I heard that $\operatorname{Spec}(\bar{\mathbb{Q}}\otimes_\mathbb{Q} \bar{\mathbb{Q}})$ is homeomorphic to $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ and so I tried to prove it in the following way:
We write $\bar{\mathbb{Q}}$ as the colimit of its finite Galois extensions. Then, $$\bar{\mathbb{Q}}\otimes_\mathbb{Q} \bar{\mathbb{Q}}=\bar{\mathbb{Q}}\otimes_\mathbb{Q} \left(\operatorname{colim} L\right)=\operatorname{colim}\bar{\mathbb{Q}}\otimes_\mathbb{Q} L = \operatorname{colim}\prod_{\sigma\in\operatorname{Gal}(L/\mathbb{Q})}\bar{\mathbb{Q}}.$$ It follows that $$\operatorname{Spec}(\bar{\mathbb{Q}}\otimes_\mathbb{Q} \bar{\mathbb{Q}})=\operatorname{Spec}\left(\operatorname{colim}\prod_{\sigma\in\operatorname{Gal}(L/\mathbb{Q})}\bar{\mathbb{Q}}\right)=\lim \coprod_{\sigma\in\operatorname{Gal}(L/\mathbb{Q})}\operatorname{Spec}\bar{\mathbb{Q}}.$$ Now, let $F$ be the forgetful functor from schemes to topological spaces. We have $$F(\operatorname{Spec}(\bar{\mathbb{Q}}\otimes_\mathbb{Q} \bar{\mathbb{Q}}))=F\left(\lim \coprod_{\sigma\in\operatorname{Gal}(L/\mathbb{Q})}\operatorname{Spec}\bar{\mathbb{Q}}\right)\overset{?}{=}\lim F\left( \coprod_{\sigma\in\operatorname{Gal}(L/\mathbb{Q})}\operatorname{Spec}\bar{\mathbb{Q}}\right)=\lim \operatorname{Gal}(L/\mathbb{Q})=\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$$ since the limits on the categories of topological spaces and of topological groups coincide (as topological spaces, of course). The problem is that I think $F$ does not preserve limits.
Can this proof be saved?
(I know that there is a similar proof of this result in the MSE, but would like to understand this one better.)
