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The set-up is the following: $\mathfrak c$ is a collection of finite groups closed under subgroups, homomorphic images, and extensions. For any group $G$, the pro-$\mathfrak c$-completion $G(\mathfrak c)$ is defined as the topological group $\varprojlim G/U$ where $U$ ranges over all normal subgroups of $G$ such that $G/U\in\mathfrak c$, with $G/U$ treated as discrete topological groups. Note that we have a group homomorphism $\eta\colon G\to G(\mathfrak c)$.

If I understand correctly, the pro-$\mathfrak c$-completion is the left adjoint to the forgetul functor from pro-$\mathfrak c$-groups to groups, as it satisfies the universal property that any group homomorphism $f\colon G\to H$ where $H$ is pro-$\mathfrak c$-group factors as $f=h\circ\eta$ where $h\colon G(\mathfrak c)\to H$ is a continuous group homomorphism (so a morphism of pro-$\mathfrak c$-groups).

It seems to me that the above implies that the free pro-$\mathfrak c$-group on a set $S$ should be the pro-$\mathfrak c$-completion of the free group on $S$. I am reading, however, that this is the case only when $S$ is finite, but not when $S$ is infinite. What am I missing?

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Adjoint functors compose: if $F_1, F_2$ are composable functors with left adjoints $G_1, G_2$, then $F_1 \circ F_2$ has left adjoint $G_2 \circ G_1$, and this is straightforward to prove. In this situation $F_1$ is the forgetful functor from groups to sets and $F_2$ is the forgetful functor from pro-c-groups to groups. Hence $G_1$ is the free group functor and $G_2$ is the pro-c-completion.

So you are right if you interpret "free pro-c-group on an infinite set" in the obvious way. However, a cursory google search found Wilson's Profinite Groups which gives a different universal property from the obvious one.

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