Show an $R$-module is a direct limit This is a scenario I've encountered in my class on $p$-adic L functions.  Let $G$ be a profinite group which is the inverse limit of a system $(G_i, f_{ij})$ of discrete finite topological groups.  For a commutative ring $R$, let $C(G_i,R)$ be the $R$-module of functions $G_i \rightarrow R$, and let $C^{\infty}(G,R)$ be the $R$-module of continuous functions from $G$ to $R$ which are everywhere locally constant (for each $x \in G$, there is a neighborhood of $x$ on which $f$ is constant; this is the same thing as saying that $f$ takes only finitely many values).  
For the projections $\pi_i:G \rightarrow G$, one takes the map $\phi_i: C(G_i,R) \rightarrow C^{\infty}(G,R)$ given by $\phi_i(\alpha) = \alpha \circ \pi_i$ (it's easy to see that $\alpha \circ \pi_i$ is actually locally constant), and for the maps $f_{ij}: G_j \rightarrow G_i$ for $i \leq j$, one takes the corresponding $R$-module homomorphisms $\tau_{ij}: C(G_i,R) \rightarrow C(G_j,R)$.  

The claim is that $C^{\infty}(G,R)$ is the direct limit of the direct system $(C(G_i,R), \tau_{ij}$)

But I don't know where to start.  With the universal mapping property, it's not obvious how to define a map from $C^{\infty}(G_i,R)$.  
 A: The problem is to show that a locally constant function $f:G \to R$ factors through some $G_i$.   The function is locally constant, so for each $g \in G$, there is a n.h. of $U$ of $g$ s.t. $f _{| U}$ is constant.
Now if we let $H_i$ denote the kernel of $G \to G_i$, then the cosets $gH_i$ form a n.h. basis of $g$ (by def'n of the pro-fintie topology on $G$).  So we may as well assume that $U$ is $g H_i$ for some $i$, which a priori depends on $g$.
To show that $f$ factors through $G_i$ for some $i$, we have to show that the $i$ in the above paragraph can actually be chosen uniformly in $g$, i.e. that there is a common value of $i$ s.t. $f_{| g H_i}$ is constant for every $g \in G$.
This is true, and is a variation of the fact (from the theory of metric spaces) that a continuous function on a compact metric space is uniformly continuous.  I will leave it to you to adapt that proof to this context.  (The general setting that encompasses the metric space context and the context of topological groups is the theory of uniform spaces, but you don't need to know about that theory to prove this result; the argument from the metric space context will just carry over.)
