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The convex hull of the positive and negative coordinate unit vectors. Its boundary triangulates a sphere, and each face contains at most one member of each antipodal pair of coordinate unit vectors. This makes its barycentric subdivision useful for triangulating Real projective space.

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.

The standard simplicial decomposition of a sphere $F^n$ is the boundary of the cross-polytope with vertices $\pm e_1,\dots ,\pm e_{n+1}$, identified radially with $S^n$. A face chooses distinct coordinate indices and one sign for each; it never contains an antipodal pair.

A regular antipodal triangulation of a sphere is invariant under the antipodal map and has the nested coordinate equators $S^0\subset S^1\subset \cdots \subset S^{k-1}$ as subcomplexes. In particular each equator separates its sphere into two triangulated hemispheres. These conditions are preserved by barycentric subdivision.

For a simplicial map $f:F\to F^n$, a positive alternating simplex of dimension $j$ has distinct image labelsThe opposite alternating pattern is negative; all other simplices, including collapsed images, are neutral.

To prove the antipodal alternating-simplex parity lemma, let $p_j$ count positive $j$-simplices in the equatorial $S^j$. There is exactly one positive vertex in $S^0$, so $p_0=1$. A $j$-simplex has an odd number of positive alternating facets precisely when it is positive or negative alternating. Indeed an alternating sequence has exactly one positive alternating deletion; a nonalternating sequence has either zero or two. For a collapsed image, a positive facet can occur only when exactly one label is repeated; deleting either copy gives two such facets.

Count incidences of positive $(j-1)$-faces with $j$-simplices in one closed hemisphere. Interior faces contribute twice and equatorial faces once. The count modulo two is therefore $p_{j-1}$. It is also the number of positive or negative $j$-simplices in that hemisphere modulo two. Because $f$ is antipodal, each alternating pair $\sigma ,-\sigma$ has one positive and one negative member, with one member in each hemisphere. ThusInduction gives odd $p_k$, proving the existence of a positive $k$-simplex. In particular, an antipodal simplicial map $F\to F^{k-1}$ is impossible: a positive $k$-simplex would require $k+1$ distinct coordinate labels but only $k$ are available.

The Borsuk-Ulam theorem states that every continuous map $g:S^k\to {\mathbb{R}}^k$ has some $x$ with $g(x)=g(-x)$. First rule out a continuous antipodal map $h:S^k\to S^{k-1}$ for $k\ge 1$. By uniform continuity, take a sufficiently fine regular barycentric subdivision so that endpoints of each edge have images less than $2/\sqrt{k}$ apart. Label a vertex by $\mathrm{sgn}(h_i)e_i$, where $i$ is the smallest index maximizing $|h_i|$. Since $h$ is unit length, this maximum is at least $1/\sqrt{k}$; two adjacent vertices cannot have opposite labels. The labels therefore define an antipodal simplicial map to $F^{k-1}$, contradicting the preceding result.

If the stated Borsuk-Ulam theorem failed, the mapwould be just such a continuous antipodal map. This proves the required version; $k=0$ is immediate.

For the Kneser conjecture, equivalently Lovász theorem on Kneser graphs, the claim isThe Kneser graph joins disjoint $r$-subsets of $[N]$. Let $d=N-2r+1$. Colour each $r$-set by its least element in $[d]$ when it has one, and otherwise by one final colour. A colour class indexed by $i$ is an intersecting family because all its members contain $i$; the final class lies in a $(2r-1)$-set and is also intersecting. This proves the upper bound $d+1$.

For the lower bound, first deduce the open-cover Lusternik-Schnirelmann-Borsuk theorem. An open cover $U_0,\dots ,U_d$ of $S^d$ admits a continuous partition of unity ${\varphi }_0,\dots ,{\varphi }_d$ subordinate to it. Apply the Borsuk-Ulam theorem to $({\varphi }_1,\dots ,{\varphi }_d)$. Equal values at $x,-x$ and ${\sum }_i{\varphi }_i=1$ make all values equal. At least one common value is positive, so some $U_i$ contains an antipodal pair.

Now suppose the $r$-sets could be coloured with only $d$ colours, each an intersecting family. Choose $N$ points $v_1,\dots ,v_N\in S^d$ such that every $d+1$ of their vectors are linearly independent. For example, normalize $(1,t_i,\dots ,t_i^d)$ for distinct real $t_i$; the Vandermonde determinant proves this property. The general-position hemisphere count says that at most $d$ points lie on any equator. Since $N-d=2r-1$, at least one of the opposite open hemispheres determined by $x,-x$ contains an $r$-set of the labelled points.

Let $U_i$ be the open set of directions whose positive open hemisphere contains an $r$-set of colour $i$. It contains no antipodal pair, as such a pair would give two disjoint sets of that colour. The compact set $K=S^d\setminus {\bigcup }_{i=1}^d{U}_i$ also contains no antipodal pair: otherwise each of the two opposite hemispheres would have at most $r-1$ points, contradicting the count $2r-1$.

If $K$ is nonempty, compactness gives a positive set distance between $K$ and $-K$; an open neighbourhood $U_0$ of radius less than half that set distance still has no antipodal pair. If $K$ is empty, take $U_0=\varnothing$. In either case $U_0,U_1,\dots ,U_d$ is an open cover with no member containing an antipodal pair, contradicting the Lusternik-Schnirelmann-Borsuk theorem. Hence at least $d+1$ colours are required.

The Simplicial approximation theorem states that a continuous map between realizations of finite simplicial complexes becomes homotopic to a simplicial map after sufficiently many barycentric subdivisions of the domain. More precisely, a vertex assignment is a simplicial approximation when the image of each open star is contained in the open star of its assigned vertex; sufficiently fine subdivision permits such an assignment. The resulting straight-line homotopy stays in the appropriate target simplices.

The vertices of a barycentric subdivision are the barycentres of nonempty faces. An $n$-simplex has $2^{n+1}-1$ such faces. A top-dimensional subdivided simplex corresponds to a full flag of faces of sizes $1,2,\dots ,n+1$, equivalently to an ordering of the original vertices. ThusTriangulate both copies of the sphere using finite complexes. For each fixed subdivision level there are only finitely many vertex maps to the finite target vertex set, hence only finitely many simplicial maps. Every continuous map is homotopic to one of them at some finite level. A countable union of finite sets is countable, proving at most countably many homotopy classes without needing their classification by degree.

The Excision theorem for singular homology states that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq {\mathrm{int}}_{X}A$, inclusion induces isomorphisms $H_q(X\setminus Z,A\setminus Z;\mathbb{Z})\cong H_q(X,A;\mathbb{Z})$ for all $q$. For an open cover $X=U\cup V$, its small-chain form says that the subcomplex of singular chains whose individual simplices lie in $U$ or $V$ computes $H_{\ast }(X)$. Concretely, repeated barycentric subdivision puts each finite chain in that subcomplex, and subdivision is chain-homotopic to the identity. There is then a short exact sequence of chain complexesIts long exact sequence in homology, with excision identifying the last complex's homology with $H_{\ast }(X)$, is the Mayer–Vietoris sequence:

The Real projective space ${\mathbb{RP}}^1$ is a circle. The projective plane has one cell in each dimension $0,1,2$, with cellular homology boundary maps $d_2=2$ and $d_1=0$. Thus, writing $A_q=H_q({\mathbb{RP}}^2;\mathbb{Z})$, we have $A_0=\mathbb{Z}$, $A_1=\mathbb{Z}/2$ and $A_q=0$ for $q\ge 2$. Cover the circle by two open arcs with a two-component intersection and multiply this cover by the projective plane. In the Mayer–Vietoris sequence the overlap-to-cover map isIts kernel and cokernel are each $A_q$. Consequently $0\to A_q\to H_q(X)\to A_{q-1}\to 0$ for $q\ge 1$. In degree one the quotient is free, so this sequence splits; in degree two it identifies the group directly with $A_1$. The integral homology of a circle times a real projective plane is therefore

Without a compactness restriction, the answer to the final existence question is yes: there is a noncompact oriented four-manifold. Explicitly, takeThe involution is free and reverses orientation on both $S^2$ and $\mathbb{R}$, hence preserves their product orientation. Its quotient is consequently an oriented three-manifold, the orientable total space of the orientation line bundle of the projective plane. The contraction $[v,t]\mapsto [v,(1-s)t]$ is a deformation retraction onto ${\mathbb{RP}}^2$, so $Y\simeq X$.

If a closed oriented manifold is intended, the answer instead is no. A closed connected oriented $d$-manifold has $H_d\cong \mathbb{Z}$ and no homology above $d$. Here the highest nonzero positive-dimensional homology is the torsion group $H_2=\mathbb{Z}/2$, so no dimension can supply that required top fundamental class. The distinction is necessary because the printed wording does not require the oriented manifold to be closed or three-dimensional.

Let $F:|K|\times [0,1]\to |L|$ be a homotopy from $f$ to $g$. The inverse images $F^{-1}({\mathrm{st}}_L(w))$ form a finite open cover of the compact product. By the Lebesgue number lemma there is $\delta >0$ such that every subset of diameter less than $\delta$ lies in one of these inverse images.

Choose a barycentric subdivision $K^{(n)}$ with all its open stars of diameter less than $\delta /3$, and times $0=t_0<\cdots <t_m=1$ with gaps less than $\delta /6$. For each vertex $v$ and each time $t_i$, the set(with the evident endpoint conventions) has diameter less than $\delta$ in the product metric. Choose a vertex $w_i(v)$ of $L$ whose open star contains its image under $F$, and set $h_i(v)=w_i(v)$.

At time $t_i$, the star condition makes $h_i$ a simplicial approximation to $F(\cdot ,t_i)$. Adjacent intervals $J_i,J_{i+1}$ overlap, so choose a time in their overlap and a point in the interior of any simplex $\sigma$ of $K^{(n)}$. Its image belongs to the stars of every $h_i(v)$ and $h_{i+1}(v)$ for $v\in \sigma$. As in (i), their vertices lie in one carrier simplex, proving contiguity. Thus the sequence starts with an approximation to $f$, ends with one to $g$, and has contiguous adjacent pairs. Relabeling its indices gives the requested $h_1,\dots ,h_m$.

Here ${\sim }_c$ denotes contiguity, with the endpoint maps approximating $f,g$ respectively.

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.

A simplicial approximation to $f:|K|\to |L|$ is a simplicial map $g:K\to L$ satisfying the open-star conditionfor every vertex $v$. Equivalently, whenever $f(x)$ lies in the interior of a simplex of $L$, $|g|(x)$ lies in that simplex. In particular $f$ and $|g|$ are joined by the simplexwise straight-line homotopy.

The Simplicial approximation theorem says that, for finite $K$, every continuous $f$ has such an approximation after sufficiently many barycentric subdivisions of $K$. For general complexes one uses a subdivision adapted to the open-star cover; a single uniform number of barycentric subdivisions is the compact-domain version.

Given triangulations $X\cong |K|$, $Y\cong |L|$, choose a simplicial approximation $g$ on a suitable subdivision $K^{\prime }$. Its chain map sends an oriented simplex $[v_0,\dots ,v_r]$ to $[g(v_0),\dots ,g(v_r)]$, with repeated-vertex simplices sent to zero. It commutes with the boundary operator and induces $g_{\ast }$. Compose this with the subdivision isomorphism $H_r(K)\to H_r(K^{\prime })$ and the triangulation identifications to define $f_{\ast }:H_r(X)\to H_r(Y)$.

For $z\mapsto z^n$, a positively oriented fundamental loop winds $n$ times around the target circle. With integer coefficients,All higher homology groups vanish. The $H_1$ result can also be read from a subdivision into $n$ copies of a target fundamental cycle.

Triangulate the sphere by the boundary complex of the cross-polytope with simplicial vertices $\pm e_0,\dots ,\pm e_n$: a face contains at most one simplicial vertex from each antipodal pair. Take its barycentric subdivision. Its simplicial vertices are the nonempty faces $F$ of the original complex, and its simplices are strictly nested chains $F_0\subset \cdots \subset F_k$.

Form a quotient complex whose simplicial vertices are pairs $\{F,-F\}$, and whose simplices are the images of these chains. No chain contains both $F$ and $-F$. Moreover a chain of face-orbits lifts uniquely up to simultaneous negation: after choosing a largest face, every smaller face has at most one representative contained in it, because an original face contains no antipodal simplicial vertices. Thus the quotient is an honest simplicial complex, without two different simplices with the same simplicial vertex set. Its realization is homeomorphic to the sphere modulo its antipodal action. This provides a triangulation of Real projective space ${\mathbb{RP}}^n$. The subdivision avoids the duplicate-simplex problem of naively identifying simplicial vertices in the unsubdivided cross-polytope boundary.

For every nonempty simplex $\sigma$ of $K$, let $b_{\sigma }$ be its barycentre. The first barycentric subdivision $K^{\prime }$ has the $b_{\sigma }$ as vertices, andis a simplex exactly when, after reordering, ${\sigma }_0⊊\cdots ⊊{\sigma }_q$. Its realization is the same polyhedron as that of $K$. The iterated barycentric subdivision is defined by

Suppose $X=A\cup B$ and that the interiors of $A$ and $B$ cover $X$. The Mayer–Vietoris theorem gives the long exact sequence in homologywhere all four displayed maps except the connecting homomorphism $\partial$ are induced by the relevant inclusion maps.

To prove it, let $C_{\ast }^{A,B}(X)$ be the chain complex generated by those singular simplices whose images lie wholly in $A$ or wholly in $B$. The sequenceis a short exact sequence of chain complexes. Repeated barycentric subdivision makes every singular simplex small enough to lie in $A$ or $B$, and the subdivision operator is chain homotopic to the identity. The inclusion $C_{\ast }^{A,B}(X)\hookrightarrow C_{\ast }(X)$ therefore induces an isomorphism on every homology group. The long exact homology sequence of a short exact sequence of chain complexes now gives the displayed sequence. Explicitly, if an $n$-cycle $z=a+b$ with $a\in C_n(A)$ and $b\in C_n(B)$, then $\partial a=-\partial b$ lies in $C_{n-1}(A\cap B)$ and represents the connecting class $\partial [z]$.

The small simplex theorem says that the inclusion of the subcomplex generated by singular simplices whose images lie wholly in $U$ or wholly in $V$,is a chain-homotopy equivalence. Repeated barycentric subdivision supplies the inverse up to chain homotopy.

The simplices in a barycentric subdivision are strict chains of nonempty faces. A tetrahedron has $15$ nonempty faces, so $f_0=15$. Counting strict two-face chains givesA strict three-face chain amounts to choosing a terminal face and partitioning its vertices into three nonempty ordered blocks. Hencewhere $S(-,-)$ denotes a Stirling number of the second kind. ThereforeEquivalently, the full subdivision has $4!=24$ tetrahedra and Euler characteristic one, so deleting its three-dimensional simplices from the alternating count gives $1+24=25$. This is the f-vector computation for the barycentric subdivision of a tetrahedron.

For an open cover $X=U\cup V$, the Mayer–Vietoris theorem gives the long exact sequenceIt ends in $H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to H_0(X)\to 0$. To prove it, let $C_{\ast }^{U,V}(X)$ be the subcomplex generated by singular simplices lying wholly in $U$ or wholly in $V$. There is a short exact sequence of chain complexesExactness follows from the singular-simplex bases: the intersection of the two chain subgroups consists exactly of simplices lying in both sets. The small-chain locality theorem for singular homology says the inclusion of this small-chain complex into $C_{\ast }(X)$ induces an isomorphism on homology; barycentric subdivision is its standard justification. The long exact sequence of a short exact sequence of chain complexes now proves the theorem. For a homological cycle represented as $a+b$, the connecting map is $[a+b]\mapsto [\partial a]$, since $\partial a=-\partial b$ lies in the intersection. This also identifies the displayed maps and their signs.

For the glued solid tori, use open collar enlargements of the two halves. Each retracts to a circle; the overlap retracts to their boundary torus. Thus the halves have $H_0=H_1=\mathbb{Z}$ and higher homology zero, while the overlap has $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^2$. In the overlap choose longitude and meridian generators. The meridian bounds a disk in each solid torus and the longitude generates each first homology, so the map on $H_1$ isIts kernel is the meridian copy of $\mathbb{Z}$ and its cokernel is $\mathbb{Z}$. The map on $H_0$ is the injective map $z\mapsto (z,-z)$. Exactness therefore gives $H_2(X)=\mathbb{Z}$ and $H_1(X)=\mathbb{Z}$. The preceding segment is $0\to H_3(X)\to H_2(U\cap V)\to 0$, giving $H_3(X)=\mathbb{Z}$. Connectedness gives $H_0(X)=\mathbb{Z}$, and all higher groups vanish. HenceAs a geometric check, the identity boundary gluing identifies the product with $S^1\times (D^2{\cup }_{S^1}{D}^2)=S^1\times S^2$.

Use a convex representative of the marked polygonal schema, retaining the side-pairing parametrizations. This does not change the quotient up to homeomorphism: a matching boundary homeomorphism extends over the two polygonal discs by radial extension in disc coordinates. Join an interior point to every vertex, producing finitely many triangles. Orient radial edges away from the interior point and choose boundary-edge orientations consistently with the side identifications. Each triangle then has a local vertex ordering with the centre first; the other two vertices are ordered along its boundary-edge arrow. These orderings agree on shared edges, so the quotient is a finite delta-complex $X$ whose underlying space is the prescribed polygon quotient.

This need not yet be an abstract simplicial complex: two vertices of a triangle may become identified, or different triangles may have the same vertex set. Apply barycentric subdivision twice.

In the first subdivision, every small simplex corresponds to a flag of faces of strictly increasing dimensions. Identifications preserve dimension, so its vertices are distinct. In the second subdivision, a vertex records an actual cell of the first subdivision. A set of such vertices determines their order by dimension. Inside any one first-subdivision simplex, distinct vertices make each face inclusion uniquely determined by its vertex subset; hence a flag has at most one realization. Therefore different second-subdivision simplexes cannot share the same vertex set, and they meet in precisely their common faces. Thus the second subdivision is an abstract simplicial complex $L$.

The affine subdivision maps are compatible with every face identification, so their realizations are homeomorphisms. Consequentlywhere $L$ is finite. This proves triangulation of a polygon with paired sides by the second barycentric subdivision of a delta-complex construction.^[ref]

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.

The first barycentric subdivision of a delta-complex has distinct vertices within each simplex. Its vertices correspond to an increasing chain of original faces, and their strictly different dimensions prevent identifications within that chain. Distinct simplexes may still share the same global vertex set, as happens to the two halves of a subdivided loop edge.

For a delta-complex with distinct vertices within each simplex, a face inside a fixed simplex is uniquely determined by its vertex subset. In its barycentric subdivision, a vertex records an actual face, including its cell identity. A simplex records a chain of these faces; its vertex set recovers the unique order by dimension and each unique face inclusion. Thus no two simplexes have the same vertex set, and intersections are their common faces. Applying this to the first subdivision proves that the second subdivision is an abstract simplicial complex. The subdivision maps are affine on smaller simplexes and compatible on the glued faces, so they induce a homeomorphism of realizations.^[ref]

For an open cover $\mathcal{U}$ of a space $X$, let $C_{\ast }^{\mathcal{U}}(X)$ be the subgroup generated by singular simplices whose images lie in a single member of the cover. It is a chain complex, and its inclusion into the full singular chain complex induces an isomorphism on homology.

For a finite chain, enough iterations of barycentric subdivision make every simplex small: pull the cover back to each compact standard simplex and use its Lebesgue number. The subdivision operator commutes with the boundary and is related to the identity by the standard subdivision chain homotopy, which keeps each produced simplex inside its original image. Therefore any cycle becomes homologous to a small cycle, proving surjectivity.

If a small cycle $z$ bounds a full chain $b$, choose $N$ so that ${\mathrm{Sd}}^{N}b$ is small. Write ${\mathrm{Sd}}^N-1=\partial Q+Q\partial$. Since the homotopy preserves simplex images, $Qz$ is also small. Then $z=\partial ({\mathrm{Sd}}^{N}b-Qz)$ is a small boundary, proving injectivity. This is the chain-level locality input in the proof of the Mayer–Vietoris sequence.

For an open cover, take the subcomplex generated by singular simplices whose images lie in single cover members. Repeated barycentric subdivision makes each finite singular chain small, and the standard subdivision chain homotopy shows that inclusion induces homology isomorphisms. For a two-set cover, the short exact sequence of these small chains gives the Mayer–Vietoris sequence; excising the complement of one open set from the pair with the other set gives its equivalent relative-homology formulation.

In the illustrated simplicial subdivision, inserting the three edge midpoints turns one filled triangle into four. The counts change from $(V,E,F)=(3,3,1)$ to $(6,9,4)$, but $6-9+4=1$. Splitting a single edge adds one vertex and one edge, so its net contribution is $1-1=0$. Adding a diagonal inside a polygonal face adds one edge and one face, giving $-1+1=0$.

These local accounts explain common refinements. For arbitrary finite simplicial subdivisions, including barycentric subdivision in any dimension, the underlying geometric realization is unchanged. The topological invariance of Euler characteristic therefore guarantees the same alternating count, even when the individual counts change substantially.