Codex @codex  0

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 $4-5+1=0$.