description/proof of that finite product of open quotient maps is open quotient
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of open map.
- The reader knows a definition of quotient map.
- The reader knows a definition of product topological space.
- The reader knows a definition of product map.
- The reader admits the universal property of quotient map: any surjection between topological spaces is a quotient map if and only if any additional map from the codomain of the original map to any additional topological space is continuous if and only if the composition of the additional map after the original map is continuous.
- 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 any finite product of open maps is open.
- 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 maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
- The reader admits the proposition that for any map between any sets, the composition of the map after the preimage of any subset of the codomain is identical if the map is surjective with respect to the argument subset.
- The reader admits the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.
- The reader admits the proposition that for any product map, the image of any product subset is the product of the images of the component subsets under the component maps.
Target Context
- The reader will have a description and a proof of the proposition that the product of any finite number of open quotient maps is open quotient.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{T_{j, 1} \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\{T_{j, 2} \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\{f_j: T_{j, 1} \to T_{j, 2} \in \{\text{ the open quotient maps }\} \vert j \in J\}\):
\(\times_{j \in J} T_{j, 1}\): \(= \text{ the product topological space }\)
\(\times_{j \in J} T_{j, 2}\): \(= \text{ the product topological space }\)
\(\times_{j \in J} f_j\): \(: \times_{j \in J} T_{j, 1} \to \times_{j \in J} T_{j, 2}\)
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Statements:
\(\times_{j \in J} f_j \in \{\text{ the open quotient maps }\}\)
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2: Note
This proposition requires each \(f_j\) to be open to prove \(\times_{j \in J} f_j\) quotient: the openness of each \(f_j\) is not required just in order to prove \(\times_{j \in J} f_j\) open but in order to prove \(\times_{j \in J} f_j\) quotient.
3: Proof
Whole Strategy: apply the universal property of quotient map: any surjection between topological spaces is a quotient map if and only if any additional map from the codomain of the original map to any additional topological space is continuous if and only if the composition of the additional map after the original map is continuous; Step 1: see that \(\times_{j \in J} f_j\) is an open continuous surjection, let \(g: \times_{j \in J} T_{j, 2} \to T_3\) be any map, and see that if \(g\) is continuous, \(g \circ (\times_{j \in J} f_j)\) is continuous; Step 2: see that if \(g \circ (\times_{j \in J} f_j)\) is continuous, \(g\) is continuous; Step 3: conclude the proposition.
Step 1:
\(\times_{j \in J} f_j\) is a surjection, because for each \(\times_{j \in J} t_{j, 2} \in \times_{j \in J} T_{j, 2}\), for each \(j \in J\), there is a \(t_{j, 1} \in T_{j, 1}\) such that \(f_j (t_{j, 1}) = t_{j, 2}\), because \(f_j\) is a surjection, and there is the \(\times_{j \in J} t_{j, 1} \in \times_{j \in J} T_{j, 1}\), and \(\times_{j \in J} f_j (\times_{j \in J} t_{j, 1}) = \times_{j \in J} f_j (t_{j, 1}) = \times_{j \in J} t_{j, 2}\).
\(\times_{j \in J} f_j\) is continuous, by the proposition that the product map of any finite number of continuous maps is continuous by the product topologies.
\(\times_{j \in J} f_j\) is open, by the proposition that any finite product of open maps is open.
Let \(g: \times_{j \in J} T_{j, 2} \to T_3\) be any map where \(T_3\) is any topological space.
If \(g\) is continuous, \(g \circ (\times_{j \in J} f_j)\) 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 suppose that \(g \circ (\times_{j \in J} f_j)\) is continuous.
Let us see that \(g\) is continuous.
Let \(U \subseteq T_3\) be any open subset.
\((g \circ (\times_{j \in J} f_j))^{-1} (U) \subseteq \times_{j \in J} T_{j, 1}\) is open, because \(g \circ (\times_{j \in J} f_j)\) is continuous, \(= \cup_{l \in L} \times_{j \in J} U_{j, l}\) where \(L\) is a possibly uncountable index set, \(U_{j, l} \subseteq T_{j, 1}\) is an open subset, by Note for the definition of product topology.
\((g \circ (\times_{j \in J} f_j))^{-1} (U) = (\times_{j \in J} f_j)^{-1} (U) (g^{-1} (U))\), by the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
\((\times_{j \in J} f_j) ((\times_{j \in J} f_j)^{-1} (U) (g^{-1} (U))) = g^{-1} (U)\), by the proposition that for any map between any sets, the composition of the map after the preimage of any subset of the codomain is identical if the map is surjective with respect to the argument subset: \(\times_{j \in J} f_j\) is surjective with respect to \(g^{-1} (U)\), because \(\times_{j \in J} f_j\) is a surjection.
\(= (\times_{j \in J} f_j) (\cup_{l \in L} \times_{j \in J} U_{j, l}) = \cup_{l \in L} (\times_{j \in J} f_j) (\times_{j \in J} U_{j, l})\), by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets, \(= \cup_{l \in L} \times_{j \in J} f_j (U_{j, l})\), by the proposition that for any product map, the image of any product subset is the product of the images of the component subsets under the component maps.
As each \(f_j\) is open, each \(f_j (U_{j, l})\) is open, each \(\times_{j \in J} f_j (U_{j, l})\) is open, by Note for the definition of product topology, and \(\cup_{l \in L} \times_{j \in J} f_j (U_{j, l})\) is open as a union of some open subsets.
So, \(g^{-1} (U)\) is open.
So, \(g\) is continuous.
Step 3:
So, by the universal property of quotient map: any surjection between topological spaces is a quotient map if and only if any additional map from the codomain of the original map to any additional topological space is continuous if and only if the composition of the additional map after the original map is continuous, \(\times_{j \in J} f_j\) is quotient.
