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A form of Beck's monadicity theorem compatible with equivalence-based monadic adjunctions is this. A functor $U:\mathcal{D}\to \mathcal{C}$ is monadic exactly when it has a left adjoint, is a conservative functor, and $\mathcal{D}$ has coequalizers of all pairs whose images under $U$ admit a split coequalizer, with $U$ preserving these coequalizers. In the strict lifting formulation, one states creation of these coequalizers. Such a pair is a functor-split coequalizer pair; the splitting is required in the base category, not necessarily upstairs.

A split coequalizer fork $A\begin{array}{c}\stackrel{r}{\to }\\ \underset{s}{\to }\end{array}B\stackrel{q}{\to }Q$ has maps $u:B\to A$ and $t:Q\to B$ such thatEvery functor preserves this coequalizer: if $hr=hs$, then $h=hru=hsu=htq$, and $q$ is split epic, so the factor through $q$ is unique. The same equations remain true after applying any functor.

For necessity, let $U^T:{\mathcal{C}}^T\to \mathcal{C}$ be the monad algebra forgetful functor. It has the free-algebra left adjoint. It reflects isomorphisms, because if an algebra map $f$ is invertible in $\mathcal{C}$, the equation $fa=bTf$ rearranges to $f^{-1}b=aT(f^{-1})$.

If two algebra maps $r,s:(A,a)\rightrightarrows (B,b)$ have an underlying split coequalizer $q:B\to Q$, then $Tq$ and $T^{2}q$ are also coequalizers. Since $qb$ equalizes $Tr,Ts$, there is a unique $c:TQ\to Q$ satisfyingAfter composing with the epimorphism $q$, the equation $c{\eta }_Q=1_Q$ reduces to the unit law for $b$. After composing with the epimorphism $T^{2}q$, the equation $cT(c)=c{\mu }_Q$ reduces to $bT(b)=b{\mu }_B$. Thus $(Q,c)$ is an algebra and $q$ an algebra map. If an algebra map $h:(B,b)\to (Z,z)$ equalizes $r,s$, its unique underlying factor $k:Q\to Z$ satisfies $kc=zTk$: this follows by precomposing with the epimorphism $Tq$. Therefore this is a coequalizer upstairs as well. This proves that the monad algebra forgetful functor creates split coequalizers. The same properties transport through the comparison equivalence of categories, with the object-lifting convention distinguished in Question 9.

For sufficiency, suppose the three conditions hold, choose $F⊣U$, and let $T=UF$ with comparison $K$. For each algebra $(X,a)$, consider the pairIts image has coequalizer $a:TX\to X$. It is split: take $r={\mu }_X$, $s=T(a)$, $q=a$, $u={\eta }_{TX}$ and $t={\eta }_X$. The monad and algebra laws giveBy hypothesis the pair has a coequalizer $q:FX\to D_X$ and $Uq$ is also a coequalizer. Identify $U{D}_X$ with $X$ using the unique coequalizer isomorphism, so $Uq=a$. Since $q$ is carried by $K$ to an algebra map from the free algebra, the transported action $b:TX\to X$ satisfiesBut $T(a)$ is a split epimorphism, with right inverse $T({\eta }_X)$, so $b=a$. Hence $(X,a)\cong K(D_X)$: the comparison is essentially surjective.

Next, for every $D\in \mathcal{D}$, put $a_D=U{\epsilon }_D$. The counit of an adjunction coequalizesThis follows from naturality of the counit. The image pair has the split coequalizer $a_D$, as above. If $q:FUD\to Q$ is the hypothesized preserved coequalizer, the induced map $h:Q\to D$ has $Uh$ invertible because both $Uq$ and $a_D$ coequalize the same pair. Conservativity makes $h$ invertible. Therefore ${\epsilon }_D$ itself is a coequalizer.

Let $f:K(D)\to K(E)$ be an algebra map, so $f{a}_D=a_{E}Tf$. Set $g={\epsilon }_{E}F(f):FUD\to E$. Under the adjunction transpose bijection, the two composites of $g$ with the displayed presentation pair correspond respectively to $f{a}_D$ and $a_{E}Tf$. They are equal, so $g$ factors uniquely through ${\epsilon }_D$, giving $h:D\to E$ with $h{\epsilon }_D=g$. Applying $U$ yieldsSince $a_D$ is split epic, $Uh=f$. Thus $K$ is full. If $Uh=Uk$, the transposes of $h{\epsilon }_D$ and $k{\epsilon }_D$ are equal, so those composites agree. The counit is epic, hence $h=k$; thus $K$ is faithful.

The comparison is full, faithful and essentially surjective, so it is an equivalence of categories. This proves monadicity. The argument also explains the role of each hypothesis: split base presentations construct all algebras, and reflection of isomorphisms turns the counit presentations into actual coequalizers upstairs.

For $t$ tetrahedra, a normal surface has a vector $x\in {\mathbb{Z}}_{\ge 0}^{7t}$ counting four triangle types and three quadrilateral types in each tetrahedron. On every glued face, the numbers of arcs of each of its three types must agree on the two sides. These give homogeneous integer linear equationsFor closed surfaces the coordinates of disks meeting an unglued boundary face must also vanish. The nonnegativity and matching equations are not the whole admissibility condition: in each tetrahedron at most one of its three quadrilateral coordinates can be nonzero. Conversely, for an admissible vector, arrange the prescribed parallel disks and glue their face arcs in order. This reconstructs an embedded normal surface, unique up to normal isotopy.

A fundamental normal surface corresponds to a nonzero admissible vector which cannot be written as $y+z$ with both $y$ and $z$ nonzero admissible vectors. These indecomposable vectors form a fundamental set: every admissible vector is their finite sum. To find them, fix which quadrilateral type is allowed in each tetrahedron. This leaves the integer points of a rational polyhedral cone $Ax=0,x\ge 0$, with some coordinates set to zero. Compute its Hilbert basis of a rational cone, for example by subdividing into rational simplicial cones and enumerating the lattice points in their fundamental parallelepipeds along with their integral ray generators, then removing decomposable generators. Repeat the finite list of quadrilateral choices, remove duplicates, and remove vectors decomposable in their compatible sector. Finiteness follows because each of these pointed rational cones has a finite Hilbert basis.

Let $E$ be the link exterior. By Alexander duality, for a two-component linkAn embedded connected closed surface separates $S^3$; its mod-two class is nonzero exactly when the two link components lie on different sides. The connectedness qualification matters: two parallel splitting spheres form a disconnected surface with zero total class. We use the criterion only for connected surfaces.

Normalize a splitting sphere by isotopy, disk compressions and removal of components lying inside tetrahedra. A compression replaces a sphere by two spheres whose mod-two classes add to the old class; at least one is still a splitting sphere. The discarded tetrahedral spheres have zero class. Thus there is a normal splitting sphere. Choose one, $S$, of least normal surface weight.

Suppose its vector is decomposable. Realize $S=A+B$ as a Haken sum, choosing a decomposition with the least possible number of intersection curves. Two elementary exchange observations are useful. First, $A$ and $B$ can be taken connected. If their total union has more than two components, perform regular exchanges connecting components until exactly two embedded groups remain, resolving any intersections within each group. Since the final sum is connected, this can be done before the last joining exchange; it uses up at least one of the old intersection curves. The resulting two normal surfaces still have total vector $x(S)$, contradicting minimality. Second, no intersection curve separates both $A$ and $B$. If it did, resolve that curve first: the two separated sides give two groups. Resolve any remaining self-intersections within each group and continue exchanges between groups only while the result remains disconnected. Just before the exchange making the final connected surface, there are again exactly two embedded normal groups with fewer intersection curves. Both assertions follow from the local four-bank resolution at an intersection circle; internal crossings are resolved before regarding a group as one embedded summand.

Every closed embedded surface in $S^3$ is orientable. Additivity of Euler characteristic in the Haken sum givesFor connected orientable closed surfaces the positive Euler characteristic is $2$, and the only zero Euler characteristic is that of a torus. Thus one summand, say $A$, is a sphere and the other $B$ is a torus. Both have smaller weight than $S$. The sphere $A$ cannot split the link by minimality, soEach intersection circle separates the sphere $A$, so the second exchange observation makes it nonseparating on $B$. In particular, these circles are essential parallel curves on the torus.

Cut $A$ along the intersection circles. There are at least two innermost disk patches. Choose one of smallest weight, $D$; then $2w(D)\le w(A)$. Its interior misses $B$, and its boundary is essential on $B$. Compress the torus along two slightly separated copies of $D$. The result is a sphere $C$, and the compression takes place in the exterior, soThus $C$ is a splitting sphere.

This alone is not quite a contradiction if the last weight inequality is equality: $C$ need not yet be normal. At the compression seam it uses the alternative, rather than regular, exchange between the disk patch and the torus. An intersection circle of normal disk sheets must cross a tetrahedral face; it cannot be a closed curve entirely in two disk interiors in one tetrahedron, which can be put in standard position with only intersection arcs. In a face, parallel arcs of the same type do not cross. At a crossing of different normal arc types, the alternative exchange joins two endpoints on the same edge, producing a returning arc. Pushing across its edge bigon strictly reduces weight. Normalize the resulting surface, retaining a nonzero-class sphere at each compression as before. We obtain a normal splitting sphere of weight strictly less than $w(S)$, a contradiction. Therefore

The algorithm is now finite. Triangulate the exterior from a link diagram, enumerate its fundamental admissible vectors with the closed-surface boundary condition, reconstruct their surfaces, and inspect the connected components and Euler characteristics. A fundamental vector yields a connected surface, since otherwise its component vectors decompose it. A connected closed surface in $S^3$ with Euler characteristic two is a sphere. Compute its mod-two class in the finite simplicial chain complex, or equivalently check which link boundary tori lie on either side. Answer yes exactly when such a sphere has nonzero class. The argument proves that every split link is detected, and every detected sphere witnesses splitting. For more than two components, $H_2(E;{\mathbb{F}}_2)\cong {\mathbb{F}}_2^{\ell -1}$ and a nonzero sphere class separates a nonempty proper subset of the components; the same least-weight argument and finite enumeration apply.

A normal surface meets each tetrahedron in disjoint disks of the four triangular and three quadrilateral types. Its arcs in a face join distinct edges, each disk meets an edge at most once, and at most one quadrilateral type occurs in a tetrahedron. These last requirements exclude, for example, a disk with an octagonal boundary.

We first establish the surgery operation needed for the sphere system. Suppose a disk $D$ disjoint from the other spheres compresses one sphere $S$. It produces spheres $S^{\prime }$ and $S^{\prime \prime }$. Small parallel copies of the three spheres bound a three-times punctured three-sphere $P$. Denote the old piece on the uncompressed side by $A$, and the two pieces on the compressed side by $B^{\prime }$ and $B^{\prime \prime }$. If both $B^{\prime }$ and $B^{\prime \prime }$ were punctured three-spheres, then $P$ glued to them would also be a punctured three-sphere, contrary to the hypothesis on the original cut manifold. Choose, say, $B^{\prime }$ which is not of this type, and retain $S^{\prime }$ alone in place of $S$.

The other new piece is $A\cup P\cup B^{\prime \prime }$. It too cannot be a punctured three-sphere. If it were, cap its external boundary spheres to make $S^3$. All the spheres separating $A$ from the attached pieces then bound three-balls by the Three-dimensional Schoenflies theorem; the component containing $A$ would be a punctured three-sphere, contradicting the hypothesis on $A$. Thus the sphere surgery preserving nontrivial complementary pieces keeps exactly $k$ spheres and keeps every complementary piece nontrivial. The assumption that all spheres separate ensures that the old and new pieces used here really are distinct sides.

Apply general position to the whole sphere system, avoiding the vertices and making it transverse to edges and faces. We describe a terminating normalization, allowing the replacement operation just proved as well as isotopy. If some component of its intersection with a tetrahedron is not a disk, choose a boundary circle innermost among circles not bounding such disk components. A disk just inside the tetrahedral boundary gives a compression; if intervening disk components occur, first push them into collars of their boundary disks. Perform the compression and retain one new sphere as above. This replaces a non-disk tetrahedral piece by disk pieces without increasing edge intersections. Repetition is finite: every compression divides the original sphere along a separating circle, and the sum of $1-\chi$ over the non-disk planar tetrahedral pieces decreases; disk components can then be put in boundary collars. A closed sphere component wholly in a tetrahedron is impossible, because it would cut off a three-ball, contrary to the complementary-piece condition.

Now each tetrahedral piece is a disk. A face circle innermost in a face allows its incident disk to be moved through that face disk, removing the circle. This reduces the number of face intersections without increasing the edge weight. A returning face arc similarly gives an edge bigon across which the surface can be pushed, reducing the number of edge intersections by two. Restore tetrahedral disk pieces by the preceding operations whenever necessary.

It remains to exclude multiple intersections of a disk with an edge, which the absence of returning face arcs alone does not exclude. Suppose a tetrahedral disk meets an edge twice. Among such disks choose a pair of intersections consecutive on the edge and with no other surface intersection between them. Such a pair exists: an intervening disk crosses the edge interval an even number of times, so passing to an innermost intervening pair eventually gives consecutive points. Join them by an arc in the disk. The disk pieces divide the tetrahedron into three-balls, and this arc together with the edge interval lies in the boundary of one of these three-balls. It therefore bounds a bigon with interior disjoint from the surface. Push across the bigon to reduce edge weight by two. At fixed weight the previous moves reduce face and non-disk complexity; the edge weight itself is a nonnegative integer. Hence the whole process terminates. A final tetrahedral disk meets each edge at most once, and its boundary separates either one vertex from three or two vertices from two. It is respectively a normal triangle or quadrilateral. Thus a normal system of exactly $k$ spheres with the required complementary-piece property exists.

For the finite decomposition, we need a bound on such normal systems, not merely an assertion that they have been normalized. Let $f_2$ be the number of faces of the fixed triangulation. In a triangular face cut by normal arcs, all but at most four complementary regions are rectangles between adjacent parallel arcs. Thus at most $4f_2$ complementary three-dimensional components contain a nonrectangular face region. Every remaining component is assembled from product regions between parallel normal disks, so is an interval bundle whose boundary components are spheres. Its connected base is either $S^2$ or ${\mathbb{RP}}^2$. The product $S^2\times I$ is a punctured three-sphere, and is excluded. The other possibility is the twisted interval bundle over ${\mathbb{RP}}^2$, namely ${\mathbb{RP}}^3$ with a three-ball removed. Each such piece contributes a distinct ${\mathbb{F}}_2$ summand to $H_1(M;{\mathbb{F}}_2)$: gluing along spheres does not identify first-homology classes, by the Mayer–Vietoris sequence. Since $k$ separating spheres give $k+1$ components, we obtain the explicit finite bound

For an arbitrary closed connected $X$, first cut out nonseparating spheres. A neighbourhood of one such sphere together with an arc joining its two sides is a punctured $S^2$-bundle over $S^1$; the bundle is the product or the twisted bundle according to the orientation of the attaching identification. Its complement is connected, so this extracts a connected sum factor. The mod-two first Betti number of the remaining capped manifold decreases by one. Hence only finitely many such extractions are possible. The remaining manifolds have only separating spheres.

Whenever a remaining factor is not a prime three-manifold, split it along a sphere into two factors neither homeomorphic to $S^3$. Viewed in the original remaining manifold, all these successive splitting spheres form a disjoint system whose pieces are not punctured three-spheres. The normalization and bound just proved prohibit infinitely many such splittings. This proves existence of a finite prime decomposition, including for $X$ without assuming orientability.

For completeness, both sphere bundles extracted above are prime three-manifolds. Their fundamental group is $\mathbb{Z}$, so a separating sphere and the van Kampen theorem express $\mathbb{Z}$ as a free product in which one capped factor has trivial fundamental group. Its punctured piece lifts homeomorphically to the universal cover $S^2\times \mathbb{R}$, identified with ${\mathbb{R}}^3\setminus \{0\}$. That compact lifted piece has one spherical boundary and must be the bounded side of this sphere in ${\mathbb{R}}^3$. It misses the origin, so the Three-dimensional Schoenflies theorem makes it a three-ball. The corresponding capped factor is $S^3$, as required.

Suppose now that $X$ is orientable. A prime factor which is not an irreducible three-manifold contains a nonseparating sphere: an essential separating sphere would already split it nontrivially. Extracting the corresponding sphere bundle shows that the factor must be $S^2\times S^1$. Thus the other nontrivial primes are irreducible.

Consider two prime decompositions. Represent each by a sphere system, including one nonseparating sphere in each $S^2\times S^1$ factor. The capped pieces are now the irreducible factors and copies of $S^3$. Make the two sphere systems transverse. To remove their intersections, choose a circle innermost on a sphere of the second system and surger the first system along its disk. Retain both surgery spheres temporarily. The old sphere and the new pair bound a three-times punctured sphere; adding the pair and then deleting the old sphere only inserts or removes spherical punctures.

This operation cannot alter an irreducible capped piece: every new sphere in that piece bounds a three-ball, so it can only split off a punctured $S^3$ piece. The other pieces were already punctured $S^3$, where the same conclusion follows from the Three-dimensional Schoenflies theorem. After small displacements the surgery removes the chosen intersection circle and introduces none. Repeat until the sphere systems are disjoint. Their union is a common refinement, and its non-$S^3$ capped irreducible pieces are precisely those of each original decomposition. Hence these factors agree, up to order and orientation-preserving homeomorphism.

Finally, the number $r$ of $S^2\times S^1$ factors is determined bywhere the sum runs over the common irreducible factors. This uses additivity of first homology under connected sum, not a claim that the fundamental group determines a manifold. Therefore the orientable prime decomposition is unique up to order and insertion or deletion of $S^3$ factors.

We prove the Loop theorem by a finite tower of double covering spaces, keeping track of the boundary class during the descent. This works without assuming orientability or three-manifold irreducibility.

Let $F$ be the boundary component containing the basepoint. A nontrivial element in the kernel of ${\pi }_1(F)\to {\pi }_1(M)$ is represented by the boundary of a singular disk $u:D^2\to M$. Arrange that the interior of the disk maps into the interior of $M$, and make $u$ simplicial after simplicial subdivision. Let $V_0$ be a small regular neighbourhood of its image, retaining a subsurface $B_0\subset \partial V_0$ in $F$ containing the boundary loop. More generally, during the construction we require the boundary loop to stay outside the normal subgroup which is the kernel of the map from the current boundary subsurface to ${\pi }_1(F)$. This records exactly the information needed at the end.

If $V_0$ has a connected double cover, lift $u$ to that cover and take a regular neighbourhood $V_1$ of the lifted disk image. Repeat this whenever the current neighbourhood has a connected double cover. This procedure is finite. Indeed, let $C_i$ be the simplicial image of the disk at stage $i$. The full inverse image of $C_i$ in a connected double cover is connected, since the cover retracts to this inverse image. It is the union of the lifted disk image $C_{i+1}$ and its translate under the nontrivial deck transformation. These two images must meet. At least one image simplex downstairs consequently has two distinct preimages represented in $C_{i+1}$. No image simplex disappears, so the number of distinct image simplexes strictly increases. This number is at most the number of simplexes of the fixed domain disk. Thus the tower proof of the loop theorem terminates.

At the top, $V_r$ has no connected double cover. Equivalently, $H^1(V_r;{\mathbb{F}}_2)=0$. Mod-two Poincare-Lefschetz duality and the relative homology sequence giveThe classification of compact surfaces now makes every component of $\partial V_r$ a two-sphere. The lifted boundary subsurface $B_r$ is planar. Its fundamental group is normally generated by its boundary circles. Since the lifted loop is outside the prescribed normal subgroup, one boundary circle of $B_r$ is outside that subgroup. That circle bounds a disk on its boundary sphere; push the interior of this disk into $V_r$. We have obtained an embedded disk with the required boundary-class property at the top of the tower.

We next descend one double cover. Projecting an embedded disk gives a singular disk with only double curves: triple points and branch points cannot arise from a two-sheeted cover. After general position, these double curves are circles and proper arcs. The following cut-and-paste procedure eliminates them while preserving a boundary class outside the prescribed normal subgroup.

First eliminate the circular double curves. Take a double circle innermost on one of its preimage sheets. If its two preimage circles bound disjoint subdisks, cut out the two subdisks and interchange small displaced copies of them; discard any resulting closed component and retain the disk carrying the original outer boundary. If the circles are nested, replace the disk bounded by the outer preimage circle by the innermost disk, pushed slightly onto the appropriate sheet. Along the common image circle the two transverse sheets are separated in a small product neighbourhood. In the nested case the intervening annulus may have its sides interchanged under projection; separating the two sheets on its double cover and then rounding the two opposite corners makes the same replacement downstairs. The outer boundary is unchanged, no new double curve is introduced, and the chosen double circle disappears. Repeating leaves only double arcs.

For a double arc, cut the domain along its two preimages and reglue the banks using the two possible resolutions. There are two disk choices. At least one has boundary outside the prescribed normal subgroup. Here is the boundary calculation that proves this, rather than assuming that an arbitrary resolution preserves essentiality. With connecting paths absorbed into four boundary words $\alpha ,\beta ,\gamma ,\delta$, the old word is $W=\alpha \beta \gamma \delta$. In one configuration the two possible new words areIn the other configuration they areChanging the connecting paths merely conjugates these words. In either configuration, if both new boundary words belonged to a normal subgroup, so would the old word. Choose a disk whose word stays outside the subgroup. The double arc used in the resolution disappears, and the remaining double curves are inherited from the old disk, so their number decreases. Remove any circular double curves again, and repeat. The resulting disk is embedded.

Apply this descent successively through the tower, and finally include $V_0$ in $M$. Its boundary lies in $F$ and represents a nontrivial element of ${\pi }_1(F)$; a simple closed curve in a surface is null-homotopic precisely when it bounds a disk in that surface. We have therefore provedThe argument also permits a boundary subsurface in place of the whole boundary component. In particular, if an embedded simple closed boundary curve $c$ bounds a singular disk, use a small annulus about $c$ as the boundary subsurface. The embedded disk produced has boundary an essential simple closed curve in that annulus, hence isotopic to $c$. Extend that isotopy through a boundary collar. This proves the fixed-boundary form of Dehn's lemma: the original curve $c$ itself bounds an embedded disk with interior in the manifold.

Now consider the knot map. Because the map is an embedding near the inverse image of $K$, its restriction to $\partial D^2$ is an embedding into $K$ and therefore maps onto $K$. No interior point can map to $K$, since it would duplicate a boundary image. Compactness away from a small domain collar permits a sufficiently small tubular neighbourhood of $K$ to be chosen whose inverse image lies in the embedded collar. Its intersection with this collar cuts off an annulus. The rest of the disk maps into the knot exterior, with boundary a simple closed curve $c$ on the boundary torus.

The annulus shows that $c$ runs once along $K$. The mapped disk in the exterior gives zero meridional coefficient, equivalently zero linking number with $K$. Thus $c$ is a longitude. Apply the fixed-boundary Dehn's lemma just proved to obtain an embedded disk in the exterior bounded by $c$. Attach the embedded collar annulus between $c$ and $K$. This gives an embedded spanning disk for $K$. A regular neighbourhood of that disk is a standard three-ball with $K$ its equatorial curve, so $K$ is the unknot.

The two-dimensional Schoenflies theorem says that a simple closed piecewise linear curve in $S^2$ separates it into two components whose closures are piecewise linear balls of dimension two. In particular, a homeomorphism of the curve to the equator extends to a piecewise linear homeomorphism of the whole sphere.

A piecewise linear ball is a space piecewise linearly homeomorphic to a standard simplex. Write $D=B_1\cap B_2$. The complementary disk to $D$ in each $\partial B_j$ is a disk by the two-dimensional Schoenflies theorem. Thus each pair $(\partial B_j,D)$ is piecewise linearly homeomorphic to a sphere with a chosen hemisphere. Choose these boundary maps to agree on $D$. Extend them to the interiors by coning from an interior point, after compatible subdivisions. The resulting maps identify $B_1$ and $B_2$ with the two halves of a standard ball, glued along its equatorial disk. ConsequentlyThe same boundary parametrization proves a useful companion fact: if a ball is indented by removing a smaller ball attached along a boundary disk, the closure of what remains is again a ball. In the standard boundary-disk model, move the indented disk through a collar neighbourhood to a flat disk.

A handle decomposition is a filtration obtained by attaching handles $D^r\times D^{n-r}$ along $\partial D^r\times D^{n-r}$ in the boundary of the preceding stage. The number $r$ is the handle index; index-zero handles are disjoint balls, and index-$n$ handles close the remaining spherical boundary components. Starting with a triangulation of $M$, thicken its vertices, then its edges, and successively its higher-dimensional simplexes. In a barycentric subdivision these thickenings have precisely the indicated handle and attaching-region product structures. Adding collars between successive attachments gives a collared handle decomposition.

For a critical level handle embedding, assign successive distinct heights $t_j$ to the handles. At height $t_j$ the horizontal slice contains the $j$th handle and its attaching region. Between handle heights, the embedded boundary of the previously attached handles travels along its product collar; there is no topological change there. The lower boundary collar meets the attaching region of the next handle, and the upper collar leaves its new boundary. The stated ambient isotopy therefore separates all topology changes into horizontal handles and all intervening motion into collars. For an embedded surface, the local changes are the appearance of a disk, attachment of a band, and disappearance of a disk.

Here is the three-dimensional conclusion, with the distinction between the bounded and unbounded regions made explicit: a piecewise linear embedded $S^2$ in ${\mathbb{R}}^3$ bounds a piecewise linear three-ball; in the one-point compactification $S^3$, both complementary closures are three-balls. The unbounded closure in ${\mathbb{R}}^3$ itself is not a compact ball.

Put the embedded sphere in critical level handle embedding position. At a regular horizontal level its intersection with the plane is a finite disjoint union of simple closed curves. An innermost such curve bounds a planar disk whose interior misses the sphere. Cut the sphere along this curve and insert two slightly displaced copies of the disk. This replaces one sphere by two spheres. Repeat at regular levels between the critical heights, using innermost curves first. The number of curves at these chosen levels decreases at each operation, so the process stops. Each resulting sphere then lies in a slab containing at most one horizontal handle.

These terminal spheres bound balls directly. With no handle the surface is a disk transported through a collar and capped at its ends, the boundary of $D^2\times I$. A minimum or maximum merely supplies one of those caps. At a band level there are two planar boundary circles on one side and one on the other. Cap the two circles, straighten each planar disk using the two-dimensional Schoenflies theorem, and straighten the joining band in its planar collar. If the circles are not nested, this gives two balls joined along a boundary disk; if they are nested, it gives a ball with one boundary-attached ball removed. The gluing three-balls along a boundary disk argument and its indentation version show in both cases that the terminal sphere bounds a ball. This also explains why the planar position of the band matters: no assertion about an arbitrary knotted band is being made.

Reverse the disk surgeries. Suppose the two resulting spheres bound balls $A$ and $B$. Their disjoint boundaries imply that the balls are disjoint or one contains the other. Shrink the small displacement of the two surgery caps back to the original planar disk. In the disjoint case, the pre-surgery sphere bounds the union of two balls along that disk. In the nested case, it bounds the larger ball with the smaller boundary-attached ball removed. Both are balls by the preceding elementary facts. Induction through the finitely many surgeries proves the Three-dimensional Schoenflies theorem. To treat the other region in $S^3$, remove a point inside the ball just constructed and apply the same argument in the resulting copy of ${\mathbb{R}}^3$.

A 3-manifold is an irreducible three-manifold when every piecewise linear embedded two-sphere bounds a ball. Suppose $p:{\mathbb{R}}^3\to M$ is a universal cover, and lift an embedded sphere $S$ to $\tilde{S}$. Since $S$ is simply connected, its lift maps homeomorphically to $S$. By the Three-dimensional Schoenflies theorem, $\tilde{S}$ bounds a compact ball. The inverse image of $S$ is locally finite, so only finitely many other lifted spheres lie inside this ball. Choose an innermost one, bounding a ball $B$ whose interior meets no lift of $S$.

The interior of $B$ maps to one component $U$ of $M\setminus S$. This restriction is a covering map onto $U$: a path in $U$ starting in the image lifts into $\mathrm{int}B$, since it cannot cross a lifted boundary sphere. The covering is proper, because a compact subset of $U$ has preimage closed in the compact ball and cannot meet $\partial B$. Near $S$, the boundary map $\partial B\to S$ is one-to-one, so this connected covering has one sheet. The map on $B$ is therefore a homeomorphism onto $U\cup S$. Thus every embedded sphere in $M$ bounds a ball, and $M$ is an irreducible three-manifold.

For a smooth complex vector bundle $E$, a connection is a complex-linear map $D:\Gamma (E)\to A^1(M;E)$ satisfying $D(fs)=df\otimes s+fDs$ for smooth complex functions $f$. Local trivializations supply flat local connections. A locally finite smooth partition of unity $({\rho }_i)$ subordinate to them gives $D={\sum }_i{\rho }_i{D}_i$; local finiteness makes the sum smooth and ${\sum }_i{\rho }_i=1$ gives the Leibniz rule. Thus connections on a vector bundle always exist on the usual paracompact smooth manifold.

Extend $D$ to bundle-valued forms by $D(\alpha \otimes s)=d\alpha \otimes s+(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge Ds$. Squaring shows that $D^2(fs)=f{D}^{2}s$, so the curvature form is the endomorphism-valued two-form $\Theta =D^2$. With a row frame $e=(e_1,\dots ,e_r)$, write $De=eA$; for coefficient columns $s=ev$, $Ds=e(dv+Av)$. Applying $D$ again and using the graded Leibniz rule givesThe cross terms $-A\wedge dv$ and $A\wedge dv$ cancel. Hence the Cartan curvature matrix equation in this convention isFor a frame change $e^{\prime }=eg$, the connection matrix and curvature transform as $A^{\prime }=g^{-1}Ag+g^{-1}dg$ and ${\Theta }^{\prime }=g^{-1}\Theta g$. Trace is invariant under conjugation, so the local forms $\mathrm{Tr}\Theta$ patch into the global trace of vector-bundle curvature.

The induced determinant connection on $\det E={\Lambda }^{r}E$ iswhere the one-form factor of each $D{s}_j$ is placed in front of the section factors. In the local determinant frame $e_1\wedge \cdots \wedge e_r$, only the diagonal terms survive, so its connection form is $\mathrm{Tr}A$. Its curvature is $d\mathrm{Tr}A$. Sinceby pairing off-diagonal terms and using $A_{ii}\wedge A_{ii}=0$, we obtainIt is closed locally by $d^2=0$, and thus closed globally.

For two connections, $B=D_1-D_0$ is a global endomorphism-valued one-form: the derivative terms cancel in their Leibniz rules. In a local frame $A_1=A_0+B$, soTaking traces cancels the mixed terms, because both factors are one-forms, and cancels $\mathrm{Tr}(B\wedge B)$. Therefore the trace curvature transgression formula isThe right-hand side is globally exact, proving independence of the de Rham cohomology class without needing a sheaf-cohomology identification.

Finally let $e$ be a holomorphic frame for the holomorphic line bundle, with $h=\Vert e{\Vert }^2>0$. Define $De=e\partial \log h$, extending by the Leibniz rule. If $e^{\prime }=eg$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andThis is exactly the line-bundle frame-change law for a connection, so the local definitions glue globally. Its $(0,1)$ part is the Dolbeault operator. Metric compatibility follows from $d\log h=\partial \log h+\bar{\partial }\log h$, or $dh=h(A+\overline{A})$ for $A=\partial \log h$. These conditions uniquely force this $A$, identifying the Chern connection. Since a scalar one-form wedges with itself to zero,This agrees with the local formula for the Chern connection on a line bundle; keeping the order of the two Dolbeault differentials fixes the sign.

An almost complex structure is a smooth endomorphism $J:TM\to TM$ with $J^2=-I$; in particular the real dimension is even. Complexifying the tangent bundle splits it into the $i$ and $-i$ eigensubbundlesinterchanged by conjugation. On a complex manifold, these are locally spanned by ${\partial }_{z^j}$ and ${\partial }_{{\bar{z}}^j}$. An integrable almost complex structure is one arising from a holomorphic coordinate atlas. Its $T^{0,1}$ sections are closed under the Lie bracket. The obstruction is the Nijenhuis tensorExpanding this expression in the two eigentypes shows that $N_J=0$ is equivalent to involutivity of $T^{0,1}$. Holomorphic coordinates make this condition necessary; the Newlander-Nirenberg theorem gives its sufficiency for smooth $J$. Integrability also gives $d=\partial +\bar{\partial }$, with ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault operator on $T^{1,0}M=T_M^{\prime }$ defines its holomorphic vector bundle structure.

A real connection on a vector bundle $\nabla$ on $TM$ preserves $J$ when $\nabla J=0$. Its complexification then preserves both eigensubbundles. Restricting to $T_M^{\prime }$ gives a complex connection $D$. Conversely any complex connection on $T_M^{\prime }$ gives such a real connection by adjoining its conjugate on $T^{0,1}$ and restricting to the conjugation-fixed real tangent bundle. Equivalently, transfer $D$ through the real isomorphism $v\mapsto (v-iJv)/2$. These operations are inverse. Preserving $J$ alone does not impose the holomorphic compatibility condition $D^{0,1}={\bar{\partial }}_{T^{\prime }}$.

For integrable $J$, the latter condition has a precise real-connection interpretation: the mixed-type torsion tensor vanishes. In holomorphic coordinates it says $D_{{\partial }_{{\bar{z}}^i}}{\partial }_{z^j}=0$. Reality also gives ${\nabla }_{{\partial }_{z^j}}{\partial }_{{\bar{z}}^i}=0$. Since these coordinate fields commute, their torsion is zero. Conversely a $J$-preserving real connection with zero mixed torsion has these two derivatives equal; they belong to opposite eigentypes, so both vanish, proving $D^{0,1}={\bar{\partial }}_{T^{\prime }}$. This is the holomorphic tangent connection and mixed torsion criterion.

Full torsion-freeness imposes the additional symmetry ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ in $D_{{\partial }_{z^i}}{\partial }_{z^j}={\Gamma }_{ij}^k{\partial }_{z^k}$, and its conjugate. Such a $J$-preserving torsion-free connection exists on any complex manifold: local holomorphic-coordinate flat connections have these properties, and a real partition of unity combines them globally. Preservation of $J$ and zero torsion persist because both conditions are affine in the connection. Conversely if $J$ is merely almost complex, a torsion-free connection preserving it forces $N_J=0$: replace brackets by ${\nabla }_{X}Y-{\nabla }_{Y}X$ in the displayed tensor and use $\nabla J=0$; all terms cancel. Thus this connection condition detects integrability, without imposing any metric condition.

A Hermitian metric is a real Riemannian metric $g$ with $g(JX,JY)=g(X,Y)$. It induces a Hermitian metric $h$ on $T_M^{\prime }$ and the fundamental two-form $\omega (X,Y)=g(JX,Y)$. The unique Chern connection satisfies both $D^{0,1}={\bar{\partial }}_{T^{\prime }}$ and metric compatibility. In a holomorphic frame with metric matrix $H$ and coefficient-column convention, its matrix is $H^{-1}\partial H$. It need not be torsion-free when transferred to $TM$.

In holomorphic coordinates its coefficients are ${\Gamma }_{ij}^k=h^{k\bar{l}}{\partial }_i{h}_{j\bar{l}}$. Their symmetry in $i,j$ is equivalent to ${\partial }_i{h}_{j\bar{l}}={\partial }_j{h}_{i\bar{l}}$, which says $\partial \omega =0$. The conjugate gives $\bar{\partial }\omega =0$. The result that a torsion-free Chern tangent connection characterizes a Kähler metric follows: the transferred Chern connection is torsion-free exactly when $\boxed{d\omega =0}$, the Kähler metric condition. In that case it equals the Levi-Civita connection, by uniqueness of the metric-compatible torsion-free connection. Equivalently a Hermitian metric is Kähler exactly when its Levi-Civita connection preserves $J$. In the reverse direction $\nabla J=0$ implies $\nabla \omega =0$, and zero torsion gives $d\omega =0$.

This explains the role of Kähler metrics: they make the real Riemannian and holomorphic connection theories coincide. A general complex manifold has torsion-free $J$-preserving connections and has Hermitian metrics, but need not have a connection satisfying both requirements simultaneously. On a Kähler manifold, the Kähler identities then link the Dolbeault Laplacian and Hodge Laplacian, giving ${\Delta }_d=2{\Delta }_{\bar{\partial }}=2{\Delta }_{\partial }$ and the resulting harmonic Hodge decomposition theorem for compact Kähler manifolds.

At a point $[X]\in {\mathbf{P}}^n$, the complex tautological line bundle has fiber $\mathbb{C}X\subseteq {\mathbb{C}}^{n+1}$. The hyperplane line bundle is its dual, $[H]=\mathcal{O}(1)$, and $\mathcal{O}(m)=\mathcal{O}(1{)}^{\otimes m}$. On $U_i=\{X_i\ne 0\}$ let $t_i=X/X_i$ be the tautological frame and $e_i$ its dual. On overlaps $e_j=(X_j/X_i)e_i$. The sheaf of holomorphic sections of a vector bundle of this dual bundle is the sheaf ${\mathcal{O}}_{{\mathbf{P}}^n}(1)$.

If a nonzero linear form $\ell$ cuts out the projective hyperplane $H$, it gives a holomorphic section $s_H$ of $\mathcal{O}(1)$ with a simple zero along $H$. Dividing a local section $s$ by $s_H$ gives a meromorphic function with no poles away from $H$ and at most a simple pole on $H$. Conversely multiplication by $s_H$ turns such a function into a holomorphic section. These mutually inverse local constructions agree on overlaps and prove the holomorphic line bundle associated to a divisor identification

The nonnegative hyperplane twist cohomology by restriction calculation givesTo derive it from the permitted vanishing, use the hyperplane exact sequence for twisting sheaveswhere $i:H\hookrightarrow {\mathbf{P}}^n$ and $H\cong {\mathbf{P}}^{n-1}$. Local multiplication by a defining equation is injective, and restriction gives exactly its cokernel. Induct on $n$, starting from a point, and then on $m$. For $m=0$, higher sheaf cohomology vanishes by assumption and global holomorphic functions are constant by compactness and the maximum principle. For $m>0$, the long exact sequence in sheaf cohomology, the inductive vanishing for $\mathcal{O}(m-1)$ and the dimension induction on $H$ give vanishing in every positive degree andSumming this recurrence gives $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$. Homogeneous degree-$m$ monomials give that many independent sections, so they form a basis. The printed extra equality to zero cannot hold in degree zero, already because $h^0({\mathbf{P}}^n,\mathcal{O})=1$; the boxed calculation gives the intended dimensions.

For the differential of the projective quotient map, the differential of a smooth map is computed by applying a tangent vector to coordinate functions. Since $x_j=X_j/X_0$,ThusUnder replacing $X$ by $\lambda X$, $d\pi$ on a constant ambient tangent vector scales by ${\lambda }^{-1}$, while a homogeneous linear form $L$ scales by $\lambda$. Therefore $d{\pi }_X(L(X){\partial }_{X_j})$ is independent of the representative. Its chart coefficients are holomorphic, so it is a global holomorphic vector field. The radial Euler vector field lies in the kernel of $d\pi$, either from the same formulas or because it differentiates a scaling orbit. Hence

Describe the Euler sequence on complex projective space fiberwise as well as in coordinates. At a line $\ell =\mathbb{C}X$, the middle bundle has fiber $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1})$ and the holomorphic tangent bundle has fiber $\mathrm{Hom}(\ell ,{\mathbb{C}}^{n+1}/\ell )$. Send a homomorphism to its composition with the quotient. The map from $\mathcal{O}$ sends a scalar to that scalar times the inclusion $\ell \hookrightarrow {\mathbb{C}}^{n+1}$. In homogeneous notation these maps areA local section of $\mathcal{O}(1)$ is a local degree-one homogeneous function on the punctured cone, so the second expression is representative-independent. Quotienting is surjective, and its kernel consists exactly of scalar inclusion maps. This proves the short exact sequence of locally free sheavesSince it is locally split, dualizing and tensoring by $\mathcal{O}(1)$ preserves exactness and gives the cotangent Euler sequence in homogeneous coordinatesOn global sections the last map identifies ${\mathbb{C}}^{n+1}$ with the basis of linear forms in $H^0(\mathcal{O}(1))$. Its kernel is zero. This proves the vanishing of global sections of the once-twisted projective cotangent bundle; left exactness gives $\boxed{h^0({\mathbf{P}}^n,{\Omega }^1(1))=0}$.

The dimension of an algebraic set is the supremum of lengths of strict chains of nonempty irreducible closed subsets. It is the maximum of the dimensions of the irreducible components. On an affine variety $X=\mathrm{Spec}A$, it equals the Krull dimension of $A$; if $X$ is irreducible, Noether normalization identifies this with the transcendence degree of its function field over $k$. Nonempty Zariski-open subsets of an irreducible algebraic variety have the same dimension, so the definition is compatible with affine charts and with birational maps. In particular, $\dim {\mathbb{A}}^n=\dim {\mathbb{P}}^n=n$ and $\dim (X\times Y)=\dim X+\dim Y$ for irreducible algebraic varieties.

The fiber dimension theorem says that a dominant morphism of algebraic varieties $f:X\to Y$ between irreducible algebraic varieties has generic fibre dimensionEvery irreducible component of every nonempty fibre over a closed point has dimension at least $r$. There is a nonempty Zariski-open subset of $Y$ on which the fibres are nonempty and are equidimensional algebraic varieties of dimension $r$. Dominance is required; for a general morphism of algebraic varieties, replace $Y$ by the closure of its image. This gives the dimension of image of a morphism formula.

For the generic fibre, the formula is the additivity of transcendence degree in the tower $k\subset k(Y)\subset k(X)$. The lower bound on closed fibres comes from cutting by at most $\dim Y$ local parameters after a finite Noether normalization of an affine target chart: the principal hypersurface dimension lemma drops dimension by at most one per equation, and the finite normalization fibre separates the finitely many target points. For the generic upper bound, take finitely many affine source charts and apply Noether normalization to each coordinate ring over $k(Y)$. Clearing denominators in $k[Y]$ makes each chart finite over ${\mathbb{A}}^r$ relative to a suitable open target chart. Its fibres have dimension at most $r$. Intersecting these finitely many target opens and using the lower bound gives generic equidimensionality of fibres.

The upper semicontinuity of local fibre dimension is a statement on the source: for a finite-type morphism of schemes, the points $x$ satisfying ${\dim }_x{X}_{f(x)}\le d$ form an open set. For a proper morphism, the maximum fibre dimension is upper semicontinuous on the target as well. The properness qualification matters; one should not assert the latter for every finite-type morphism of schemes.

A first application is dimension counting for plane sections and intersections. On an irreducible affine variety, imposing $q$ polynomial equations gives each nonempty component algebraic codimension at most $q$, by repeated principal ideal theorem. On a smooth variety this underlies the intersection dimension bound on a smooth variety. A second application is to images: if a morphism of algebraic varieties has finite nonempty fibres, its image closure has the same dimension as the source. If it is also proper, that image is closed. Thus an isolated fibre point forces dominance in equal dimensions when the irreducible source and target have equal dimension.

For the cubic-surface application, projective lines in ${\mathbb{P}}^3$ form the Grassmannian $G=\mathrm{Gr}(2,4)$ of dimension four. Cubic forms form ${\mathbb{P}}^{19}$, since there are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ degree-three monomials in four variables. The cubic surface line incidence varietyhas fibre ${\mathbb{P}}^{15}$ over every $L$: restricting a cubic to a projective line gives four independently prescribable coefficients. It is therefore an irreducible projective-space bundle of dimension $4+15=19$. The other projection $\pi :I\to {\mathbb{P}}^{19}$ is a proper morphism, so its image $Z$ is closed and irreducible.

To show $Z={\mathbb{P}}^{19}$, dimension counting alone is not enough; exhibit a fibre with an isolated point. TakeA nearby projective line is the graph $x_2=a{x}_0+b{x}_1$, $x_3=c{x}_0+d{x}_1$. Restriction givesIt vanishes only when $a=b=c=d=0$. Thus $L_0$ is an isolated reduced point of the fibre. Applying the lower bound in the fiber dimension theorem to $I\to Z$ gives $0\ge 19-\dim Z$. Since $Z\subseteq {\mathbb{P}}^{19}$, its dimension is exactly 19, and closedness forces $Z={\mathbb{P}}^{19}$. This is the incidence proof that a cubic surface contains a line. The auxiliary cubic $F_0$ need not be smooth: the argument proves existence for every cubic form, and hence in particular for every smooth cubic surface.