Simplicial Complexes and Simplicial Homology

Subtract the two expressions for $x$ to obtain ${\sum }_{i=0}^n({\lambda }_i-{\mu }_i)v_i=0$, and subtract the two sum conditions to obtain ${\sum }_{i=0}^n({\lambda }_i-{\mu }_i)=0$.

The previous step is an affine dependence relation among $v_0,\dots ,v_n$ whose coefficients sum to $0$. Since the vertices are affinely independent, every coefficient vanishes, so ${\lambda }_i-{\mu }_i=0$ for all $i$.

Let $x\in \mathrm{\mid }\sigma \mathrm{\mid }\cap \mathrm{\mid }\tau \mathrm{\mid }$. Viewed as a barycentric-coordinate function on the full vertex set of $K$, the support of $x$ is contained in $\sigma$ because $x\in \mathrm{\mid }\sigma \mathrm{\mid }$, and it is contained in $\tau$ because $x\in \mathrm{\mid }\tau \mathrm{\mid }$. Hence $\mathrm{supp}(x)\subseteq \sigma \cap \tau$, so $x\in \mathrm{\mid }\sigma \cap \tau \mathrm{\mid }$.

Conversely, if $x\in \mathrm{\mid }\sigma \cap \tau \mathrm{\mid }$, then $\mathrm{supp}(x)\subseteq \sigma \cap \tau$, so in particular $\mathrm{supp}(x)\subseteq \sigma$ and $\mathrm{supp}(x)\subseteq \tau$. Therefore $x\in \mathrm{\mid }\sigma \mathrm{\mid }\cap \mathrm{\mid }\tau \mathrm{\mid }$.

Steps 1.1 and 1.2 prove the two inclusions, so the two sets are equal.

If $K$ has no nonempty simplices, then $\mathrm{\mid }K\mathrm{\mid }=\mathrm{\varnothing }$, which is compact and Hausdorff. Otherwise $K$ has finitely many vertices; write them as $v_1,\dots ,v_N$. For each simplex $\sigma$ of $K$, the subset $\mathrm{\mid }\sigma \mathrm{\mid }\subseteq \mathrm{\mid }K\mathrm{\mid }$ identifies with a Euclidean simplex in $[0,1{]}^N$, cut out by finitely many linear equations and inequalities, so $\mathrm{\mid }\sigma \mathrm{\mid }$ is compact.

The realization $\mathrm{\mid }K\mathrm{\mid }$ is the union of the finitely many subsets $\mathrm{\mid }\sigma \mathrm{\mid }$ over the nonempty simplices $\sigma$ of $K$, and this union is empty in the case handled at the start of step 1.1. Therefore step 1.1 makes $\mathrm{\mid }K\mathrm{\mid }$ a finite union of compact sets and hence compact.

For each simplex $\sigma$, let $O_{\sigma }\subseteq [0,1{]}^N$ be open with $U\cap \mathrm{\mid }\sigma \mathrm{\mid }=O_{\sigma }\cap \mathrm{\mid }\sigma \mathrm{\mid }$. Since there are only finitely many simplices, a subset $U\subseteq \mathrm{\mid }K\mathrm{\mid }$ is weakly open exactly when $U=({\bigcup }_{\sigma }{O}_{\sigma })\cap \mathrm{\mid }K\mathrm{\mid }$. Thus the weak topology on $\mathrm{\mid }K\mathrm{\mid }$ agrees with the subspace topology from $[0,1{]}^N$. The cube $[0,1{]}^N$ is Hausdorff, so $\mathrm{\mid }K\mathrm{\mid }$ is Hausdorff as a subspace.

Steps 2.1 and 2.2 give compactness and Hausdorffness.

If $\sigma =\{v_0,\dots ,v_n\}$ is a simplex of $K$ and $x\in \mathrm{\mid }\sigma \mathrm{\mid }$ has barycentric coordinates $x={\sum }_{i=0}^n{\lambda }_i{v}_i$, then $\mathrm{\mid }f\mathrm{\mid }(x)={\sum }_{i=0}^n{\lambda }_{i}f(v_i)\mathrm{.}$ Thus the restriction $\mathrm{\mid }f\mathrm{\mid }{\mathrm{\mid }}_{\mathrm{\mid }\sigma \mathrm{\mid }}:\mathrm{\mid }\sigma \mathrm{\mid }\to \mathrm{\mid }f(\sigma )\mathrm{\mid }$ is the affine map determined by the vertex map $f$, so it is continuous.

For every barycentric function $\alpha$, the identity vertex map leaves every coefficient unchanged, so $\mathrm{\mid }{\mathrm{id}}_K\mathrm{\mid }(\alpha )=\alpha$. Likewise $\mathrm{\mid }g\circ f\mathrm{\mid }(\alpha )(u)={\sum }_{w\in g^{-1}(u)}{\sum }_{v\in f^{-1}(w)}\alpha (v)=\mathrm{\mid }g\mathrm{\mid }(\mathrm{\mid }f\mathrm{\mid }(\alpha ))(u)$, so realizations preserve composition.

If $\sigma$ and $\tau$ meet, then they meet along $\mathrm{\mid }\sigma \cap \tau \mathrm{\mid }$, and the affine formulas from step 1.1 agree there because they are both determined by the same vertex map $f$. Since $\mathrm{\mid }K\mathrm{\mid }$ carries the weak topology with respect to its simplices, these simplexwise affine maps patch to a continuous map $\mathrm{\mid }f\mathrm{\mid }:\mathrm{\mid }K\mathrm{\mid }\to \mathrm{\mid }L\mathrm{\mid }$.

Steps 2.1 and 1.2 give continuity and the identity/composition laws.

Every odd permutation is a product of an odd number of transpositions, and each transposition switches the two orientation classes by definition.

After composing an odd number of such sign reversals, the final ordering represents the opposite orientation class, so its oriented simplex is the negative of the original one.

It is enough to compare two orderings that differ by one adjacent transposition, because adjacent transpositions generate the symmetric group.

Let $w=[v_0,\dots ,v_i,v_{i+1},\dots ,v_n]$ and $w^{\mathrm{\prime }}=[v_0,\dots ,v_{i+1},v_i,\dots ,v_n]=-w$. For $j<i$ or $j>i+1$, deleting the $j$th vertex from $w$ and $w^{\mathrm{\prime }}$ leaves two orderings of the same face that still differ by one adjacent transposition, so the corresponding face terms differ by a minus sign. The face obtained from $w$ by deleting $v_i$ is exactly the face obtained from $w^{\mathrm{\prime }}$ by deleting $v_i$ in position $i+1$, and their coefficients are $(-1{)}^i$ and $(-1{)}^{i+1}$. Likewise the face obtained from $w$ by deleting $v_{i+1}$ is the face obtained from $w^{\mathrm{\prime }}$ by deleting the entry in position $i$, again with opposite coefficients. Hence every term of $\mathrm{\partial }{w}^{\mathrm{\prime }}$ is the negative of the corresponding term of $\mathrm{\partial }w$, so $\mathrm{\partial }{w}^{\mathrm{\prime }}=-\mathrm{\partial }w$.

Therefore equivalent oriented representatives have the same boundary value, so the boundary formula is well defined on orientation classes.

Expanding ${\mathrm{\partial }}_{n-1}{\mathrm{\partial }}_n[v_0,\dots ,v_n]$ produces the sum of all codimension-two faces obtained by deleting two vertices, once by deleting $v_i$ then $v_j$ and once by deleting $v_j$ then $v_i$.

The two appearances of the same codimension-two face have opposite signs because the exponents differ by $1$. Hence every codimension-two face cancels with its partner, and the full sum is $0$.

The boundary maps are homomorphisms, so vanishing on every oriented simplex implies ${\mathrm{\partial }}_{n-1}{\mathrm{\partial }}_n=0$ on all of $C_n(K)$.

By the boundary formula, ${\mathrm{\partial }}_1[v_0,v_1]=[v_1]-[v_0]$. Applying the augmentation gives $\epsilon ([v_1]-[v_0])=1-1=0$.

In degrees $n\ge 1$, the ordinary simplicial differentials already satisfy ${\mathrm{\partial }}_{n-1}{\mathrm{\partial }}_n=0$, so adding $\epsilon$ at degree $0$ preserves the chain-complex condition.

If $f(v_0),\dots ,f(v_n)$ are pairwise distinct, then deleting one vertex before applying $f$ or after applying $f$ produces the same oriented face. Therefore $\mathrm{\partial }{f}_{\mathrm{\#}}[v_0,\dots ,v_n]=f_{\mathrm{\#}}\mathrm{\partial }[v_0,\dots ,v_n]$ term by term.

If some image vertices repeat, then $f_{\mathrm{\#}}[v_0,\dots ,v_n]=0$. In $f_{\mathrm{\#}}\mathrm{\partial }[v_0,\dots ,v_n]$, the faces whose remaining image vertices still repeat map to $0$, while the two faces obtained by deleting one of a repeated pair map to the same oriented simplex with opposite signs and cancel. Hence $f_{\mathrm{\#}}\mathrm{\partial }[v_0,\dots ,v_n]=0=\mathrm{\partial }{f}_{\mathrm{\#}}[v_0,\dots ,v_n]$.

Steps 1.1 and 1.2 cover all simplices, so $\mathrm{\partial }{f}_{\mathrm{\#}}=f_{\mathrm{\#}}\mathrm{\partial }$.

The previous lemma shows that each $f_{\mathrm{\#}}$ is a chain map, so simplicial homology may be applied degreewise to obtain homomorphisms $f_{\ast }$.

On every oriented simplex, the identity simplicial map induces the identity chain map, and $(g\circ f{)}_{\mathrm{\#}}=g_{\mathrm{\#}}\circ f_{\mathrm{\#}}$ by direct inspection of the defining formula. Therefore the induced homology maps satisfy $({\mathrm{id}}_K{)}_{\ast }=\mathrm{id}$ and $(g\circ f{)}_{\ast }=g_{\ast }\circ f_{\ast }$.

Thus simplicial homology defines a functor from simplicial complexes and simplicial maps to graded abelian groups.

Let $x={\sum }_i{\lambda }_i{v}_i$ lie in a simplex $\sigma =\{v_0,\dots ,v_n\}$ of $K$. Contiguity says that the vertices $f(v_i)$ and $g(v_i)$ together span a simplex of $L$, so for each $t\in [0,1]$ the barycentric combination $H(x,t):={\sum }_i(1-t){\lambda }_{i}f(v_i)+{\sum }_{i}t{\lambda }_{i}g(v_i)$ lies in $\mathrm{\mid }L\mathrm{\mid }$.

On each simplex of $K$, the formula in step 1.1 is affine in both $x$ and $t$, so it is continuous there. If a point lies on a common face of two simplices, the same barycentric formula is obtained from either side, so the simplexwise formulas patch to a continuous map $H:\mathrm{\mid }K\mathrm{\mid }\times [0,1]\to \mathrm{\mid }L\mathrm{\mid }$.

At $t=0$ the formula gives $\mathrm{\mid }f\mathrm{\mid }(x)$, and at $t=1$ it gives $\mathrm{\mid }g\mathrm{\mid }(x)$. Thus $H$ is a homotopy from $\mathrm{\mid }f\mathrm{\mid }$ to $\mathrm{\mid }g\mathrm{\mid }$.

Let $z$ be an $n$-cycle. Its support generates a finite subcomplex $K_z$ of $K$. Choose a total ordering of the finite vertex set of $K_z$, and use the increasing vertex order as the preferred oriented generator of every simplex of $K_z$. On these generators define $P_k[v_0,\dots ,v_k]:={\sum }_{i=0}^k(-1{)}^i[f(v_0),\dots ,f(v_i),g(v_i),\dots ,g(v_k)],$ omitting a summand when its displayed vertices are not pairwise distinct, and extend linearly. This is a well-defined homomorphism on $C_k(K_z)$ because it is defined on a chosen free basis, and contiguity makes every nondegenerate summand a simplex of $L$.

For each preferred generator of $C_k(K_z)$, expand the boundary of the $i$th prism simplex from step 1.1. Consecutive interior faces cancel, the two outer faces give $g_{\mathrm{\#}}-f_{\mathrm{\#}}$, and the remaining faces give $-P\mathrm{\partial }$. Terms with repeated image vertices cancel in the corresponding normalized formula. Hence $\mathrm{\partial }P+P\mathrm{\partial }=g_{\mathrm{\#}}-f_{\mathrm{\#}}$ on the chains of $K_z$.

Applying step 2.1 to the cycle $z$ gives $g_{\mathrm{\#}}z-f_{\mathrm{\#}}z=\mathrm{\partial }(Pz),$ because $\mathrm{\partial }z=0$. Thus $f_{\mathrm{\#}}z$ and $g_{\mathrm{\#}}z$ represent the same homology class. Every homology class has such a finitely supported cycle representative, so $f_{\ast }=g_{\ast }$ in every degree.

Define $h_{-1}:\mathbb{Z}\to C_0({\mathrm{\Delta }}^n)$ by $h_{-1}(1)=[a]$. For an oriented simplex $[v_0,\dots ,v_k]$, set $h_k[v_0,\dots ,v_k]=[a,v_0,\dots ,v_k]$ if $a\mathrm{\notin }\{v_0,\dots ,v_k\}$ and set $h_k[v_0,\dots ,v_k]=0$ if $a\in \{v_0,\dots ,v_k\}$. Because adjoining $a$ to a face of a simplex still gives a face of ${\mathrm{\Delta }}^n$, each $h_k$ is well defined.

If $a\mathrm{\notin }\{v_0,\dots ,v_k\}$, then the extra face created by applying $\mathrm{\partial }{h}_k$ and deleting $a$ is exactly $[v_0,\dots ,v_k]$, while every other face cancels with the corresponding term in $h_{k-1}\mathrm{\partial }$. If $a$ is already among the vertices, then $h_k$ is $0$ and the same cancellation leaves the identity term. Thus $\mathrm{\partial }h+h\mathrm{\partial }=\mathrm{id}$ on the augmented complex.

The family $(h_k{)}_{k\ge -1}$ is therefore a contracting homotopy, so the augmented simplicial chain complex of ${\mathrm{\Delta }}^n$ is contractible.

The previous lemma gives a contracting homotopy for the augmented simplicial chain complex of ${\mathrm{\Delta }}^n$.

A contractible chain complex is acyclic, so every reduced simplicial homology group of ${\mathrm{\Delta }}^n$ vanishes.

If $x\in \mathrm{\mid }\sigma \mathrm{\mid }$ and $u$ is a vertex of the simplex $\sigma$, then the straight-line barycentric homotopy inside the Euclidean simplex $\mathrm{\mid }\sigma \mathrm{\mid }$ joins $x$ to $u$. Hence every point of $\mathrm{\mid }K\mathrm{\mid }$ lies in the same connected component as any vertex of a simplex supporting it, and all vertices of one simplex lie in the same connected component of $\mathrm{\mid }K\mathrm{\mid }$.

If vertices $v$ and $w$ are joined by an edge path $v=v_0,\dots ,v_m=w$, then $[w]-[v]={\sum }_{i=1}^m\mathrm{\partial }[v_{i-1},v_i]$, so vertices in the same edge-path component define the same class in $H_0^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)$.

Fix a vertex $v$ of $K$, let $E(v)$ be the set of vertices joined to $v$ by edge paths, and let $K(v)$ be the subcomplex whose simplices have all vertices in $E(v)$. By step 1.1, every simplex that contains one vertex of $E(v)$ has all its vertices in $E(v)$, so for each simplex $\sigma$ the intersection $\mathrm{\mid }K(v)\mathrm{\mid }\cap \mathrm{\mid }\sigma \mathrm{\mid }$ is either $\mathrm{\mid }\sigma \mathrm{\mid }$ or $\mathrm{\varnothing }$. Hence $\mathrm{\mid }K(v)\mathrm{\mid }$ is open and closed in the weak topology. It is connected because every point of $\mathrm{\mid }K(v)\mathrm{\mid }$ lies in a simplex whose vertices are edge-path connected to $v$, so step 1.1 and concatenation of those edge paths connect the point to $v$. Therefore $\mathrm{\mid }K(v)\mathrm{\mid }$ is exactly the connected component of $\mathrm{\mid }K\mathrm{\mid }$ containing $v$. In particular, the connected components of $\mathrm{\mid }K\mathrm{\mid }$ are exactly the realizations of the edge-path components of the vertices, and if $\mathrm{\mid }K\mathrm{\mid }\mathrm{\ne }\mathrm{\varnothing }$ every connected component contains a vertex.

Let ${\pi }_0(\mathrm{\mid }K\mathrm{\mid })$ be the set of connected components of $\mathrm{\mid }K\mathrm{\mid }$. Since the vertices of every simplex lie in one component by step 1.1, the assignment sending a vertex $u$ to the basis vector $e_{[u]}$ of the free abelian group ${⨁}_{C\in {\pi }_0(\mathrm{\mid }K\mathrm{\mid })}\mathbb{Z}{e}_C$ extends to a homomorphism $C_0(K)\to {⨁}_{C\in {\pi }_0(\mathrm{\mid }K\mathrm{\mid })}\mathbb{Z}{e}_C$. Boundaries of edges map to $0$, so this homomorphism factors through $\mathrm{\Phi }:H_0^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)\to {⨁}_{C\in {\pi }_0(\mathrm{\mid }K\mathrm{\mid })}\mathbb{Z}{e}_C$. If $\mathrm{\mid }K\mathrm{\mid }=\mathrm{\varnothing }$, then $C_0(K)=0$ and both groups are zero. Otherwise step 2.1 shows that every connected component contains a vertex, so $\mathrm{\Phi }$ is surjective.

Choose one vertex $a_C$ in each nonempty connected component $C$. Every class in $H_0^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)$ is represented by a finite $0$-chain $z={\sum }_u{n}_u[u]$. By step 2.1, two vertices lie in the same connected component exactly when they are edge-path connected, so step 1.2 gives $[u]-[a_C]\in B_0(K)$ for every $u\in C$. Hence in $H_0^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)$ one has $[z]={\sum }_C\left({\sum }_{u\in C}{n}_u\right)[a_C]$. If $\mathrm{\Phi }([z])=0$, then every component sum ${\sum }_{u\in C}{n}_u$ is zero, so $[z]=0$. Thus $\mathrm{\Phi }$ is injective.

Therefore $\mathrm{\Phi }$ is an isomorphism, so $H_0^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)$ is the free abelian group on the connected components of $\mathrm{\mid }K\mathrm{\mid }$.

Every nonempty simplex of $K$ lies in exactly one summand $K_{\alpha }$, so for each $n\ge 0$ one has $C_n(K)\cong {⨁}_{\alpha \in A}{C}_n(K_{\alpha })$, and under this identification the boundary operator acts componentwise.

Therefore for each $n\ge 0$ the cycle groups, boundary groups, and homology groups split componentwise, giving $H_n^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)\cong {⨁}_{\alpha \in A}{H}_n^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K_{\alpha })$. For $n<0$, both sides are zero by definition. This proves the statement for every $n$.

For each $n$, the chain group $C_n(K)$ is free abelian on the oriented $n$-simplices of $K$, so $\mathrm{rank}{C}_n(K)=f_n(K)$. Since $K$ is finite, only finitely many of these groups are nonzero.

Therefore the simplicial chain complex of $K$ is a bounded chain complex of finite-rank free abelian groups, and its homology groups are free of finite rank by hypothesis. The finite-free Euler-Poincare theorem applies and gives ${\sum }_{n\ge 0}(-1{)}^n\mathrm{rank}{C}_n(K)={\sum }_{n\ge 0}(-1{)}^n\mathrm{rank}{H}_n^{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}}(K)$.

Replace $\mathrm{rank}{C}_n(K)$ by $f_n(K)$ using step 1.1, and replace the left-hand side by $\chi (K)$ by definition. This yields the stated formula.