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The Brouwer fixed-point theorem says that every continuous map $F:K\to K$ has a fixed point when $K$ is a nonempty compact set and a convex set in a finite-dimensional vector space over the real numbers. Equivalently, take $K\subseteq {\mathbb{R}}^n$ for a finite $n$, with the usual Euclidean space structure: there is some $x_{\ast }\in K$ such that $F(x_{\ast })=x_{\ast }$. The conclusion guarantees at least one fixed point, without asserting uniqueness or telling us how quickly to find one. ^[ref]

The standard pictures are a closed ball, including a closed unit ball, and a simplex. A filled disk qualifies; its bounding circle alone does not. A probability simplex also qualifies even though its dimension is smaller than that of its ambient Euclidean space. The hypotheses have concrete meanings: being a convex set keeps every line segment between allowed points inside $K$, and compactness prevents escape or loss of a limiting point. In Euclidean space, the Heine-Borel theorem identifies compact sets with closed sets that are bounded sets. Crucially, every output must belong to the same $K$ as the input.

For a fully computable example, let $D=\{(u,v):u^2+v^2\le 1\}$ be the filled unit disk. Apply a counterclockwise quarter-turn rotation, shrink by $0.4$, and add a translation:This affine map is a continuous map. Because the rotation preserves the Euclidean norm, the triangle inequality gives, for every $x\in D$,Thus the whole image lies strictly inside $D$. More precisely, it is a disk of radius $0.4$ centred at $(0.3,0.1)$. This containment check is what makes the Brouwer fixed-point theorem applicable; a formula being a continuous map by itself would not suffice.

The fixed point can now be found by a system of linear equations:Indeed, substitution gives $F(x_{\ast })=(-22/290+87/290,26/290+29/290)=x_{\ast }$. Its Euclidean norm is $\sqrt{290}/58\approx 0.293610$, safely inside $D$. The figure shows arrows from sample inputs to their outputs; an arrow disappears at the fixed point because the displacement $F(x)-x$ vanishes there.

Repeated iterations of a map happen to succeed for this example. Starting at $x_0=(1,0)$ and setting $x_{k+1}=F(x_k)$ gives $x_1=(0.3,0.5)$, $x_2=(0.1,0.22)$, and $x_3=(0.212,0.14)$. Subtracting the fixed point equation yields $x_{k+1}-x_{\ast }=0.4R_{90}(x_k-x_{\ast })$, so the Euclidean distance to $x_{\ast }$ shrinks by exactly $0.4$ each step. This extra fact follows from the contraction mapping property and the Banach fixed-point theorem. It is additional information beyond the Brouwer fixed-point theorem.

The distinction becomes visible with two simpler continuous maps of the same disk. The pure rotation $G(u,v)=(-v,u)$ has exactly one fixed point, $(0,0)$, since $u=-v$ and $v=u$ force both to vanish. Yet iterations of a map starting at $(1,0)$ here cycle through $(0,1)$, $(-1,0)$, and $(0,-1)$ forever. Their Euclidean distance from $(0,0)$ stays one, so there is no convergence in a metric space. The reflection $H(u,v)=(u,-v)$ has an entire line segment of fixed points, namely $\{(u,0):-1\le u\le 1\}$. Starting off that line segment produces an alternating pair. Existence, uniqueness, and convergence of iterations of a map are separate questions.

Removing hypotheses changes the answer. Restrict the same quarter-turn rotation to the circle $S^1=\{(u,v):u^2+v^2=1\}$. It remains a continuous map into that compact set, but its only possible fixed point, $(0,0)$, has been removed. The circle is not a convex set. Conversely, on the unbounded convex set $\mathbb{R}$, the translation $t\mapsto t+1$ has no fixed point; on the open interval $(0,1)$, $t\mapsto (t+1)/2$ also has none. Neither domain is a compact set. The finite-dimensional closed ball statement should likewise not be extended to arbitrary infinite-dimensional vector spaces, where a closed ball need not be a compact set.

The one-dimensional version explains the connection to the intermediate value theorem. For a continuous map $f:[a,b]\to [a,b]$, put $h(t)=f(t)-t$. Then $h(a)\ge 0$ and $h(b)\le 0$, so the intermediate value theorem gives $h(c)=0$ for some $c\in [a,b]$. In higher dimensions, the Brouwer fixed-point theorem supplies a corresponding existence principle when this scalar sign argument is unavailable. Its practical value is ensuring that a proposed equation $x=F(x)$ has a solution once a suitable domain is found; it supplies no general promise of a fast algorithm.

A useful application is positive matrix eigenvector from simplex normalization: rescale the output of a strictly positive matrix to make its entries sum to one. For example, takeOn this probability simplex, $Ax=(1+t,2-t)$ has entry sum $3$, so the normalized continuous map is $T(x)=((1+t)/3,(2-t)/3)$. Both entries are positive and sum to one, keeping the image inside the probability simplex. Its fixed point equation is $t=(1+t)/3$, hence $t=1/2$. Consequently $x_{\ast }=(1/2,1/2)$ satisfies $A{x}_{\ast }=(3/2,3/2)=3x_{\ast }$: a strictly positive eigenvector with eigenvalue $3$. In larger dimensions, the Brouwer fixed-point theorem guarantees the corresponding positive eigenvector without requiring us to solve a system of linear equations in advance. The fixed point expresses preservation of relative weights, rather than preservation of their unnormalized size.

Proofs can be approached through the no-retraction theorem or Sperner's lemma. In the latter approach, increasingly fine triangulations locate approximate fixed points; compactness supplies a convergent subsequence, and continuous maps turn its limit into an exact fixed point. ^[ref]

A primary historical reference is L. E. J. Brouwer's 1911 paper Über Abbildung von Mannigfaltigkeiten, whose digitized original is hosted by Zenodo under DOI 10.1007/BF01456812. This gives a direct route to the early literature behind the Brouwer fixed-point theorem. ^[ref]

Every closed combinatorial surface has geometric realization homeomorphic to exactly one member of the following list: the sphere $S^2$; the connected sum of surfaces of $g\ge 1$ tori; or the connected sum of surfaces of $k\ge 1$ real projective planes. Every member admits a finite triangulation. The orientable family has Euler characteristic $2-2g$, including $g=0$ for the sphere; the nonorientable surface family has Euler characteristic $2-k$. Thus orientability and Euler characteristic determine the homeomorphism type. The first two nonorientable cases are the real projective plane and the Klein bottle. For a finite disconnected simplicial complex satisfying the same local conditions, the unordered list of its connected components' types determines its type.

For two connected closed surfaces, remove an open disc from each and identify the resulting boundary circles by a homeomorphism. The resulting closed surface is their connected sum, defined up to homeomorphism independently of these choices. A triangulation version removes one triangular face from each and glues their boundary three-cycles by a simplicial homeomorphism.

At a seam vertex the two vertex links of a simplex are paths; gluing their matching endpoints produces a cycle graph, proving the local surface condition. Removing a triangular face subtracts one from each Euler characteristic, while boundary gluing reduces both the vertex count and the edge count by three. Therefore $\chi (M\#N)=\chi (M)+\chi (N)-2$. If the result were orientable, either punctured summand would inherit consistent choices of orientation of a simplex, and these would extend over its capping face. Thus a nonorientable summand forces a nonorientable surface result. The classification theorem for closed combinatorial surfaces then gives $T^2\#{\mathbb{RP}}^2\cong {\mathbb{RP}}^2\#{\mathbb{RP}}^2\#{\mathbb{RP}}^2$.

Relative to a triangulation, a normal surface meets every tetrahedron in disjoint normal disks: four types of triangles cutting off vertices and three types of quadrilaterals separating opposite pairs of edges. A normal disk meets any edge at most once; its face arcs join distinct edges. Two different quadrilateral types cannot coexist in a tetrahedron.

The weight of a normal surface counts its intersections with the edges of the triangulation. It is a positive linear function of its coordinates: each corner of a normal disk contributes the reciprocal of the number of tetrahedral sectors around its edge. Thus a nontrivial decomposition into compatible coordinate vectors gives both summands strictly smaller weight.

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.

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.

Take the double cover of $S^2$ branched at $2g+2$ points, with deck involution $\iota$. One explicit construction cuts the sphere along $g+1$ disjoint arcs pairing the branch points and cross-glues two copies along the cuts. Its Euler characteristic is $2\chi (S^2)-(2g+2)=2-2g$, so it is an oriented closed surface of genus $g$. Locally at a branch point the projection is $z\mapsto z^2$ and the involution is $z\mapsto -z$; thus the involution is nonidentity and orientation-preserving, also when $g=0$.

The invariant rational cohomology of a finite surface action identifies with the cohomology of its quotient. For this action, triangulate the quotient with the branch points as vertices and lift the triangulation. Quotient cochains identify with invariant cochains, and averaging over the two deck transformations makes taking invariants exact over $\mathbb{Q}$. Hence the invariant part of $H^1(S;\mathbb{Q})$ is $H^1(S^2;\mathbb{Q})=0$. Since ${\iota }^2=1$, its eigenvalues are $\pm 1$ and it is therefore $-I$ on the $2g$-dimensional $H^1$. Its actions on $H^0$ and $H^2$ are both the identity. ThusHomotopy invariance and the result just proved show that every map homotopic to this involution has fixed points, for every genus.

Let $\pi :E\to B$ be an oriented real vector bundle of rank $r$ over a paracompact base. A Thom class is a class $U\in H^r(E,E\setminus B;\mathbb{Z})$ restricting on every oriented fiber pair $({\mathbb{R}}^r,{\mathbb{R}}^r\setminus 0)$ to its positive generator. The Thom isomorphism theorem states that this class exists uniquely with those fiber restrictions and thatHere $B$ is identified with the zero section. Equivalently use the disk/sphere bundle pair. The compact-support version is $H_c^q(B;\mathbb{Z})\cong H_c^{q+r}(E;\mathbb{Z})$, with fiberwise compact support in the Thom construction. If the bundle is not oriented, the integral assertion requires the corresponding orientation local system.

For a closed co-oriented embedded submanifold $Y$ of codimension $r$, co-orientation means an orientation of its normal bundle. A tubular neighborhood identifies a neighborhood of $Y$ with that bundle. Its Thom class, transferred by excision, is a class in $H^r(M,M\setminus Y;\mathbb{Z})$. For compact $Y$ this maps into the direct limit defining compactly supported cohomology, giving $\boxed{{\epsilon }_Y\in H_c^r(M;\mathbb{Z})}$. The construction is independent of the tubular neighborhood and uses the chosen co-orientation. In an oriented ambient manifold it is the Poincare dual of the oriented submanifold. Here “closed submanifold” has the usual compact-without-boundary meaning; if it meant merely a closed subset and $Y$ were noncompact, the supported relative class would still exist but need not have compact support.

We next prove the fixed-point implication without imposing an unprinted orientability assumption on $M$. SetUse a finite triangulation of the closed smooth manifold. The elementary Hopf trace identity says that the alternating trace of a chain endomorphism equals its alternating trace on homology. To verify it, split each finite-dimensional chain space as boundaries, a complement representing homology, and a complement mapping isomorphically onto the preceding boundaries. The traces on these two boundary summands cancel in adjacent degrees, leaving exactly the homology traces. The cohomology traces equal the homology traces because the induced maps are dual over $\mathbb{Q}$.

If $f$ has no fixed points, compactness gives $\delta ={\min }_{x}d(x,f(x))>0$. Choose a triangulation $K$ of mesh less than $\delta /4$ and a sufficiently fine subdivision $K^{\prime }$ admitting a simplicial approximation $F:K^{\prime }\to K$ uniformly within $\delta /4$ of $f$. The subdivision chain map $s:C_{\ast }(K)\to C_{\ast }(K^{\prime })$ sends each simplex to simplices inside it. For a vertex $v$ of such a small simplex, $d(v,F(v))>3\delta /4$, so $F(v)$ cannot lie in the original coarse simplex. Consequently $F_{\#}s$ has zero diagonal coefficient on every coarse simplex. Its trace vanishes in every degree, while it induces $f_{\ast }$ on homology. The trace identity yields $L(f)=0$. Taking the contrapositive provesThis is the Lefschetz fixed-point theorem, established here rather than invoked. For oriented $M$ the same identity is the graph-diagonal formula for the Lefschetz number: $L(f)=⟨(1,f{)}^{\ast }{\epsilon }_{\Delta },[M]⟩$, which vanishes when the graph misses the diagonal. Finally, homotopic maps induce the same cohomology maps, so their Lefschetz numbers agree.

Use integral homology groups. The sphere is $A=S^3$; each disk meets $A$ in its boundary circle, while $B\cap C=\{0\}$. In particular, the two disks also meet away from the sphere. These spaces have compatible triangulations, so the Mayer–Vietoris sequence applies to their unions as subcomplexes; equivalently use small regular open neighborhoods with the same homotopy types.

First let $Y=A\cup B$. Its intersection is $A\cap B=S^1$, and $B$ is contractible. The Mayer–Vietoris sequence givesFor example, the degree-two portion is $0\to H_2(Y)\to H_1(S^1)\to 0$, and the map from $H_0(S^1)$ into $H_0(A)\oplus H_0(B)$ is injective.

Now $Y\cap C=\partial C\bigsqcup \{0\}\simeq S^1\bigsqcup \{\text{point}\}$. It has $H_1=\mathbb{Z}$ and $H_0={\mathbb{Z}}^2$. Exactness gives $H_3(X)=H_3(Y)=\mathbb{Z}$ andThis sequence splits: lift a generator of the quotient $\mathbb{Z}$ and extend its lift by integer multiples. Hence $H_2(X)={\mathbb{Z}}^2$. In degree one the sequence becomesThe middle map has kernel generated by $(1,-1)$ and cokernel $\mathbb{Z}$. There are no higher groups, either by the same sequence or by the three-dimensional triangulation. Thus the homology of two disks meeting inside a sphere givesThe first-homology generator records a path from the sphere to the common interior point through one disk and back through the other; omitting that interior intersection would incorrectly remove this generator.

For an open cover $X=A\cup B$, the Mayer–Vietoris theorem gives the long exact sequenceThe same sequence holds for a union of simplicial subcomplexes. All homology groups here have integer coefficients.

An explicit triangulation of $X_q$ is as follows. Take vertices $b_0,b_1,b_2$ forming a target circle, a circle of $3q$ vertices $a_0,\dots ,a_{3q-1}$, and a cone vertex $c$. With subscripts on $a$ interpreted modulo $3q$ and on $b$ modulo three, include trianglesfor every $j$, together with all their faces. The first two types triangulate the mapping cylinder of the degree-$q$ covering $a_j\mapsto b_j$; the last type fills the inner circle by a disc. A filled disc with an attached annular collar is still a disc, and identifying the outer collar boundary by this covering gives exactly the source quotient. Distinct inner vertices ensure this is a genuine simplicial complex, not just a one-cell presentation.

Let $A$ be the mapping cylinder and $B$ the coned disc. Then $A$ retracts to the target circle, $B$ is contractible and $A\cap B$ is the inner circle. Its inclusion into $A$ induces multiplication by $q$ on $H_1$. The exact sequence therefore reduces toConnectedness and the absence of higher-dimensional simplices giveThis is a two-dimensional Moore space with torsion in degree one.

The suspension of a connected complex is the union of two cones whose intersection retracts to the original complex. Reduced Mayer-Vietoris givesIn particular its $H_1$ vanishes. For the wedge of two connected polyhedra, choose neighborhoods retracting onto the two summands with contractible intersection at the wedge point. ThenDecompose each prescribed $G_i$ into cyclic summands. Realize each infinite cyclic summand in degree $i$ by $S^i$, and each finite nontrivial summand $\mathbb{Z}/q$ by $S^{i-1}{X}_q$, meaning $i-1$ successive suspensions. Take their finite wedge, or a point if there are no summands. The suspension formula shifts the torsion to degree $i$ and creates no other positive-degree homology, while the wedge formula adds summands. Thus the resulting connected polyhedron satisfies $\boxed{H_0(X)=\mathbb{Z},H_i(X)=G_i(1\le i\le n),H_i(X)=0(i>n)}$. The finite triangulations above, their simplicial suspensions and wedges provide an actual polyhedron.

A Heegaard splitting of a closed oriented three-manifold is a decomposition $Y=H_0{\cup }_{\Sigma }{H}_1$ into two handlebodies, with their common boundary the Heegaard surface $\Sigma$. With boundary present, the corresponding pieces are compression bodies, whose negative boundaries account for $\partial Y$.

For the given triangulation, take the barycentric subdivision. A regular neighbourhood $H_0$ of the original one-skeleton is a handlebody: thicken vertices to balls and edges to one-handles, then contract a spanning tree. The complementary region is a regular neighbourhood $H_1$ of the dual one-skeleton, whose vertices are tetrahedron centres and whose edges cross triangular faces. It too is a handlebody. Both graphs are connected, and their common boundary supplies the Heegaard splitting. If the original triangulation has $v$ vertices and $e$ edges, its Heegaard surface has genus $e-v+1$; the dual count gives the same number by $v-e+f-t=0$.

Choose an oriented meridian of a knot $\mu$ and the Seifert longitude $\lambda$ supplied by a Seifert surface for the null-homologous knot. For relatively prime integers $p,q$, rational Dehn surgery removes the interior of a tubular neighborhood and attaches a solid torus with its meridian of a solid torus on the unoriented slope $p\mu +q\lambda$. The choices $(p,q)$ and $(-p,-q)$ describe the same slope; $q=0$ is the original meridional filling. The boundary gluing reverses boundary orientation so that the oriented three-manifold extends across the filling.

For integral coefficients $n_i$, attach two-handles to $B^4$ along the components of the framed link, with their indicated Seifert framing shifts. The boundary operation removes $S^1\times D^2$ and inserts $D^2\times S^1$, with meridian $n_i{\mu }_i+{\lambda }_i$. Thus the compact oriented surgery trace satisfiesFor a finite rational coefficient, use a negative continued fractionReplace that component by an integrally framed chain of successive meridians with coefficients $a_0,\dots ,a_k$. Repeated slam-dunk moves give back $p/q$. Perform this replacement for every rationally framed component, leaving meridional fillings out. The resulting integral framed link has the same filled boundary, so its surgery trace proves that every such rational filling bounds a compact oriented four-manifold.

Every complete polygon edge pairing gives a compact connected surface with empty manifold boundary. Here is the local reason at the vertices. Replace each polygon corner by a small link interval. The polygon edge pairings identify all endpoints of these link intervals in pairs. In a given vertex equivalence class, the resulting graph is connected: the endpoint identifications generating that vertex equivalence class also connect its corner intervals. Every link graph vertex has degree of a vertex two, so the graph is a circle. Its topological cone is a disc, giving the required local surface model. Edge and face CW cell interiors have the usual disc neighbourhoods as well. A finite triangulation compatible with the polygon edge pairings realizes the result as a finite simplicial complex, giving a Hausdorff, second-countable space.

For every such surface, $\chi =V-2\ge -1$. If it is orientable, the classification of compact surfaces gives $\chi =2-2g$, so $g=0$ or $1$. If it is a nonorientable surface, write $k\ge 1$ for its nonorientable genus; then $\chi =2-k\ge -1$ gives $k\le 3$. The exhaustive list is therefore

To prove that every listed possibility actually occurs, the following six-letter boundary words specify complete pairings of a hexagon:The vertex counts follow by matching starts and ends of equally directed arrows, just as in part (b). The first and third words pair every letter with its inverse, giving orientable surfaces. Each other word contains a letter twice with the same sign, giving a nonorientable surface: an orientation on the face would induce equal, rather than opposite, boundary directions along that glued pair. With the displayed Euler characteristics, the classification of compact surfaces identifies the examples exactly as listed. All five types occur, and no others occur.

The classification theorem for closed combinatorial surfaces states that the geometric realization of every closed combinatorial surface is homeomorphic to exactly one of the following:

With ${\Sigma }_0=S^2$, the complete distinguishing data are orientability and the Euler characteristic:Here $\chi (K)=V-E+F$, where $V,E,F$ count the vertices of a simplicial complex, edges, and triangles of a finite triangulation. Thus two such connected closed surfaces are homeomorphic if and only if they have the same orientability and Euler characteristic. In particular $N_1$ is the real projective plane and $N_2$ is the Klein bottle. Every listed type admits a finite triangulation. For disconnected finite closed combinatorial surfaces, apply the theorem to each of the finitely many connected components; the resulting unordered collection of types determines the homeomorphism type.

Use compatible edge identifications: choose a bijection between the three boundary vertices of $\tau$ and those of $\sigma$, and extend it linearly on each edge. This defines a simplicial homeomorphism of the two boundary circles. This compatibility is the usual meaning of gluing the two boundary triangles; independently chosen edge reversals need not agree at their endpoints. The classification needed here is 3(a)(iv); the printed reference to 3(a)(iii) is a source error.

Delete the two triangular simplexes but retain all their proper faces of a simplex, and identify the two resulting boundary simplicial subcomplexes by this map. The result is a finite abstract simplicial complex $L$: vertices from different sides are identified only in the boundary, and the only shared edges are boundary edges. There is no triangle consisting of all three boundary vertices after the two faces have been removed. Consequently no other simplexes become duplicates. Its geometric realization has the specified quotient topology and is homeomorphic to $Z$.

We check the closed combinatorial surface conditions. Away from the seam, vertex links of a simplex remain cycle graphs. At a boundary vertex of $X$ or $Y$, removing the face changes its link from a cycle graph to a graph path, with endpoints corresponding to its two boundary edges. Gluing the boundaries identifies the two paths' respective endpoints, making one cycle graph. Each seam edge now meets one triangle from $X$ and one from $Y$, and all other edges retain their original two incident triangles. Thus every vertex link is a cycle graph and every simplex lies in a triangle. The simplicial complex is connected, since the two connected punctured surfaces meet along their boundary circle. Hence $L$ is a closed combinatorial surface.

This construction is the connected sum of surfaces $T^2\#{\mathbb{RP}}^2$. For completeness, compute its Euler characteristic directly. Removing a triangular face while retaining its edges and vertices subtracts one, soGluing identifies three vertex pairs and three edge pairs, givingFor the explicit triangulations above, this is also $62-189+126=-1$.

Finally $Z$ is a nonorientable surface. If $Z$ were orientable, its restriction would make $Y$ orientable. The consistent choices of orientation of a simplex on its triangles extend across a capping disc: choose the cap's orientation of a simplex so that the induced directions on the common boundary oppose. Capping $Y$ by $\sigma$ would therefore make the real projective plane orientable, a contradiction. The obstruction to orientability in the real projective plane can also be seen from a Möbius band neighbourhood of a projective line.

By the classification theorem for closed combinatorial surfaces, a nonorientable surface with Euler characteristic $-1$ has $2-k=-1$, hence $k=3$. Therefore

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

We construct both the required triangulation and its polygonal schema. Subdivide the square $Q=[0,1{]}^2$ into a $4\times 4$ grid. Insert the centre of each small square and join it by line segments to its four corners, making four triangles per small square. Identify opposite sides of $Q$ by translations: $(0,t)\sim (1,t)$ and $(t,0)\sim (t,1)$. This is the standard quotient space presentation of the torus.

The grid and centre vertices, with these identifications, define an abstract simplicial complex $K_1$. Every triangle contains its own small-square centre and two consecutive grid vertices, so no triangle has identified vertices and no distinct triangles acquire the same vertex set. Only intended common sides or vertices are shared. Thus this is a genuine finite triangulation, whose geometric realization is homeomorphic to $T^2$. There are $32$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic zero.

Choose the simplex $\tau$ represented in the interior of $Q$ by the triangle with verticesRemove its relative interior. Before identifying the square's sides, the remainder is a closed disc with a triangular hole, hence an annulus. Cut this annulus along the line segment from $(0,0)$ to $(1/4,1/4)$, treating the two banks as distinct boundary arcs. The cut space is a closed disc; traversing its boundary follows the four outer edges, one bank, the three inner edges, and the other bank. It therefore has nine marked boundary sides.

A homeomorphism from this cut disc to a convex nine-sided polygon can be chosen to carry its nine marked sides to the nine polygon sides. One justification is to choose a boundary homeomorphism with the prescribed marked vertices and extend it over the disc by radial extension after disc parametrizations. Transport the identifications to this convex polygon. With labels recording traversal direction, its boundary word isIdentify equally labelled sides in their arrow directions: the two $a$ sides and two $b$ sides restore the torus pairings, while the two $c$ sides undo the cut. Leave $d,e,f$ unpaired. Their union is the triangular boundary of the removed simplex. The resulting quotient space is exactlyThis is the polygonal presentation of a punctured surface. The word supplies all nine sides and every gluing; inverse exponents reverse the boundary traversal direction.

A triangulation of $S^k$ is regular in this antipodal-combinatorics convention when the antipodal map preserves its simplices and its nested coordinate equators $S^0\subset S^1\subset \cdots \subset S^{k-1}$ are subcomplexes. The two closed hemispheres at each stage are then triangulated balls. Barycentric subdivision preserves these conditions while making the mesh of a simplicial complex arbitrarily small. This convention is distinct from other meanings of regular triangulation in polyhedral geometry. Cambridge extremal-combinatorics notes use this convention.