Meaning of epimorphism from full Galois group to some group

Question

My problem has two parts: let $\;G:=\operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q)\;$ be the full Galois group of the rationals and $\;K\;$ be some finite group, then:

(1) Does having an epimorphism (of groups, in the meanwhile) $\;G\to K\;$ means there exists a finite Galois fields extension $\;F/\Bbb Q\;$ such that $\;K=\operatorname{Gal}(F/\Bbb Q)\;$ ? I sense this must be true, but I cannot justify it clearly, maybe because I still don't understand completely the concept of profinite groups;

1') As above, but requiring the surjection to be of topological groups. I'm not sure whether this makes any actual difference as $\;K\;$ is finite, though. As I mentioned, I still don't fully grasp the concept of profinite groups, so maybe (1) above is true only if the morphism includes the topological part;

2) Now, my actual problem is to fully justify that there can't be a factoring of the epimorphism $\;f: G\to \Bbb Z/2\Bbb Z\;$, when we identify the latter group with $\operatorname{Gal}(\Bbb Q(i)/\Bbb Q)\;$, through $\;C_4=\Bbb Z/4\Bbb Z\;$ .

My idea, which I cannot make formal because of (1-1') above, is that such a factoring would imply that there exists a cyclic extension of order four of $\;\Bbb Q\;$ which contains $\;\Bbb Q(i) \;$ and thus contains $\;i\;$ , which is impossible...

I think fully understanding (1)-(1') would in fact solve, or almost, my problem (2). Any ideas will be sincerely appreciated.

Answer 1

If your map is continuous, its kernel is normal and closed, hence of the form $Gal(\overline{\mathbb{Q}}/F)$, where $F/\mathbb{Q}$ is finite Galois (I think you can find all the results you need and their proofs in the book of Pat Morandi,"Fields and Galois theory") Hence, you get $K\simeq Gal(\overline{\mathbb{Q}}/\mathbb{Q})/Gal(\overline{\mathbb{Q}}/F)\simeq Gal(F/\mathbb{Q})$.

Edited after Whatsup comment.

Answer 2

I understand question 2 as: Show that there is no Galois extension $F$ of $\mathbb{Q}$ with Galois group $C_4$ and such that $F$ contains $\mathbb{Q}(i)$. This is a special case of a general cohomological fact: A quadratic extension $K(\sqrt a)$ of a field $K$ of characteristic not $2$ embeds in a $C_4$-Galois extension if and only if $a$ is a sum of two squares in $K$ (see for instance J.-P. Serre, Topics in Galois Theory, Theorem 1.2.4); Another useful equivalent condition is that the quaternion algebra $(a,-1/K)$ splits. Obviously, $-1$ is not a sum of two squares in $\mathbb{Q}$.