Let $G$ be a profinite group, written as the projective limit $\operatorname{lim}(G_i)$ of a diagram of finite groups $G_i$.
Let $H$ be a group which is equipped with maps $f_i : H \to G_i$, compatible with the diagram. This induces a universal map $u : H \to G$.
What are conditions on $H$ or $f_i$ which ensure the map $u$ is surjective?
In particular, I am not sure if $u$ is surjective when the $f_i$ are surjective. Under this assumption, I can certainly find for each finite segment of an element of $G$---viewing elements of $G$ as sequences of elements from each $G_i$ which are sent to each other via the maps in the diagram---an element of $H$ which is sent to it by $u$. But it seems like one would need extra assumptions to find preimages for a general element of $G$.
Every group $H$ has a profinite completion $\widehat{H}$, namely the profinite group given by the cofiltered limit over all finite quotients $H_i$ of $H$. By construction, all of the maps $f_i : H \to H_i$ are surjective, but the corresponding map $H \to \widehat{H}$ is almost never surjective.
For example, when $H = \mathbb{Z}$, $\widehat{H}$ is the profinite integers
$$\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$$
where the product ranges over all primes and $\mathbb{Z}_p$ denotes the $p$-adic integers. This is a gigantic group, much bigger than $\mathbb{Z}$, and contains all sorts of weird elements such as $1! + 2! + 3! + 4! + \dots $ (which converges with respect to a suitable topology).