Is U the Quotient Topology for Continuous Functions between Topological Spaces?

[beetle2]

Let (X; T ) be a topological space. Given the set Y and the function f : X \rightarrow Y, define

U := {H\inY \mid f^{-1}(H)\in T}

Show that U is the finest topology on Y with respect to which f is continuous.

Homework Equations

The Attempt at a Solution

I was wondering is this implying that U is the Quotient topology?

 

[Dick]

No, it's not a quotient topology. You aren't identifying any points. Suppose you give Y a finer topology than U? Use the topological definition of continuity.

 

[beetle2]

Thanks

 

[Hurkyl]

You aren't identifying any points.

Yes he is, via the equivalence relation P~Q iff f(P)=f(Q).

 

[beetle2]

So if it is a quotient topology will I need to prove the iff statement

 

[beetle2]

I mean do I have to show that if there is an equivalence relation I need to show that there is a function which is a homeomorhism.

 

[Dick]

Yes he is, via the equivalence relation P~Q iff f(P)=f(Q).

What is ~ supposed to mean?! f:R->R with the usual topology. f(x)=x^2. You are defining a topology on Y. Of course, f(1)=f(-1). How is 1~(-1)?

 

[Hurkyl]

What is ~ supposed to mean?!

~ is a symbol denoting an equivalence relation, which I subsequently defined.

For any function f : X -> Y, Ker(f) is the set of pairs (a,b) for which f(a)=f(b).

Theorem (First isomorphism theorem): Ker(f) is an equivalence relation on X. The corresponding set of equivalence classes is canonically bijective with the image of f


For any surjective continuous function f:X -> Y, it's fair to ask if the canonical bijection X/Ker(f) -> Y is a homeomorphism, in which case it would be fair to say that Y has the quotient topology.

 

[Dick]

~ is a symbol denoting an equivalence relation, which I subsequently defined.

For any function f : X -> Y, Ker(f) is the set of pairs (a,b) for which f(a)=f(b).

Theorem (First isomorphism theorem): Ker(f) is an equivalence relation on X. The corresponding set of equivalence classes is canonically bijective with the image of f


For any surjective continuous function f:X -> Y, it's fair to ask if the canonical bijection X/Ker(f) -> Y is a homeomorphism, in which case it would be fair to say that Y has the quotient topology.

Ok, so it's a quotient in some sense on XxY if I get your drift. I still think this is a bit of a distraction from the gloriously simple approach of just picking a topology strictly finer than U and then showing f is not continuous in that topology.

 

[Hurkyl]

I wasn't trying to suggest an approach to the problem; I just didn't want him to be misinformed about quotient topologies. (Incidentally, for the OP, the definition Wikipedia gives of "quotient topology" is exactly the topology you wrote... with the extra condition that f is supposed to be surjective)

 

[boneill3]

Thanks for your help guys. In my textbook is says that the quotient topology is the finest. However in my tutorials when ever they talked about quotient topologies they always mentioned something about an equivalence relation. I think this question was meant to make me think about it which I did. Once again thanks

 

[Dick]

I wasn't trying to suggest an approach to the problem; I just didn't want him to be misinformed about quotient topologies. (Incidentally, for the OP, the definition Wikipedia gives of "quotient topology" is exactly the topology you wrote... with the extra condition that f is supposed to be surjective)

I see what you mean now. Thanks for clarifying.