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For chain maps $f,g:C_{\ast }\to D_{\ast }$, a chain homotopy from $g$ to $f$ is a family of group homomorphisms $s_q:C_q\to D_{q+1}$ satisfyingIf $z$ is a chain cycle, then $(f-g)(z)=d^{D}s(z)$ is a chain boundary. Consequently $[f(z)]=[g(z)]$, so $\boxed{f_{\ast }=g_{\ast }}$ on every homology group.

Order the vertices of the simplex as $v_0,\dots ,v_n$. Its simplicial chain complex hasand zero groups otherwise. An arbitrary ordering represents the sign of the permutation times the increasing ordering. Its boundary operator isIn $d^2$ each face obtained by deleting the vertices in positions $r<t$ occurs twice. Deleting $r$ first gives sign $(-1{)}^r(-1{)}^{t-1}$, whereas deleting $t$ first gives $(-1{)}^t(-1{)}^r$. These signs are opposite, so all terms cancel. The case $d_0{d}_1=0$ follows directly from $d_0=0$. Thus $\boxed{d^2=0}$.

For the homology calculation, add the augmentation $\epsilon :S_0\to \mathbb{Z}$ sending every vertex to $1$. Regard $\mathbb{Z}$ as the group in degree $-1$ of an augmented chain complex. Define the simplicial cone chain contraction byIf the simplex does not contain $v_0$, expansion of its coned boundary gives $ds(\sigma )=\sigma -sd(\sigma )$. If it contains $v_0$, the only face on which $s$ is nonzero is the face omitting $v_0$, and $sd(\sigma )=\sigma$. In degree zero this uses the augmentation, and in degree $-1$ it says $\epsilon s_{-1}=1$. HenceEvery positive-degree chain cycle is therefore a chain boundary. In degree zero, the same identity says $z-\epsilon (z)[v_0]=d{s}_0(z)$, so the augmentation induces an isomorphism $H_0\to \mathbb{Z}$, with inverse $1\mapsto [v_0]$. Thus, including $n=0$,This is a direct chain homotopy calculation and uses no result about cellular homology.

Represent a class in $H_k(K)$ by a chain cycle $z\in Z_k(K)$. Since every simplex of $K=M\cup N$ belongs to at least one of the two simplicial subcomplexes, split the chain in the simplicial chain complex asBecause $\partial z=0$, we haveThe left side lies in $C_{k-1}(M)$ and the right side in $C_{k-1}(N)$, so this common chain is supported in $L=M\cap N$. It is a cycle because ${\partial }^2=0$. The connecting homomorphism is thereforeChanging the decomposition or the representative changes $\partial m$ only by a boundary in $L$, which is why the construction descends to homology classes.

Because $L$ is a simplicial subcomplex of $K$, every face of a simplex of $L$ also belongs to $L$. The simplicial boundary operator therefore satisfiesso $C_{\bullet }(L)$ is a chain subcomplex of the simplicial chain complex $C_{\bullet }(K)$.

Define a map on the quotient groups byIf $c^{\prime }=c+\ell$ with $\ell \in C_k(L)$, then $\partial c^{\prime }-\partial c=\partial \ell \in C_{k-1}(L)$, so this definition is independent of the representative. Moreover,Thus $C_{\bullet }(K,L)$ is the relative simplicial chain complex.

Use the orientationsThe simplicial chain complex haswithThe map ${\partial }_2$ is injective, so $H_2(K)=0$. The complex is connected, so $H_0(K)\cong \mathbb{Z}$ and $\mathrm{rank}{\partial }_1=3$. Hence $\ker {\partial }_1$ has rank $5-3=2$; quotienting by the rank-one primitive subgroup $\mathrm{im}{\partial }_2$ givesThere are no higher simplices. ThereforeThe result agrees with the Euler characteristic $4-5+1=0$.

Use the cellular chain complex or simplicial chain complex over $\mathbb{Q}$, rather than the infinite singular chain complex. Put $r_i=\mathrm{rank}{\partial }_i$, with $r_0=r_{d+1}=0$. The rank-nullity theorem gives $\dim \ker {\partial }_i=\dim C_i-r_i$; quotienting the chain cycles by the chain boundaries then gives $b_i=\dim C_i-r_i-r_{i+1}$. Each boundary operator's rank occurs in adjacent degrees with opposite signs, soFiniteness makes this cancellation an ordinary finite sum. This is the algebra behind the cell-count formula for Euler characteristic.