Proof:

For each $x \in Z_n(L)$,

$$H_n(\psi\circ\phi)(\overline{x}) = \overline{\psi_n\circ\phi_n(x)} = H_n(\psi)(\overline{\phi_n(x)}) = H_n(\psi)\circ H_n(\phi)(\overline{x}).$$

We shall at some point as we go along, drop the primes and denote both sets of boundary maps by $\partial _n$ or even $\partial$. Observe that if $z \in$   ker$\;\phi_n$ then

$$\phi_{n-1}(\partial _n z) = \partial _{n-1} \phi_n(z) = 0,$$

whereby we conclude that $\partial _n$ maps ker$\;\phi_n$ into ker $\;\phi_{n-1}$ and we get a chain complex

$$\begin{CD} @> >>$$   ker$$\;\phi_{n+1} @> {\partial _{n+1}} >>$$   ker$$\;\phi_{n} @> \partial _n >>$$   ker$$\;\phi_{n-1} @> >> \end{CD}$$

which we denote by ker$\;\phi$. Likewise we get the chain complex

$$\begin{CD} @> >>$$   Im$$\;\phi_{n+1} @> {\partial _{n+1}} >>$$   Im$$\;\phi_{n} @> \partial _n >>$$   Im$$\;\phi_{n-1} @> >> \end{CD}$$

which we denote by Im$\;\phi$. It is clear from (29.13) that $\partial _n$ maps Im$\;\phi_n$ into Im $\;\phi_{n-1}$.