Why is $g \mapsto (gU)$ onto the profinite completion? Let $\mathcal{C}$ denote a variety of finite groups (closed to subgroups, finite direct products, quotients).
Let $G$ be a compact group, and assume there is a local basis $\mathcal{B}$ of normal subgroups of $G$ around the identity. Moreover, assume that $\cap \mathcal{B} = \{e\}$, is trivial, and that for $N \in \mathcal{B}$ we have $G/N \in \mathcal{C}$.
Define the pro$-\mathcal{C}$ completion $\hat{G} = \underset{\longleftarrow}{\text{lim}}G/N$ over these open normal subgroups in $\mathcal{B}$.
Now the map $h:G \to \hat{G}$, $g \mapsto (gU)_{U \in \mathcal{B}}$, is continuous and a group homomorphism. Moreover, it is compatible with projections from $G$ onto $G/U$. That is, if $L_U: \hat{G} \to G/U$ denotes the restriction of the projection $\prod G/V \to G/U$, then we have $\pi_U = L_U \circ h$. From the universal property of inverse limits it should hold that this is the unique such map - then it should be true that this map is onto, as $G$ is compact. How can I see that it is surjective directly?
 A: The local basis $\mathcal{B}$ is a directed set (ordered by reverse inclusion), so let's use it to define a net. Let $x\in \hat{G}$. For each $N\in\mathcal{B}$, choose $g_N\in G$ such that $h(g_N)_N = x_N$. By compactness, the net $(g_N)_{N\in \mathcal{B}}$ has a convergent subnet $r:\Lambda \to \mathcal{B}$, which converges to some limit $g$. 
Choose $N\in \mathcal{B}$. Since $g_{r(\lambda)}\to g$, there is $\lambda_0\in\Lambda$ such that $g_{r(\lambda)}\in gN$ for $\lambda \ge \lambda_0$. 
By finality of $r$, for any $N$, there is $\mu \in \Lambda$ with $r(\mu)\subseteq N$. Then by directedness of $\Lambda$, choose $\nu\ge \mu,\lambda_0$. Then $g_{r(\nu)}\subseteq gN$, so 
$$h(g)_N = gN = g_{r(\nu)}N = g_{r(\nu)}r(\nu)N = x_{r(\nu)}N = x_N,$$
with the first equality being the definition of $h$, the second being due to $\nu \ge \lambda_0$, the third being because $r(\nu)\subseteq N$, since $\nu\ge \mu$, the fourth being choice of $g_{r(\nu)}$, and the fifth being due to the explicit construction of the inverse limit.
Thus $h(g)=x$, so since $x$ was arbitrary, $h$ is surjective, as desired.
