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Continuous functions in topology

  • Thread starter ehrenfest
  • Start date
  • #1
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1. 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?


2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

  • #2
Hurkyl
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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.
 
  • #3
matt grime
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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?
 
  • #4
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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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