333: Subset of Subspace of Adjunction Topological Space Is Open Iff Projections of Preimage of Subset Are Open with Condition|T.B.P.

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A description/proof of that subset of subspace of adjunction topological space is open iff projections of preimage of subset are open with condition.

Topics

About: topological space

Starting Context

The reader knows a definition of adjunction topological space obtained by attaching topological space via continuous map to topological space.
The reader knows a definition of subspace topology.
The reader admits the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
The reader admits the proposition that for any map between sets, any subset is contained in the preimage of the image of the subset.

Target Context

The reader will have a description and a proof of the proposition that for any adjunction topological space and its any subspace, any subset of the subspace is open if and only if the projections of the preimage of the subset under the quotient map onto the attaching-origin space and the attaching-destination space are open on the projections of the preimage of the subspace under the quotient map with the condition that the attaching-origin space projection accord with the attaching-destination space projection with respect to the attaching map.

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: Description

For any topological spaces,

T_{1}, T_{2}

, any subset,

S \subseteq T_{1}

, any continuous map,

f : S \to T_{2}

, the adjunction topological space,

T_{2} \cup_{f} T_{1}

where

f_{2} : T_{1} \to T_{1} + T_{2}

and

f_{3} : T_{2} \to T_{1} + T_{2}

are the inclusions and

f_{4} : T_{1} + T_{2} \to T_{2} \cup_{f} T_{1}

is the quotient map, and any subspace,

T_{3} \subseteq T_{2} \cup_{f} T_{1}

, any subset,

S^{'} \subseteq T_{3}

, is open if and only if

{f_{4}}^{- 1} (S^{'}) \cap T_{1}

is open on the subspace of

T_{1}

{T_{1}}^{'} = {f_{4}}^{- 1} (T_{3}) \cap T_{1}

, with an open set,

U^{'} \subseteq T_{1}

, such that

{f_{4}}^{- 1} (S^{'}) \cap T_{1} = U^{'} \cap {T_{1}}^{'}

and

{f_{4}}^{- 1} (S^{'}) \cap T_{2}

is open on the subspace of

T_{2}

{T_{2}}^{'} = {f_{4}}^{- 1} (T_{3}) \cap T_{2}

, with an open set,

U^{″} \subseteq T_{2}

, such that

{f_{4}}^{- 1} (S^{'}) \cap T_{2} = U^{″} \cap {T_{2}}^{'}

, with the condition that

U^{'} \cap S = f^{- 1} (U^{″} \cap f (S))

2: Proof

Suppose that

{f_{4}}^{- 1} (S^{'}) \cap T_{1}

is open on

{T_{1}}^{'}

and

{f_{4}}^{- 1} (S^{'}) \cap T_{2}

is open on

{T_{2}}^{'}

with the prescribed condition. Let us think of

U^{‴} := f_{4} (f_{2} (U^{'}) \cup f_{3} (U^{″}))

. The condition,

U^{'} \cap S = f^{- 1} (U^{″} \cap f (S))

, means that for any

[p] \in U^{‴}

such that

p \in f (S), [p] = f^{- 1} (p) \cup {p}

f^{- 1} (p) \cup {p} \subseteq f_{2} (U^{'} \cap S) \cup f_{3} (U^{″} \cap f (S))

, because there is a

p^{'} \in f_{2} (U^{'} \cap S) \cup f_{3} (U^{″} \cap f (S))

such that

f_{4} (p^{'}) = [p]

, which means that

p^{'} \in f_{2} (U^{'} \cap S)

and

f (p^{'}) = p

or that

p^{'} \in f_{3} (U^{″} \cap f (S))

and

p^{'} = p

; for the former case,

p^{'} \in U^{'} \cap S = f^{- 1} (U^{″} \cap f (S))

f (p^{'}) = p \in f (f^{- 1} (U^{″} \cap f (S))) \subseteq U^{″} \cap f (S)

by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set; for the latter case,

p = p^{'} \in U^{″} \cap f (S)

; for any

p^{″} \in f^{- 1} (p) \cup {p}

p^{″} \in f^{- 1} (p)

p^{″} = p

; for the former case,

p^{″} \in U^{'} \cap S = f_{2} (U^{'} \cap S)

by the condition; for the latter case,

p^{″} \in U^{″} \cap f (S) = f_{3} (U^{″} \cap f (S))

.

The openness of

U^{‴}

T_{2} \cup_{f} T_{1}

is nothing but the openness of

{f_{4}}^{- 1} (U^{‴}) \cap T_{1}

T_{1}

and the openness of

{f_{4}}^{- 1} (U^{‴}) \cap T_{2}

T_{2}

{f_{4}}^{- 1} (U^{‴}) = f_{2} (U^{'}) \cup f_{3} (U^{″})

, because while

f_{2} (U^{'}) \cup f_{3} (U^{″}) \subseteq {f_{4}}^{- 1} (U^{‴})

is true by the proposition that for any map between sets, any subset is contained in the preimage of the image of the subset, also

{f_{4}}^{- 1} (U^{‴}) \subseteq f_{2} (U^{'}) \cup f_{3} (U^{″})

holds, because for any

p \in {f_{4}}^{- 1} (U^{‴})

, if

p \in T_{1} ∖ S

p \in T_{2} ∖ f (S)

f_{4} (p) = [p] = {p} \in U^{‴}

, and there is a

p^{'} \in f_{2} (U^{'}) \cup f_{3} (U^{″})

such that

f_{4} (p^{'}) = [p]

, which only

p^{'} = p

satisfies, so,

p \in f_{2} (U^{'}) \cup f_{3} (U^{″})

; if

p \in S

p \in f (S)

f_{4} (p) = [p^{'}] \in U^{‴}

where

p^{'} \in f (S), [p^{'}] = f^{- 1} (p^{'}) \cup {p^{'}}

, so, by the claim in the last paragraph,

p \in f^{- 1} (p^{'}) \cup {p^{'}} \subseteq f_{2} (U^{'} \cap S) \cup f_{3} (U^{″} \cap f (S)))

. So,

{f_{4}}^{- 1} (U^{‴}) \cap T_{1} = f_{2} (U^{'})

, open on

T_{1}

, and

{f_{4}}^{- 1} (U^{‴}) \cap T_{2} = f_{3} (U^{″})

, open on

T_{2}

. So,

U^{‴}

is open on

T_{2} \cup_{f} T_{1}

S^{'} = U^{‴} \cap T_{3}

, because for any

[p] \in S^{'}

, if

p \in T_{1}

p \in {f_{4}}^{- 1} (S^{'}) \cap T_{1} = U^{'} \cap {T_{1}}^{'} \subseteq U^{'}

, so,

f_{4} (f_{2} (p)) = [p] \in U^{‴} = f_{4} (f_{2} (U^{'}) \cup f_{3} (U^{″}))

; if

p \in T_{2}

p \in {f_{4}}^{- 1} (S^{'}) \cap T_{2} = U^{″} \cap {T_{2}}^{'} \subseteq U^{″}

, so,

f_{4} (f_{3} (p)) = [p] \in U^{‴} = f_{4} (f_{2} (U^{'}) \cup f_{3} (U^{″}))

; of course

[p] \in T_{3}

; for any

[p] \in U^{‴} \cap T_{3}

, if

p \in T_{1}

p \in ({f_{4}}^{- 1} (U^{‴}) \cap T_{1}) \cap ({f_{4}}^{- 1} (T_{3}) \cap T_{1}) = f_{2} (U^{'}) \cap {T_{1}}^{'} = U^{'} \cap {T_{1}}^{'} = {f_{4}}^{- 1} (S^{'}) \cap T_{1}

, so,

[p] = f_{4} (p) \in S^{'}

; if

p \in T_{2}

p \in ({f_{4}}^{- 1} (U^{‴}) \cap T_{2}) \cap ({f_{4}}^{- 1} (T_{3}) \cap T_{2}) = f_{3} (U^{″}) \cap {T_{2}}^{'} = U^{″} \cap {T_{2}}^{'} = {f_{4}}^{- 1} (S^{'}) \cap T_{2}

, so,

[p] = f_{4} (p) \in S^{'}

.

So, by the definition of subspace topology,

S^{'}

is open.

Suppose that

S^{'}

is open. By the definition of subspace topology, there is an open set,

U^{‴} \subseteq T_{2} \cup_{f} T_{1}

, such that

S^{'} = U^{‴} \cap T_{3}

.

Define

U^{'} := {f_{4}}^{- 1} (U^{‴}) \cap T_{1}

and

U^{″} := {f_{4}}^{- 1} (U^{‴}) \cap T_{2}

, both open by the definition of quotient topology and the definition of topological sum. They satisfy the condition,

U^{'} \cap S = f^{- 1} (U^{″} \cap f (S))

, because for any

p \in U^{'} \cap S

f_{4} (p) = [f (p)] \in U^{‴}

{f_{4}}^{- 1} ([f (p)]) \cap T_{2} = {f (p)} \subseteq U^{″}

, so,

f (p) \in U^{″}

, so,

p \in f^{- 1} (U^{″}) = f^{- 1} (U^{″} \cap f (S))

; for any

p \in f^{- 1} (U^{″} \cap f (S))

f (p) \in U^{″} \cap f (S)

f_{4} (f (p)) = [p] \in U^{‴}

, so,

p \in {f_{4}}^{- 1} ([p]) \subseteq {f_{4}}^{- 1} (U^{‴}) \cap T_{1} = U^{'}

, but as

p \in S

p \in U^{'} \cap S

{f_{4}}^{- 1} (S^{'}) \cap T_{1} = U^{'} \cap {T_{1}}^{'}

, because for any

p \in {f_{4}}^{- 1} (S^{'}) \cap T_{1}

f_{4} (p) = [p] \in S^{'} \subseteq U^{‴}

, so,

p \in {f_{4}}^{- 1} (U^{‴})

, but as

p \in T_{1}

p \in {f_{4}}^{- 1} (U^{‴}) \cap T_{1} = U^{'}

, and of course,

p \in {T_{1}}^{'}

; for any

p \in U^{'} \cap {T_{1}}^{'}

f_{4} (p) \in f_{4} (U^{'} \cap {T_{1}}^{'}) \subseteq f_{4} (U^{'}) \cap f_{4} ({T_{1}}^{'}) \subseteq U^{‴} \cap T_{3} = S^{'}

p \in {f_{4}}^{- 1} (S^{'}) \cap T_{1}

{f_{4}}^{- 1} (S^{'}) \cap T_{2} = U^{″} \cap {T_{2}}^{'}

, because for any

p \in {f_{4}}^{- 1} (S^{'}) \cap T_{2}

f_{4} (p) = [p] \in S^{'} \subseteq U^{‴}

;

p \in {f_{4}}^{- 1} (U^{‴}) \cap T_{2} = U^{″}

; of course,

p \in {T_{2}}^{'}

; for any

p \in U^{″} \cap {T_{2}}^{'}

f_{4} (p) \in f_{4} (U^{″} \cap {T_{2}}^{'}) \subseteq f_{4} (U^{″}) \cap f_{4} ({T_{2}}^{'}) \subseteq U^{‴} \cap T_{3} = S^{'}

p \in {f_{4}}^{- 1} (S^{'}) \cap T_{2}

U^{‴} = f_{4} (f_{2} (U^{'}) \cup f_{3} (U^{″}))

, because for any

[p] \in U^{‴}

p \in {f_{4}}^{- 1} ([p]) \cap T_{1} \subseteq U^{'}

p \in {f_{4}}^{- 1} ([p]) \cap T_{2} \subseteq U^{″}

, so,

[p] \in f_{4} (f_{2} (U^{'}) \cup f_{3} (U^{″}))

; for any

[p] \in f_{4} (f_{2} (U^{'}) \cup f_{3} (U^{″}))

[p] \in f_{4} (f_{2} (U^{'}))

[p] \in f_{4} (f_{3} (U^{″}))

, so, there is a

p \in U^{'}

such that

f_{4} (f_{2} (p)) = [p]

or a

p \in U^{″}

such that

f_{4} (f_{3} (p)) = [p]

, so,

[p] \in U^{‴}

3: Note

The point is that only the openness of

{f_{4}}^{- 1} (S^{'}) \cap T_{1}

{T_{1}}^{'}

and the openness of

{f_{4}}^{- 1} (S^{'}) \cap T_{2}

{T_{2}}^{'}

do not guarantee the openness of

S^{'}

, because

U^{‴}

T_{2} \cup_{f} T_{1}

is being required.

References

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2022-08-14