently $K=k\circ {\psi }_1$ has its entire first jet equal to identity on the curve, including the tangential normal shear.

Disk families reduce to frames. For a smooth compact family of disk embeddings $e_z$, choose a metric by [F5] and put $p_z=e_z(0)$. The geodesic ODE [F3] and the inverse function theorem give exponential coordinate disks at all centers with one uniformly positive radius. For uniformly small $r>0$, put $E_z(w)={\exp }_{p_z}^{-1}(e_z(w))$ near the source origin. The embeddings ${\exp }_{p_z}(E_z(rqw)\mathrm{/}q)$ for $0<q\le 1$ extend smoothly at $q=0$ to ${\exp }_{p_z}(r(d{e}_z{)}_{0}w)$, because $E_z(rqw)\mathrm{/}q=r{\int }_0^1(d{E}_z{)}_{srqw}wds$. All images remain in the exponential disks; for positive $q$ they are compositions of restriction, coordinate diffeomorphism, dilation and inverse coordinate diffeomorphism, and at zero their linear derivatives are invertible. This formula is metric-defined and agrees on overlaps for interval and square parameter families. Thus shrinking the disks and then applying this deformation reduces families to their centers and positive linear frames. The positive scale and positive triangular part of such a frame are contractible, leaving a unit tangent direction. Shrinking changes the base disk along a path; its monodromy transport only conjugates endpoint maps, and conjugation takes twists to twists about transported curves.

Two-parameter lifting without disk-group contractibility. For a smooth family $e_{s,t}$, extend each disk embedding slightly beyond its source boundary locally in parameters. On its compact graph in the product of parameter space and surface prescribe $V_{s,t}(e_{s,t}(w))={\mathrm{\partial }}_s{e}_{s,t}(w)$. Local smooth extensions exist in product charts. A finite cover and a partition of unity patch them, since all values agree on the graph; a cutoff keeps support away from protected surface collars. Integrate in $s$, with $t$ stationary, using [F3]. This yields smooth ambient diffeomorphisms carrying $e_{0,t}$ to $e_{s,t}$. If the disk family is constant in $s$ on a protected parameter edge, extend it constantly on a parameter collar so the lifted velocity vanishes there. The same construction lifts paths in $t$. Thus for a based loop homotopy constant at $t=0,1$, the endpoint diffeomorphisms of its lifts vary by a smooth isotopy fixing the disk pointwise. Two lifts of one path differ by a family fixing the disk at every time; their endpoints give the same relative mapping class. This proves the required independence and homotopy invariance of monodromy, using no assertion about disk-group connectivity.

Kernel generators are twists. The planar surface retracts onto a finite graph, so center loops are words of simple loops. Its tangent bundle is trivial, and the remaining frame loop has integer rotation number. A center push along a simple loop is the product of two twists on its annular neighborhood: in coordinates $(\theta ,u)\in S^1\times [0,2]$, add $2\pi tu$ for $u\le 1$ and $2\pi t(2-u)$ for $u\ge 1$, smoothing the two pieces while preserving the central circle. At $t=1$ the two halves give $T_{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}}{T}_{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}^{-1}$. A full rotation of the disk frame is a twist about its boundary. The construction is away from other collars and descends to the disk complement. The disk-family, lifting, and explicit push constructions therefore express every kernel element as a twist word, completing the planar induction in step 5.1. Only surjectivity onto the kernel was needed, not injectivity of a Birman sequence.

Remove the remaining collar germ. Choose the unique small angular lift of $K-\mathrm{i}\mathrm{d}$ near the curve. Uniform Taylor estimates give $K-\mathrm{i}\mathrm{d}=O(u^2)$ and $dK-I=O(\mathrm{\mid }u\mathrm{\mid })$. Let ${\rho }_{\delta }$ equal one for $\mathrm{\mid }u\mathrm{\mid }\le \delta$, zero for $\mathrm{\mid }u\mathrm{\mid }\ge 2\delta$, with derivative bounded by $C\mathrm{/}\delta$. The maps $L_s=\mathrm{i}\mathrm{d}+s{\rho }_{\delta }(K-\mathrm{i}\mathrm{d})$ have derivative $I+O(\delta )$, including the cutoff derivative term. For small $\delta$ lift the collar to ${\mathbb{R}}^2$ and extend the displacement by zero in the normal direction. The derivative of this periodic displacement has norm less than $1\mathrm{/}2$. Applying [F7] to the displacement on each straight line segment gives its Lipschitz constant at most $1\mathrm{/}2$. Thus the lift is injective by the estimate $\mathrm{\mid }{L}_s(x)-L_s(y)\mathrm{\mid }\ge \mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}2$, and is surjective: for each target $z$, iteration of $x\mapsto z-(L_s(x)-x)$ has successive differences decreasing by a factor at most $1\mathrm{/}2$, hence is Cauchy in ${\mathbb{R}}^2$ and converges to the required preimage. Its local inverses are smooth. Equivariance under the angular period descends this diffeomorphism to the annulus, and extension by identity gives a diffeomorphism of the surface. Since $L_1=K$ on a smaller collar, $L_1^{-1}K$ is identity there. Cutting along that collar lowers genus by one and adds two fixed boundary collars. Induction on genus, based on step 8.1, proves generation by twist words. In closed genus zero first fix a disk using the disk-motion construction and its frame adjustment; the complement is the disk case of step 4.1. Thus the sphere case is also smooth, without the topological Alexander trick.

Heegaard splitting. Suppose $X$ is connected. Apply [F1] with empty faces and choose a regular level between indices one and two. Its lower sublevel $H$ consists of zero-handles and one-handles. It is connected: two- and three-handles have connected attaching regions and cannot join components, whereas $X$ is connected. Contract a spanning tree of the finite zero/one-handle graph. Each tree edge joins two balls through a tube; straighten its attaching disks in ball collars, identifying the union with a ball. Iteration gives a ball with $g$ one-handles, the standard handlebody $H_g$. Reading the upper sublevel upside down gives another connected handlebody. Their common boundary has Euler characteristic $2-2g$, so their genera agree. Hence $X=H_g{\cup }_f(-H_g)$, with $f:\mathrm{\partial }{H}_g\to \mathrm{\partial }{H}_g$ orientation preserving. The identifications are the disk-and-tube constructions just given, not an application of three-manifold classification. By step 9.1, $f$ is smoothly isotopic to a finite twist word.

Local trace of a twist. Suppose $Y=H_g{\cup }_q(-H_g)$ and $c$ is a curve on the splitting surface. Push it into a product collar and let $(\mu ,\lambda )$ be the meridian and surface-parallel longitude of a solid-torus neighborhood. Cutting the splitting annulus and regluing by $(\theta ,u)\mapsto (\theta \pm 2\pi b(u),u)$ adds one signed longitude circuit to a normal meridian. The replacing solid torus therefore has meridian $\mu \pm \lambda$. Conversely this filling cancels the shear, because the coordinate change is identity at the annulus ends. This is integral surgery with coefficient $\pm 1$ relative to the surface framing. The sign is chosen to realize the specified twist on the chosen side of the collar. Attaching a four-dimensional two-handle $D^2\times D^2$ to $Y\times [0,1]$ with that framing removes $S^1\times \mathrm{int}{D}^2$ from the outgoing boundary and inserts $D^2\times S^1$, exactly this torus replacement. Its oriented boundary is $(-Y)\bigsqcup Y^{\mathrm{\prime }}$, where $Y^{\mathrm{\prime }}$ has gluing $T_c^{\pm 1}q$. Seams are product charts and corners are rounded away from the incoming face by [F5].

Construct the filling. The double $H_g{\cup }_{\mathrm{i}\mathrm{d}}(-H_g)$ bounds the rounded product $H_g\times [0,1]$. Attach the finitely many two-handle traces of step 11.1 for a twist word for $f$. The outgoing boundary is $H_g{\cup }_f(-H_g)$ up to smooth isotopy of the gluing map. Such an isotopy induces a diffeomorphism of glued manifolds: in a collar use $(x,s)\mapsto (f_{\beta (s)}{f}_0^{-1}(x),s)$ with $\beta$ constant at its ends, and extend by identity. Composing with step 10.1 gives the required orientation-preserving identification $j:\mathrm{\partial }W\to X$. For $g=0$ the starting product is a rounded four-ball, and step 9.1 already handles its sphere gluing map.

Components and markings. Components of a manifold are open; compactness gives finitely many components of $X$. Perform step 12.1 with each given orientation and take the finite disjoint union. Empty $X$ uses empty $W$. For an actual capped manifold $X=M\cup C$, pull the marked submanifolds and their common collars back by $j$. The specified cap parametrizations and decorations are then exactly retained. A product collar of $\mathrm{\partial }W$ presents their common seam as the prescribed corner between the two boundary faces; no diffeomorphism is extended across a cap or across $W$. This is marked boundary existence, with no extra assertion about relative handle trading.

Every input therefore has a single-copy compact oriented smooth filling, so its class is zero in ${\mathrm{\Omega }}_3^{SO}$ by [F6]. The integral conclusion follows from the constructed filling, without rationalization, integral characteristic-number detection or a stable-stem calculation. AC covers the inherited countable choice and the auxiliary smooth choices; for a set of inputs it can additionally select one filling from each nonempty set of fillings. No canonical or natural selection is asserted.