Did I do this correct? Suppose p: X → Y is a surjective map from a topological space X to a set Y. Verifythat the quotient topology on Y defined to be: {V S Y|p (V) is open in X} is atopology.Definition:If X is a space and A is a set and if p: X → A is a surjective map, then there existsexactly one topology T on A relative to which p is a quotient map; it is called thequotient topology induced by p.Proof:Let J = {V C Y\p¬'(V) is open in X}.1) p-1(Y) = X is open in X.Therefore, Y E JT.p-²(Ø) = Ø is open in X.Therefore, Ø E T.2) Let U,V E T, then p-(U) and p-(V) are open in X.Then, p-1(U) n p-'(V) is open in X.Then, p-(UnV) = p-'(U) np-(V) is open in X.Therefore, Un VET.3) Let V, E T, where j€ J is an index set.Then, p-1(V;) is open in X.Then, p-(Uje, V;) = UjejP¯(V,).Therefore, Ujej V, ET.Therefore, the quotient topology on Y defined by T is a topology.

Answered: Image /qna-images/answer/6676b913-8bde-48c5-a8f5-2b5cf330608a.jpg

Answered: Suppose p: X → Y is a surjective map…

Suppose p: X → Y is a surjective map from a topological space X to a set Y. Verify that the quotient topology on Y defined to be: {V S Y|p(V) is open in X} is a topology.

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter5: Rings, Integral Domains, And Fields

Section5.4: Ordered Integral Domains

Problem 1E: Complete the proof of Theorem 5.30 by providing the following statements, where and are arbitrary...

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Question

Did I do this correct?

Suppose p: X → Y is a surjective map from a topological space X to a set Y. Verify
that the quotient topology on Y defined to be: {V S Y|p (V) is open in X} is a
topology.
Definition:
If X is a space and A is a set and if p: X → A is a surjective map, then there exists
exactly one topology T on A relative to which p is a quotient map; it is called the
quotient topology induced by p.
Proof:
Let J = {V C Y\p¬'(V) is open in X}.
1) p-1(Y) = X is open in X.
Therefore, Y E JT.
p-²(Ø) = Ø is open in X.
Therefore, Ø E T.
2) Let U,V E T, then p-(U) and p-(V) are open in X.
Then, p-1(U) n p-'(V) is open in X.
Then, p-(UnV) = p-'(U) np-(V) is open in X.
Therefore, Un VET.
3) Let V, E T, where j€ J is an index set.
Then, p-1(V;) is open in X.
Then, p-(Uje, V;) = UjejP¯(V,).
Therefore, Ujej V, ET.
Therefore, the quotient topology on Y defined by T is a topology.

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