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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.

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.

An element of the underlying vertex set of an abstract simplicial complex; a simplex is a finite subset of these vertices. In a barycentric subdivision, the vertices are nonempty simplices of the original complex.