Continuous functions in topology

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Homework Help Overview

The discussion revolves around the concept of continuity in topology, specifically examining the implications of the definition of a continuous function between topological spaces. The original poster questions whether continuity necessitates that a function be surjective, based on the requirement that the pre-image of every open set in the codomain must also be open in the domain.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reason through the definition of continuity and its implications for surjectivity, questioning the validity of their logic. Some participants suggest constructing counterexamples to explore the original poster's claim further.

Discussion Status

Participants are actively engaging with the original poster's question, with some offering guidance on how to approach the problem through counterexamples. The discussion is exploring different interpretations of the continuity definition without reaching a consensus.

Contextual Notes

There is an emphasis on the definition of topology and the properties of open sets, with references to specific examples and the nature of functions in the context of continuity.

ehrenfest
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Homework Statement


In topology, a f: X -> Y is continuous when

U is open in Y implies that f^{-1}(U) is open in X

Doesn't that mean that a continuous function must be surjective i.e. it must span all of Y since every point y in Y is in an open set and that open set must have a pre-image open in X, so there must be some x in X s.t. f(x) = y?

What is wrong with my logic?


Homework Equations





The Attempt at a Solution

 
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If you think it isn't true, the try and construct a counterexample. :smile:

If it's true, this might help you see why. If it's false, analysis of the counterexample might help you spot the flaw.
 
So you're asking about the case when there is no x in X st... That would seem to be the empty set. Now what was the definition of topology?
 
I think matt grime answered it, but my counter example is:

f:R->R
f(x) = abs(x)

the set (-4, -2) is open but when when you pull it back you get the null set, which is open in the usual topology on R. I think I see now!
 

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