Proof:

(i) Let $\tilde G$ be the coproduct (direct sum) of the abelian groups $\{G_{\alpha}\}$ and we regard (for simplifying notations) the groups $G_{\alpha}$ as being subgroups of $\tilde G$ and $i_{\alpha} : G_{\alpha}\longrightarrow G$ the inclusion maps. Declare $x_{\alpha}\in G_{\alpha}$ and $x_{\beta}\in G_{\beta}$ as being equivalent if there exists $\gamma \in \Lambda$ such that $\gamma \geq \alpha$, $\gamma \geq \beta$ and

$$f_{\alpha\gamma}(x_{\alpha}) = f_{\beta\gamma}(x_{\beta}). \eqno(40.4)$$

Lemma (40.1) states that this is a well defined equivalence relation. We denote by $\sim$ the equivalence relation just defined and define $N$ to be the subgroup generated by

$$\{ x_{\alpha} - x_{\beta}\;/\; x_{\alpha} \sim x_{\beta} \}.$$

Finally let $G = {\tilde G}/N$ and $\eta:{\tilde G}\longrightarrow G$ be the quotient map. We claim that $G$ is the inductive limit with respect to the maps $f_{\alpha}$ given by the composition

$$\begin{CD} G_{\alpha} @> i_{\alpha}>> {\tilde G} @> \eta >> {\tilde G}/N. \end{CD} \eqno(40.5)$$

We now check the conditions (1) and (2) in definition (40.2). For $\alpha \leq \beta$ we derive from

$$(f_{\beta\beta}\circ f_{\alpha\beta})(x_{\alpha}) = f_{\alpha\beta}(x_{\alpha}).$$

the useful piece of information

$$f_{\alpha\beta}(x_{\alpha}) \sim x_{\alpha},\quad x_{\alpha}\in G_{\alpha}. \eqno(40.6)$$

Hence $f_{\alpha\beta}(x_{\alpha}) - x_{\alpha} \in N$ whereby we conclude

$$\eta(f_{\alpha\beta}(x_{\alpha})) = \eta(x_{\alpha}),$$

which in turn implies $f_{\beta}\circ f_{\alpha\beta} = f_{\alpha}$. Turning to the universal property (2) assume given an abelian group $H$ and a family of group homomorphisms $g_{\alpha} : G_{\alpha}\longrightarrow H$ such that

$$g_{\beta}\circ f_{\alpha\beta} = g_{\alpha},\quad \alpha\leq \beta.\eqno(40.7)$$

We first use the defining property of the coproduct to get a group homomorphism $\phi: {\tilde G}\longrightarrow H$ such that the following diagram commutes:

$$\xymatrix{ G_{\alpha} \ar[rr]^{i_{\alpha}}\ar[rd]_{g_{\alpha}} & & {\tilde G} \ar[ld]^{\phi}\\ & H }$$

That is $\phi\circ i_{\alpha} = g_{\alpha}$. From (40.4) and (40.7) we get $g_{\alpha}(x_{\alpha}) = g_{\beta}(x_{\beta}),$ or in view of the fact that we have identified $G_{\alpha}$ as a subgroup of ${\tilde G}$, $\phi(x_{\alpha}) = \phi(x_{\beta}).$ Hence there is a group homomorphism $\psi:{\tilde G}/N\longrightarrow H$ such that

$$\psi\circ\eta = \phi. \eqno(40.8)$$

Upon applying this to an arbitrary $x_{\alpha}\in G_{\alpha}$ we get using (40.5) that $\psi\circ f_{\alpha} = g_{\alpha}$ for every $\alpha \in \Lambda$. The homomorphism satisfying (40.8) is unique since the elements $\{f_{\alpha}(x_{\alpha})/\alpha\in \Lambda$    and $x_{\alpha} \in G_{\alpha}\}$ generate the group ${\tilde G}/N$.

We now prove (ii) which we shall use in the next lecture. Since $x \in N$, there exists a finite set of indices $\alpha_1, \beta_1, \alpha_2, \beta_2,\dots,\alpha_k, \beta_k$ such that

$$x = \sum_{j = 1}^k (x_{\alpha_j} - x_{\beta_j}) \eqno(40.9)$$

where $x_{\alpha_j} \sim x_{\beta_j}$ for each $j$. Thus for each $j$ there is a $\gamma_j$ exceeding both $\alpha_j$ and $\beta_j$ such that $f_{\alpha_j\gamma_j}(x_{\alpha_j}) = f_{\beta_j\gamma_j}(x_{\beta_j}).$ Since (40.9) spells out a relation in the direct sum of the groups $G_{\beta}$, it decomposes into a bunch of equations namely

$$x = \sum_{\alpha_i = \alpha} x_{\alpha_i} - \sum_{\beta_i = \alpha} x_{\beta_i}$$

$$= \sum_{\alpha_i = \lambda} x_{\alpha_i} - \sum_{\beta_i = \lambda} x_{\beta_i},\quad \lambda \neq \alpha$$ 0

The index $\lambda$ runs through a finite subset of $\alpha_1, \beta_1,\dots,\alpha_k, \beta_k$. Taking $\delta$ to be sufficiently large and applying $f_{\alpha\delta}$ to the first and $f_{\lambda\delta}$ to the second of the above displayed equations and adding we get

$$f_{\alpha\delta}(x) = \sum_{j = 1}^k (f_{\alpha_i\delta}(x_{\alpha_j}) - f_{\beta_i\delta}(x_{\beta_j}))\eqno(40.10)$$

Using lemma (40.1) we see that if $\delta$ is sufficiently large each of the summands on the right hand side of (40.10) is in $N$ and so $f_{\alpha\delta}(x) = 0$ as asserted.