Proof:

We must first show that displayed formula (29.18) gives a well-defined map since several choices are being made. First, for $k_n \in Z_n(K)$ surjectivity of $g_n$ shows that there exists $x_n \in G_n$ such that $g_n(x_n) = k_n$. Applying the boundary map $\partial _n$ we see that $g_{n-1}(\partial x_n) = \partial k_n = 0$ which, by virtue of exactness of (29.16) and injectivity of $f_{n-1}$, shows there is a unique $y_{n-1} \in L_{n-1}$ such that

$$f_{n-1}(y_{n-1}) = \partial x_n. \eqno(29.19)$$

We have to now show that $y_{n-1}$ is a cycle in $L$. This is clear if $n = 1$ and so we assume $n\geq 2$. Applying the boundary map to (29.19) gives $f_{n-2}(\partial y_{n-1}) = 0$ from which we conclude, since $f_{n-2}$ is injective, that $y_{n-1}\in Z_{n-1}(L)$. Hence the assignment

$$k_n\mapsto \overline{y_{n-1}},\quad k_n \in Z_n(K) \eqno(29.20)$$

is well defined once we show that it is independent of the choice of $x_n \in g_n^{-1}(k_n)$.

Second, we suppose that for a given $k_n \in Z_n(K)$, $x_n^{\prime}$ and $x_n^{\prime\prime}$ are two members of $g_n^{-1}(k_n)$ then

$$x_n^{\prime} - x_n^{\prime\prime}\in$$   ker$$\;g_n =$$   im$$\;f_n.$$

So there is a $u_n\in L_n$ such that $x_n^{\prime} - x_n^{\prime\prime} = f_n(u_n)$. On the other hand, for these two choices there exist $y_{n-1}^{\prime}$ and $y_{n-1}^{\prime\prime}$ in $L_{n-1}$ such that (29.19) holds and so

$$f_{n-1}(y_{n-1}^{\prime}) - f_{n-1}(y_{n-1}^{\prime\prime}) = \pa... ...n^{\prime} - x_n^{\prime\prime}) = \partial f_n(u_n) = f_{n-1}(\partial u_n).$$

Injectivity of $f_{n-1}$ implies $y_{n-1}^{\prime}$ and $y_{n-1}^{\prime\prime}$ differ by a boundary and so define the same homology class.

Third, we must show that the same homology class results if we begin with two homologous cycles $k_n^{\prime}$ and $k_n^{\prime\prime}$. In this there exists $v_{n+1} \in K_{n+1}$ and $x_{n+1}\in G_{n+1}$ such that

$$k_n^{\prime} - k_n^{\prime\prime} = \partial v_{n+1} = \partial g_{n+1}(x_{n+1}) = g_n(\partial x_{n+1}).$$

Let $x_n^{\prime}$ and $x_n^{\prime\prime}$ be chosen from $g_n^{-1}(k_n^{\prime})$ and $g_n^{-1}(k_n^{\prime\prime})$ respectively so that $g_n(x_n^{\prime} - x_n^{\prime\prime} - \partial x_{n+1}) = 0.$ By exactness of (29.16) there is a $w_n\in L_n$ such that $x_n^{\prime} - x_n^{\prime\prime} - \partial x_{n+1} = f_n(w_n).$ Applying $\partial$ to this and and recalling (29.19) we see that the corresponding cycles $y_{n-1}^{\prime}$ and $y_{n-1}^{\prime\prime}$ satisfy

$$f_{n-1}(y_{n-1}^{\prime} - y_{n-1}^{\prime\prime}) = \partial f_n(w_n) = f_{n-1}(\partial w_n).$$

Since $f_{n-1}$ is injective we see that the cycles $y_{n-1}^{\prime}$ and $y_{n-1}^{\prime\prime}$ are homologous.