 Solution

A chain homotopy from $g$ to $f$ consists of homomorphisms $s_n:C_n\to C_{n+1}^{\prime }$ such thatfor every $n$. If $z\in C_n$ is a cycle, $d_{n}z=0$, so $f_n(z)-g_n(z)=d_{n+1}^{\prime }{s}_n(z)$ is a boundary. Hence $f_n(z)$ and $g_n(z)$ represent the same homology class. Every homology class has a cycle representative, proving $\boxed{f_{\ast }=g_{\ast }}$ in every degree.