I have a question. I don't know if is trivial or even it makes sense, but is something I could not understand. For example, consider the following exercise of Ribes-Zalesskii:
Exercise 9.1.1 Let $G = A \ast B$ be a free product of abstract groups. Prove that $G_{\widehat{C}} = A_{\widehat{C}} \bigsqcup B_{\widehat{C}}$.
In this case, C can be considered as a variety of groups finite groups, $(\cdot)_{\widehat{C}}$ means the pro-C completion and $\bigsqcup$ means the free pro-C product.
He gives two hints. First, the obvious hint is to use the universal property. My question arises on the second hint: "use Corollary 3.1.6".
This corollary says that
If $G = L \ast H$ then
(a) The pro-C topology of $G$ induces on $H$ its full pro-C topology
(b) If $G$ is residually C, then $H$ is closed in the pro-C topology of $G$.
There is a lot of other results like this: if $G$ has something hypothesis, then the pro-C topology induces the full pro-C topology on some subgroup.
What is the relevance of this fact? I mean, in the case of Exercise 9.1.1 I think (correct me if I'm wrong) it ensures that the pro-C completion of $A$ and $B$ will be closed subgroups in the pro-C completion of $G$ and this is important to decomposition as free pro-C product.
But I cannot see how is it explicit used to solve this problem. It seems essentially a problem solved with universal property and I cannot see the point where this is relevant.
Trying to be more clear, I would like to know exactly where this corollary is fundamental to solve this exercise. I don't need necessarily a proof of the exercise, I want just want to see why the result is necessary. I think is just to ensure that the pro-C completions of $A$ em $B$ will be pro-C groups but I'm not sure about that.
Indeed, the importance of the corollary in the context of Exercise 9.1.1 is that it ensures that the pro-C completions of $A$ and $B$ will be closed subgroups in the pro-C completion of $G$. This fact is crucial for the decomposition of $G_{\widehat{C}}$ as the free pro-C product of $A_{\widehat{C}}$ and $B_{\widehat{C}}$. Let me explain this more explicitly.
In order to show that $G_{\widehat{C}} = A_{\widehat{C}} \bigsqcup B_{\widehat{C}}$, we need to show that the pro-C completion of $G$ has the desired decomposition, which means that $A_{\widehat{C}}$ and $B_{\widehat{C}}$ should be closed subgroups of $G_{\widehat{C}}$. This is where Corollary 3.1.6 comes into play.
Part (a) of Corollary 3.1.6 says that the pro-C topology of $G$ induces on $H$ (in our case, $A$ and $B$) its full pro-C topology. This implies that the pro-C completion of $A$ and $B$ will be closed subgroups in the pro-C completion of $G$, as we wanted.
Part (b) of Corollary 3.1.6 further states that if $G$ is residually C, then $H$ (in our case, $A$ and $B$) is closed in the pro-C topology of $G$. This fact reinforces the importance of the pro-C completion of $A$ and $B$ being closed subgroups in the pro-C completion of $G$.
Now that we know that $A_{\widehat{C}}$ and $B_{\widehat{C}}$ are closed subgroups of $G_{\widehat{C}}$, we can proceed to show that $G_{\widehat{C}}$ is indeed the free pro-C product of $A_{\widehat{C}}$ and $B_{\widehat{C}}$ using the universal property.
To summarize, the relevance of Corollary 3.1.6 in Exercise 9.1.1 is to ensure that the pro-C completions of $A$ and $B$ will be closed subgroups in the pro-C completion of $G$. This fact is essential for establishing the decomposition of $G_{\widehat{C}}$ as the free pro-C product of $A_{\widehat{C}}$ and $B_{\widehat{C}}$.