Zero-th simplicial homology is free on connected components

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.

To prove injectivity, let a class be represented by a finite $0$-chain $z={\sum }_u{n}_u[u]$ with $\mathrm{\Phi }([z])=0$. Only finitely many connected components meet its support. For each of those components $C$, select one vertex $a_C$ from the support of $z$ in $C$; this is a finite selection. 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 supported $u\in C$. Hence $[z]={\sum }_C\left({\sum }_{u\in C}{n}_u\right)[a_C]$. Every coefficient in this finite sum is zero because $\mathrm{\Phi }([z])=0$. Thus $[z]=0$ and $\mathrm{\Phi }$ 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 $\mathrm{\mid }K\mathrm{\mid }$.