 Solution

Chain maps $f,g:C_{\ast }\to C_{\ast }^{\prime }$ are chain homotopic when maps $h_n:C_n\to C_{n+1}^{\prime }$ satisfyFor a cycle $z$, $f(z)-g(z)=\partial h(z)$ is a boundary, so $f_{\ast }=g_{\ast }$ on homology.

Fix a vertex $e_0$ of ${\Delta }^n$ and define the cone operator by adjoining $e_0$ to an oriented simplex, with zero when it is already present. The simplicial boundary formula giveson the reduced chain complex. Thus its identity is null-homotopic andfor every $k$; equivalently, $H_0({\Delta }^n)\cong \mathbb{Z}$ and higher homology vanishes.

For the 2-skeleton $K$ of ${\Delta }^6$,It is connected and simply connected, so $H_0\cong \mathbb{Z}$ and $H_1=0$. Euler characteristic then givesFor $\sigma =(0123)(456)$, no vertex or edge is fixed. The only invariant 2-simplex is $[456]$, on which the 3-cycle preserves orientation, so the chain traces are $0,0,1$. The Lefschetz trace formula givestherefore