description/proof of that for group topological space, if there is 'groups - homomorphisms' isomorphism homeomorphism from space onto topological group, space is topological group
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological group.
- The reader knows a definition of %category name% isomorphism.
- The reader knows a definition of homeomorphism.
- The reader admits the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
- The reader admits the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
- The reader admits the proposition that for any map, the map preimage of the empty subset is the empty set.
Target Context
- The reader will have a description and a proof of the proposition that for any group topological space, if there is any 'groups - homomorphisms' isomorphism homeomorphism from the space onto any topological group, the space is a topological group.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G_1\): \(\in \{\text{ the groups }\} \cap \{\text{ the topological spaces }\}\)
\(G_2\): \(\in \{\text{ the topological groups }\}\)
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Statements:
\(\exists f: G_1 \to G_2 \in \{\text{ the 'groups - homomorphisms' isomorphisms }\} \cap \{\text{ the homeomorphisms }\}\)
\(\implies\)
\(G_1 \in \{\text{ the topological groups }\}\)
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2: Note
Some people may deem this obvious, but the point is that it is about that there is a 'groups - homomorphisms' isomorphism homeomorphism, not about there are a 'groups - homomorphisms' isomorphism and a homeomorphism: the 'groups - homomorphisms' isomorphism needs to be a homeomorphism.
3: Proof
Whole Strategy: Step 1: see that the multiplication and the inversion on \(G_1\) are continuous; Step 2: see that \(G_1\) is Hausdorff; Step 3: conclude the proposition.
Step 1:
Let \(m_1: G_1 \times G_1 \to G_1\) be the multiplication on \(G_1\).
Let \(m_2: G_2 \times G_2 \to G_2\) be the multiplication on \(G_2\).
\(m_1 = f^{- 1} \circ m_2 \circ (f, f)\), because for each \(g_1, g'_1 \in G_1\), \(f^{- 1} \circ m_2 \circ (f, f) (g_1, g'_1) = f^{- 1} \circ m_2 \circ (f (g_1), f (g'_1)) = f^{- 1} (f (g_1) f (g'_1)) = f^{- 1} (f (g_1 g'_1))\), because \(f\) is a group homomorphism, \(= g_1 g'_1 = m_1 (g_1, g'_1)\).
\((f, f)\) is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
\(m_2\) is continuous, because \(G_2\) is a topological group.
\(f^{- 1}\) is continuous, because \(f\) is a homeomorphism
\(m_1 = f^{- 1} \circ m_2 \circ (f, f)\) is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
Let \(i_1: G_1 \to G_1\) be the inversion on \(G_1\).
Let \(i_2: G_2 \to G_2\) be the inversion on \(G_2\).
\(i_1 = f^{- 1} \circ i_2 \circ f\), because for each \(g_1 \in G_1\), \(f^{- 1} \circ i_2 \circ f (g_1) = f^{- 1} (f (g_1)^{- 1}) = f^{- 1} (f (g_1^{- 1}))\), because \(f\) is a group homomorphism, \(= g_1^{- 1} = i_1 (g_1)\).
\(f\) is continuous, because \(f\) is a homeomorphism
\(i_2\) is continuous, because \(G_2\) is a topological group.
\(f^{- 1}\) is continuous, because \(f\) is a homeomorphism
\(i_1 = f^{- 1} \circ i_2 \circ f\) is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
Step 2:
Let us see that \(G_1\) is Hausdorff.
Let \(g_1, g'_1 \in G_1\) be any such that \(g_1 \neq g'_1\).
\(f (g_1), f (g'_1) \in G_2\) such that \(f (g_1) \neq f (g'_1)\), because \(f\) is a bijection.
There are some open neighborhoods of \(f (g_1), f (g'_1)\), \(U_{f (g_1)}, U_{f (g'_1)} \subseteq G_2\), such that \(U_{f (g_1)} \cap U_{f (g'_1)} = \emptyset\), because \(G_2\) is Hausdorff.
\(f^{- 1} (U_{f (g_1)}), f^{- 1} (U_{f (g'_1)}) \subseteq G_1\) are some open neighborhoods of \(g_1, g'_1\), because \(f\) is a homeomorphism, and \(f^{- 1} (U_{f (g_1)}) \cap f^{- 1} (U_{f (g'_1)}) = f^{- 1} (U_{f (g_1)} \cap U_{f (g'_1)})\), by the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets, \(= f^{- 1} (\emptyset) = \emptyset\), by the proposition that for any map, the map preimage of the empty subset is the empty set.
So, \(G_1\) is Hausdorff.
Step 3:
By Step 1 and Step 2, \(G_1\) is a topological group.
