description/proof of that for group with topology with continuous operations (especially, topological group), inversion map, multiplication-by-element-from-left-or-right map, and conjugation-by-element map are homeomorphisms
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of topological space.
- The reader knows a definition of continuous map.
- The reader knows a definition of conjugation for group by element.
- The reader knows a definition of homeomorphism.
- The reader admits the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.
- The reader admits the proposition that for any group, the conjugation by any element is a 'groups - homomorphisms' isomorphism.
- The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any finite multiplication map is continuous.
Target Context
- The reader will have a description and a proof of the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.
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:
\(G\): \(\in \{\text{ the groups }\}\) with any topology such that the group operations are continuous
\(g_0\): \(\in G\)
\(f_i\): \(: G \to G, g \mapsto g^{-1}\)
\(f_l\): \(: G \to G, g \mapsto g_0 g\)
\(f_r\): \(: G \to G, g \mapsto g g_0\)
\(f_c\): \(: G \to G, g \mapsto g_0 g {g_0}^{-1}\)
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Statements:
\(f_i, f_l, f_r, f_c \in \{\text{ the homeomorphisms }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f_i\) is a bijection and take the inverse map, \({f_i}^{-1}\); Step 2: see that \(f_i\) and \({f_i}^{-1}\) are continuous; Step 3: see that \(f_l\) is a bijection and take the inverse map, \({f_l}^{-1}\); Step 4: see that \(f_l\) and \({f_l}^{-1}\) are continuous; Step 5: see that \(f_r\) is a bijection and take the inverse map, \({f_r}^{-1}\); Step 6: see that \(f_r\) and \({f_r}^{-1}\) are continuous; Step 7: see that \(f_c\) is a bijection and take the inverse map, \({f_c}^{-1}\); Step 8: see that \(f_c\) and \({f_c}^{-1}\) are continuous.
Step 1:
Let us see that \(f_i\) is a bijection.
Let \(g, g' \in G\) be any such that \(g \neq g'\).
Let us suppose that \(g^{-1} = g'^{-1}\).
\(g' g^{-1} g = g' g'^{-1} g\), so, \(g = g'\), a contradiction.
So, \(g^{-1} \neq g'^{-1}\), so, \(f_i\) is a injection.
Let \(g' \in G\) be any.
\({g'^{-1}}^{-1} = g'\), so, \(f_i\) is a surjection.
So, \(f_i\) is a bijection, so, there is the inverse, \({f_i}^{-1}: G \to G\).
\({f_i}^{-1}: g \mapsto g^{-1}\).
Step 2:
\(f_i\) is continuous, by the supposition.
\({f_i}^{-1}\) is continuous, because it equals \(f_i\).
So, \(f_i\) is a homeomorphism.
Step 3:
Let us see that \(f_l\) is a bijection.
Let \(g, g' \in G\) be any such that \(g \neq g'\).
Let us suppose that \(g_0 g = g_0 g'\).
\({g_0}^{-1} g_0 g = {g_0}^{-1} g_0 g'\), so, \(g = g'\), a contradiction.
So, \(g_0 g \neq g_0 g'\), so, \(f_l\) is a injection.
Let \(g' \in G\) be any.
\(g_0 {g_0}^{-1} g' = g'\), so, \(f_l\) is a surjection.
So, \(f_l\) is a bijection, so, there is the inverse, \({f_l}^{-1}: G \to G\).
\({f_l}^{-1}: g \mapsto {g_0}^{-1} g\).
Step 4:
\(f_l\) is the map induced by fixing the 1st argument of the multiplication map, \(: G \times G \to G\), which is continuous as the multiplication map is continuous, by the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.
\({f_l}^{-1}\) is the map induced by fixing the 1st argument of the multiplication map, \(: G \times G \to G\), which is continuous likewise.
So, \(f_l\) is a homeomorphism.
Step 5:
Let us see that \(f_r\) is a bijection.
Let \(g, g' \in G\) be any such that \(g \neq g'\).
Let us suppose that \(g g_0 = g' g_0\).
\(g g_0 {g_0}^{-1} = g' g_0 {g_0}^{-1}\), so, \(g = g'\), a contradiction.
So, \(g g_0 \neq g' g_0\), so, \(f_r\) is a injection.
Let \(g' \in G\) be any.
\(g' {g_0}^{-1} g_0 = g'\), so, \(f_r\) is a surjection.
So, \(f_r\) is a bijection, so, there is the inverse, \({f_r}^{-1}: G \to G\).
\({f_r}^{-1}: g \mapsto g {g_0}^{-1}\).
Step 6:
\(f_r\) is the map induced by fixing the 2nd argument of the multiplication map, \(: G \times G \to G\), which is continuous as the multiplication map is continuous, by the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.
\({f_r}^{-1}\) is the map induced by fixing the 2nd argument of the multiplication map, \(: G \times G \to G\), which is continuous likewise.
So, \(f_r\) is a homeomorphism.
Step 7:
\(f_c\) is a bijection, by the proposition that for any group, the conjugation by any element is a 'groups - homomorphisms' isomorphism.
So, there is the inverse, \({f_c}^{-1}: G \to G\).
\({f_c}^{-1}: g \mapsto {g_0}^{-1} g g_0\).
Step 8:
The 3-multiplication map, \(: G \times G \times G \to G, (g_1, g_2, g_3) \mapsto g_1 g_2 g_3\), is continuous, by the proposition that for any group with any topology with any continuous operations (especially, topological group), any finite multiplication map is continuous.
\(f_c\) is the map induced by fixing the 1st argument and the 3rd argument of the 3-multiplication map, which is continuous as the 3-multiplication map is continuous, by the proposition that for any continuous map from any product topological space into any topological space, the induced map with any set of some components of the domain fixed is continuous.
\({f_c}^{-1}\) is the map induced by fixing the 1st argument and the 3rd argument of the 3-multiplication map, which is continuous likewise.
So, \(f_c\) is a homeomorphism.
