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In the following one is referred to the book.
At this page, the author defines the Krull topology on a Galois group (not necessarily finite); on the 22nd page of the same book, the author defines the product topology on a profinite group.
Since a Galois group is also a profinite group, and as these two topologies are actually the same, a fact the author chooses to leave as an exercise, one asks why the two topologies coincide. More precisely,

Let $G={\rm Gal}(\Omega/k)$ be a Galois group, which is also a profinite group $\varprojlim ({\rm Gal}(\Omega/L))$, where $\Omega$ is a Galois extension, and $L$ is a finite Galois extension, of $k$. Then why are the Krull topology on $G$, and the product topology on $G$, viewed as a profinite group, the same?

I have pondered upon the relations between these two topologies for some time, but cannot explicitly formulate an isomorphism, or rather, the reason why the map is an isomorphism. So please help in clarifying their relations.
Thanks in advance.

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I believe this page, under "Inverse limit structure", describes the homeomorphism: planetmath.org/InfiniteGaloisTheory.html –  Adeel Nov 4 '12 at 11:06
    
Thanks for the link: I know this map in advance, and the question is: why is that map bicontinuous? I think this is the essence of the construction, but failed to prove it. Thanks again. –  awllower Nov 4 '12 at 11:24
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When you are trying to prove two topologies are the same, you could show a basic open set in one topology is open in the other topology (or the closed sets). Since we're dealing with topologies that are compact Hausdorff, recall too that a continuous bijection of compact Hausdorff spaces is a homeomorphism; you might try that idea on the identity map $G \rightarrow G$ where $G$ has a differently defined topology on the left and the right. Then the identity being a homeomorphism means the topologies are equal. –  KCd May 20 at 19:19

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