Categorical proof of a property of direct limits of abelian groups

Question

There is a very useful property of direct limits, which reads (as in an appendix in Bredon):

Suppose given a direct system $\{G_{\alpha}, f_{\beta,\alpha}\}$ of abelian groups. Let $A$ be an abelian group and $h_{\alpha}: G_{\alpha} \to A$ homomorphisms such that $\beta > \alpha \implies h_{\beta} \circ f_{\beta, \alpha}=h_{\alpha}$. Then there exists an unique homomorphism $h: \lim\limits_{\longrightarrow} G_{\alpha} \to A$ such that $h \circ i_{\alpha}=h_{\alpha}$ for all $\alpha$. Moreover:

1. $im(h)=\{a \in A \mid a=h_{\alpha} (g) \text{ for some g and } \alpha\}=\bigcup im(h_{\alpha})$,
2. $\ker(h)=\{g \in \lim\limits_{\longrightarrow} G_{\alpha} \mid \exists {\alpha} \text{ and } g_{\alpha} \in G_{\alpha} \text{ such that }g=i_{\alpha}(g_{\alpha}) \text{ and } h_{\alpha}(g_{\alpha})=0\}=\bigcup i_{\alpha} (\ker h_{\alpha}).$

The proof given in Bredon utilizes the explicit definition of the direct limit as $\bigoplus G_{\alpha}/ \sim$ (since it even states the universal property as a proposition, above).

However, given the universal property as the definition, how could one prove $1$ and $2$ above without resorting to the explicit form of the direct limit, only using the universal property? I searched on Rotman (introduction to homological algebra), and he also proves (an analogue of) the second part of the proposition above using the explicit definition.

Answer 1

Let $G = \lim\limits_{\longrightarrow} G_\alpha$ and $G' = \bigcup i_\alpha(G_\alpha)$. The maps $i_\alpha$ map $G_\alpha$ into $G'$ and are compatible with the $f_{\beta,\alpha}$. Let $T$ be a set with maps $\eta_\alpha : G_\alpha \to T$ that are compatible with the $f_{\beta,\alpha}$. There exists a unique map $\phi : G\to T$ such that for all $\alpha$, $\phi\circ i_\alpha = \eta_\alpha$. Define a map $f : G'\to T$ by setting $f(g) = \eta_\alpha(g_\alpha)$ if $g = i_\alpha(g_\alpha)$. If $f$ is well-defined, then it is the unique map such that for all $\alpha$, $f\circ i_\alpha = \eta_\alpha$. To see that $f$ is well-defined, suppose $i_\alpha(g_\alpha) = i_\beta(g_\beta)$ for some $\alpha$ and $\beta$. Then $$\eta_\alpha(g_\alpha) = \phi(i_\alpha(g_\alpha)) = \phi(i_\beta(g_\beta)) = \eta_\beta(g_\beta).$$ Hence $f$ is well-defined, and consequently $G'$ is a direct limit of the $G_\alpha$. Thus $G'$ is in set bijection with $G$; since, in addition $G'\subset G$, then $G' = G$. Now $h(G) = h(G') = \bigcup hi_\alpha(G_\alpha) = \bigcup h_\alpha(G_\alpha)$, proving the first part.

For the second part, note that for every $\alpha$, $\operatorname{ker}(h) \cap i_\alpha(G_\alpha) = i_\alpha(\operatorname{ker}(h_\alpha))$. As $G = \bigcup i_\alpha(G_\alpha)$ and $\operatorname{ker}(h)\subset G$, then $\operatorname{ker}(h) = \bigcup\, (\operatorname{ker}(h)\cap i_\alpha(G_\alpha)) = \bigcup i_\alpha(\operatorname{ker}(h_\alpha))$.