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Proof that Galois theory covering space

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Galois theory is a branch of abstract algebra that studies field extensions and their automorphisms. One of the key concepts in Galois theory is the notion of a covering space, which is a topological space that maps onto another space in a continuous and surjective way.

To prove that Galois theory covers space, we can consider the fundamental theorem of Galois theory, which states that there is a one-to-one correspondence between intermediate fields of a Galois extension and subgroups of the Galois group of the extension. This correspondence allows us to understand the structure of the field extension in terms of the symmetries of its automorphisms.

Furthermore, the Galois correspondence also tells us that the fixed field of a subgroup of the Galois group corresponds to the subfield fixed by the automorphisms in that subgroup. This means that the intermediate fields of a Galois extension can be understood as the fibers of a covering space, where each fiber corresponds to a subgroup of the Galois group.

Overall, the concept of covering space in Galois theory provides a powerful tool for understanding the structure of field extensions and their automorphisms. By studying the symmetries of the Galois group and their corresponding subfields, we can gain insight into the algebraic properties of the extension and its relationship to the base field.
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Theorem: If $$G$$ is a finite group of automorphisms of a covering space $$X \rightarrow Y$$, then the Galois group of the covering space is isomorphic to a subgroup of $$G$$.

Proof: Let $$p: X \rightarrow Y$$ be a covering space with group of covering transformations $$G$$. Consider the projection map $$\pi : G \rightarrow Aut(X)$$, where $$\pi(\sigma)(x) = \sigma(x)$$ for all $$\sigma \in G$$ and $$x \in X$$.

Claim 1: $$\pi(\sigma)$$ is an automorphism of $$X$$ for all $$\sigma \in G$$.

Proof of Claim 1: $$\pi(\sigma)$$ is clearly bijective since $$\sigma$$ is bijective. To show that $$\pi(\sigma)$$ is a homeomorphism, it is enough to show that $$\pi(\sigma)$$ is continuous. But this is clear since $$\sigma$$ is a homeomorphism.

Claim 2: The kernel of $$\pi$$ is the group of covering transformations.

Proof of Claim 2: If $$\pi(\sigma) = id_X$$, then $$\sigma(x) = x$$ for all $$x \in X$$. Thus, $$\sigma$$ is a covering transformation. Conversely, if $$\sigma$$ is a covering transformation, then $$\sigma(x) = x$$ for all $$x \in X$$. Thus, $$\pi(\sigma) = id_X$$.

Claim 3: The image of $$\pi$$ is the group of deck transformations.

Proof of Claim 3: If $$\pi(\sigma) = \tau$$ is a deck transformation, then $$\tau(x) = \sigma(x)$$ for all $$x \in X$$. Thus, $$\sigma$$ is a covering transformation that takes the fiber over $$y \in Y$$ to the fiber over $$\tau(y)$$. Conversely, if $$\sigma$$ is a covering transformation that takes the fiber over $$y \in Y$$ to the fiber over $$\tau(y)$$, then $$\pi(\sigma) = \tau$$.

By Claims 1, 2, and 3, we have the following exact sequence:
$$1 \rightarrow \text{Covering transformations} \rightarrow G \rightarrow \text{Deck transformations} \rightarrow 1$$

Thus, the Galois group of the covering space, which is isomorphic to the group of deck transformations, is isomorphic to a subgroup of the group of covering transformations, which is itself isomorphic to $$G$$.
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Collins Benton
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