description/proof of that for group with topology with continuous operations (especially, topological group) and closed subgroup, cosets of subgroup quotient topological space is Hausdorff and classification map is open
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of closed subset of topological space.
- The reader knows a definition of left or right cosets of subgroup quotient topological space of group with topology with continuous operations (especially, topological group).
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of open map.
- The reader admits the proposition that for any group, any subgroup, and the left or right cosets of the subgroup quotient set, the composition of the preimage after the classification map of any subset is the subgroup multiplied by the subset from left or right.
- The reader admits the proposition that for any group, any \(2\) elements, and any subgroup, the left or right cosets of the subgroup by the elements are same if and only if the product of the inverse of an element and the other element or the product of an element and the inverse of the other element is contained in the subgroup.
- The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of \(1\) such that the element multiplied from left by the neighborhood of \(1\) and multiplied from right by the inverse of the neighborhood of \(1\) is contained in the neighborhood of the element.
- The reader admits the proposition that for any group, any \(2\) subsets, and any symmetric subset, if the intersection of the 1st subset and the product of the 2nd subset and the symmetric subset is empty, the intersection of the product of the 1st subset and the symmetric subset and the 2nd subset is empty.
- The reader admits the proposition that for any group, any element, and any possibly uncountable number of subsets, the product of the element and the intersection of the subsets is the intersection of the products of the element and the subsets.
- The reader admits the proposition that for any group, any subgroup, and any \(2\) subsets, if the intersection of the 1st subset and the product of the 2nd subset and the subgroup is empty, the intersection of the product of the 1st subset and the subgroup and the product of the 2nd subset and the subgroup is empty.
Target Context
- The reader will have a description and a proof of the proposition that for any group with topology with continuous operations (especially, topological group) and any closed subgroup, the left or right cosets of the subgroup quotient topological space is Hausdorff and the classification map is open.
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:
\(G'\): \(\in \{\text{ the groups with topologies with continuous operations }\}\)
\(G\): \(\in \{\text{ the closed subgroups of } G'\}\)
\(G' / \sim_{G, l}\): \(= \text{ the left cosets of } G \text{ quotient topological space of } G'\)
\(G' / \sim_{G, r}\): \(= \text{ the right cosets of } G \text{ quotient topological space of } G'\)
\(f_l\): \(: G' \to G' / \sim_{G, l}\), \(= \text{ the classification map }\)
\(f_r\): \(: G' \to G' / \sim_{G, r}\), \(= \text{ the classification map }\)
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Statements:
\(G' / \sim_{G, l}, G' / \sim_{G, r} \in \{\text{ the Hausdorff topological spaces }\}\)
\(\land\)
\(f_l, f_r \in \{\text{ the open maps }\}\)
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2: Proof
Whole Strategy: Step 1: see that for each open \(U' \subseteq G'\), \({f_l}^{-1} \circ f_l (U') = U' G\), and conclude that \(f_l\) is open; Step 2: see that for each open \(U' \subseteq G'\), \({f_r}^{-1} \circ f_r (U') = G U'\), and conclude that \(f_r\) is open; Step 3: for each \(g'_1 G \neq g'_2 G\), see that \({g'_1}^{- 1} g'_2 \in G' \setminus G\), take a symmetric neighborhood of \(1 \in G'\), \(N'_1\), such that \(N'_1 {g'_1}^{- 1} g'_2 N'_1 \subseteq G' \setminus G \), and see that \(f_l (g'_1 N'_1 G)\) and \(f_l (g'_2 N'_1 G)\) are some disjoint nonempty neighborhoods of \(g'_1 G\) and \(g'_2 G\); Step 4: for each \(G g'_1 \neq G g'_2\), see that \(g'_1 {g'_2}^{- 1} \in G' \setminus G\), take a symmetric neighborhood of \(1 \in G'\), \(N'_1\), such that \(N'_1 g'_1 {g'_2}^{- 1} N'_1 \subseteq G' \setminus G\), and see that \(f_r (G N'_1 g'_1)\) and \(f_r (G N'_1 g'_2)\) are some disjoint nonempty neighborhoods of \(G g'_1\) and \(G g'_2\).
Step 1:
Let \(U' \subseteq G'\) be any open subset.
\({f_l}^{-1} \circ f_l (U') = U' G\), by the proposition that for any group, any subgroup, and the left or right cosets of the subgroup quotient set, the composition of the preimage after the classification map of any subset is the subgroup multiplied by the subset from left or right.
\(U' G \subseteq G'\) is open, because it is \(\cup_{g \in G} U' g\) and each \(U' g \subseteq G'\) is open, by 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
Then, \(f_l (U') \subseteq G' / \sim_{G, l}\) is open, by the definition of quotient topology.
So, \(f_l\) is open.
Step 2:
Let \(U' \subseteq G'\) be any open subset.
\({f_r}^{-1} \circ f_r (U') = G U'\), by the proposition that for any group, any subgroup, and the left or right cosets of the subgroup quotient set, the composition of the preimage after the classification map of any subset is the subgroup multiplied by the subset from left or right.
\(G U' \subseteq G'\) is open, because it is \(\cup_{g \in G} g U'\) and each \(g U' \subseteq G'\) is open, by 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.
Then, \(f_r (U') \subseteq G' / \sim_{G, r}\) is open, by the definition of quotient topology.
So, \(f_r\) is open.
Step 3:
Let \(g'_1 G, g'_2 G \in G' / \sim_{G, l}\) be any such that \(g'_1 G \neq g'_2 G\).
\({g'_1}^{- 1} g'_2 \notin G\), by the proposition that for any group, any \(2\) elements, and any subgroup, the left or right cosets of the subgroup by the elements are same if and only if the product of the inverse of an element and the other element or the product of an element and the inverse of the other element is contained in the subgroup.
So, \({g'_1}^{- 1} g'_2 \in G' \setminus G\).
As \(G\) is closed, \(G' \setminus G\) is open, and \(G' \setminus G \subseteq G'\) is an open neighborhood of \({g'_1}^{- 1} g'_2\).
There is a symmetric neighborhood of \(1 \in G'\), \(N'_1 \subseteq G'\), such that \(N'_1 {g'_1}^{- 1} g'_2 N'_1 \subseteq G' \setminus G\), by the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of \(1\) such that the element multiplied from left by the neighborhood of \(1\) and multiplied from right by the inverse of the neighborhood of \(1\) is contained in the neighborhood of the element.
So, \(N'_1 {g'_1}^{- 1} g'_2 N'_1 \cap G = \emptyset\).
\({g'_1}^{- 1} g'_2 N'_1 \cap N'_1 G = \emptyset\), by the proposition that for any group, any \(2\) subsets, and any symmetric subset, if the intersection of the 1st subset and the product of the 2nd subset and the symmetric subset is empty, the intersection of the product of the 1st subset and the symmetric subset and the 2nd subset is empty.
\(g'_1 ({g'_1}^{- 1} g'_2 N'_1 \cap N'_1 G) = g'_1 \emptyset = \emptyset\), but the left hand side is \(g'_1 {g'_1}^{- 1} g'_2 N'_1 \cap g'_1 N'_1 G\), by the proposition that for any group, any element, and any possibly uncountable number of subsets, the product of the element and the intersection of the subsets is the intersection of the products of the element and the subsets, \(= g'_2 N'_1 \cap g'_1 N'_1 G\).
\(g'_2 N'_1 G \cap g'_1 N'_1 G = \emptyset\), by the proposition that for any group, any subgroup, and any \(2\) subsets, if the intersection of the 1st subset and the product of the 2nd subset and the subgroup is empty, the intersection of the product of the 1st subset and the subgroup and the product of the 2nd subset and the subgroup is empty.
\(f_l (g'_1 N'_1 G) \subseteq G' / \sim_{G, l}\) is a neighborhood of \(g'_1 G \in G' / \sim_{G, l}\), because there is an open neighborhood of \(1\), \(U'_1 \subseteq G'\), such that \(U'_1 \subseteq N'_1\), and \(f_l (g'_1 U'_1 G) \subseteq f_l (g'_1 N'_1 G)\) while \(f_l (g'_1 U'_1 G)\) is an open neighborhood of \(g'_1 G\), because \(g'_1 U'_1 G\) is open and \(f_l\) is open and \(g'_1 = g'_1 1 1 \in g'_1 U'_1 G\) and \(f_l (g'_1) = g'_1 G\): \(g'_1 U'_1 G = \cup_{g \in G} g'_1 U'_1 g\), and \(g'_1 U'_1\) is open and \(g'_1 U'_1 g\) is open, by 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.
\(f_l (g'_2 N'_1 G) \subseteq G' / \sim_{G, l}\) is a neighborhood of \(g'_2 G \in G' / \sim_{G, l}\), likewise.
\(f_l (g'_1 N'_1 G) \cap f_l (g'_2 N'_1 G) = \emptyset\), because if \([g'] \in f_l (g'_1 N'_1 G) \cap f_l (g'_2 N'_1 G)\), \(f_l (g''_1) = [g'] = f_l (g''_2)\) for a \(g''_1 \in g'_1 N'_1 G\) and a \(g''_2 \in g'_2 N'_1 G\), which would mean that \(g''_1 G = g''_2 G\), which would imply that \({g''_1}^{-1} g''_2 = g \in G\), by the proposition that for any group, any \(2\) elements, and any subgroup, the left or right cosets of the subgroup by the elements are same if and only if the product of the inverse of an element and the other element or the product of an element and the inverse of the other element is contained in the subgroup, so, \(g''_2 = g''_1 g\), but as \(g''_1 \in g'_1 N'_1 G\), \(g''_2 \in g'_1 N'_1 G\), because while \(g''_1 = g'_1 n'_1 \widetilde{g}\) for an \(n'_1 \in N'_1\) and a \(\widetilde{g} \in G\), \(g''_2 = g''_1 g = g'_1 n'_1 \widetilde{g} g = g'_1 n'_1 (\widetilde{g} g)\) while \(\widetilde{g} g \in G\), a contradiction against \(g'_2 N'_1 G \cap g'_1 N'_1 G = \emptyset\).
So, \(G' / \sim_{G, l}\) is Hausdorff.
Step 4:
Let \(G g'_1, G g'_2 \in G' / \sim_{G, r}\) be any such that \(G g'_1 \neq G g'_2\).
\(g'_1 {g'_2}^{- 1} \notin G\), by the proposition that for any group, any \(2\) elements, and any subgroup, the left or right cosets of the subgroup by the elements are same if and only if the product of the inverse of an element and the other element or the product of an element and the inverse of the other element is contained in the subgroup.
So, \(g'_1 {g'_2}^{- 1} \in G' \setminus G\).
As \(G\) is closed, \(G' \setminus G\) is open, and \(G' \setminus G \subseteq G'\) is an open neighborhood of \(g'_1 {g'_2}^{- 1}\).
There is a symmetric neighborhood of \(1 \in G'\), \(N'_1 \subseteq G'\), such that \(N'_1 g'_1 {g'_2}^{- 1} N'_1 \subseteq G' \setminus G\), by the proposition that for any group with any topology with any continuous operations (especially, topological group), any element, and any neighborhood of the element, there is a symmetric neighborhood of \(1\) such that the element multiplied from left by the neighborhood of \(1\) and multiplied from right by the inverse of the neighborhood of \(1\) is contained in the neighborhood of the element.
So, \(N'_1 g'_1 {g'_2}^{- 1} N'_1 \cap G = \emptyset\).
\(N'_1 g'_1 {g'_2}^{- 1} \cap G N'_1 = \emptyset\), by the proposition that for any group, any \(2\) subsets, and any symmetric subset, if the intersection of the 1st subset and the product of the 2nd subset and the symmetric subset is empty, the intersection of the product of the 1st subset and the symmetric subset and the 2nd subset is empty.
\((N'_1 g'_1 {g'_2}^{- 1} \cap G N'_1) g'_2 = \emptyset g'_2 = \emptyset\), but the left hand side is \(N'_1 g'_1 {g'_2}^{- 1} g'_2 \cap G N'_1 g'_2\), by the proposition that for any group, any element, and any possibly uncountable number of subsets, the product of the element and the intersection of the subsets is the intersection of the products of the element and the subsets, \(= N'_1 g'_1 \cap G N'_1 g'_2\).
\(G N'_1 g'_1 \cap G N'_1 g'_2 = \emptyset\), by the proposition that for any group, any subgroup, and any \(2\) subsets, if the intersection of the 1st subset and the product of the 2nd subset and the subgroup is empty, the intersection of the product of the 1st subset and the subgroup and the product of the 2nd subset and the subgroup is empty.
\(f_r (G N'_1 g'_1) \subseteq G' / \sim_{G, r}\) is a neighborhood of \(G g'_1 \in G' / \sim_{G, r}\), because there is an open neighborhood of \(1\), \(U'_1 \subseteq G'\), such that \(U'_1 \subseteq N'_1\), and \(f_r (G U'_1 g'_1) \subseteq f_r (G N'_1 g'_1)\) while \(f_r (G U'_1 g'_1)\) is an open neighborhood of \(G g'_1\), because \(G U'_1 g'_1\) is open and \(f_r\) is open and \(g'_1 = 1 1 g'_1 \in G U'_1 g'_1\) and \(f_r (g'_1) = G g'_1\): \(G U'_1 g'_1 = \cup_{g \in G} g U'_1 g'_1\), and \(U'_1 g'_1\) is open and \(g U'_1 g'_1\) is open, by 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.
\(f_r (G N'_1 g'_2) \subseteq G' / \sim_{G, r}\) is a neighborhood of \(G g'_2 \in G' / \sim_{G, r}\), likewise.
\(f_r (G N'_1 g'_1) \cap f_r (G N'_1 g'_2) = \emptyset\), because if \([g'] \in f_r (G N'_1 g'_1) \cap f_r (G N'_1 g'_2)\), \(f_r (g''_1) = [g'] = f_r (g''_2)\) for a \(g''_1 \in G N'_1 g'_1\) and a \(g''_2 \in G N'_1 g'_2\), which would mean that \(G g''_1 = G g''_2\), which would imply that \(g''_1 {g''_2}^{-1} = g \in G\), by the proposition that for any group, any \(2\) elements, and any subgroup, the left or right cosets of the subgroup by the elements are same if and only if the product of the inverse of an element and the other element or the product of an element and the inverse of the other element is contained in the subgroup, so, \(g''_1 = g g''_2\), but as \(g''_2 \in G N'_1 g'_2\), \(g''_1 \in G N'_1 g'_2\), because while \(g''_2 = \widetilde{g} n'_1 g'_2\) for an \(n'_1 \in N'_1\) and a \(\widetilde{g} \in G\), \(g''_1 = g g''_2 = g \widetilde{g} n'_1 g'_2 = (g \widetilde{g}) n'_1 g'_2\) while \(g \widetilde{g} \in G\), a contradiction against \(G N'_1 g'_1 \cap G N'_1 g'_2 = \emptyset\).
So, \(G' / \sim_{G, r}\) is Hausdorff.
