Exactness of (29.17):

We first check the exactness at the junction $H_n(G)$. Since (29.15) implies $H_n(g)\circ H_n(f) = 0$, it suffices to prove ker$\;H_n(g)\subset$   im$\;H_n(f)$. So let $g_n(x_n) = \partial _{n+1}k_{n+1}$ for some $x_n \in Z_n(G)$ and $k_{n+1} \in K_{n+1}$. Since $g_{n+1}$ is surjective we can find $x_{n+1}\in G_{n+1}$ such that $k_{n+1} = g_{n+1}(x_{n+1})$ and

$$g_n(x_n) = \partial _{n+1}g_{n+1}(x_{n+1}) = g_n(\partial _{n+1}x_{n+1}),$$

from which we conclude there exists $y_n\in L_n$ such that $x_n - \partial _{n+1}x_{n+1} = f_n(y_n).$ Applying the operator $\partial _n$ and using injectivity of $f_{n-1}$ we see that $y_n\in Z_n(L)$ and the result is established.

We now turn to the exactness at the junction $H_n(K)$. It is clear from (29.18) that $\delta_n(\overline{g_nx_n}) = 0$ for any $x_n \in Z_n(G)$ so that $\delta_n\circ H_n(g) = 0$. To prove ker$\;\delta_n(g)\subset$   im$\;H_n(g)$ let $k_n \in Z_n(K)$ such that $\delta_n(\overline{k_n}) = 0$. Equation (29.18) then implies, for any $x_n \in g_n^{-1}(k_n)$ there is $l_n \in L_n$ such that

$$f_{n-1}^{-1}\partial _nx_n = \partial _nl_n.$$

From this we get $x_n - f_n(l_n) = x^{\prime}_n \in Z_n(G)$. Applying $g_n$ we see that $k_n = g_n(x_n) = g_n(x_n^{\prime})$ whereby we conclude $\overline{k_n} \in$   im$\;H_n(g)$.

Finally we come to the exactness at the junction $H_{n-1}(L)$. From (29.18) follows $H_{n-1}(f)\circ \delta_n = 0$. To show ker$\;H_{n-1}(f)\subset$   im$\;\delta_n$, pick a cycle $l_{n-1}$ such that $f_{n-1}(l_{n-1})$ is a boundary say $\partial _nx_n$ for some $x_n \in G_n$. Applying $g_{n-1}$ to the equation

$$f_{n-1}l_{n-1} = \partial _nx_n$$

gives a cycle $k_n = g_n(x_n) \in Z_n(K)$. From (29.18) we infer $\delta_n(\overline{k_n}) = \overline{l_{n-1}}$ and this suffices for a proof.