Does a Continuous Function Need to Be One-to-One?

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Discussion Overview

The discussion revolves around the properties of continuous functions, specifically whether a continuous function must be one-to-one. Participants explore the implications of continuity in the context of mappings and preimages in mathematical analysis.

Discussion Character

  • Conceptual clarification
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions if a continuous function must be one-to-one, suggesting that it does not need to be.
  • Another participant agrees that a function does not need to be one-to-one and suggests using the term "preimage" instead of "inverse image."
  • A participant provides an example using the function f(x) = x², illustrating that it is not one-to-one and confirming that the preimage of an open set remains open.
  • Some participants share personal anecdotes related to misunderstandings of the concept of preimages and inverses in their academic experiences.

Areas of Agreement / Disagreement

Participants generally agree that a continuous function does not need to be one-to-one, but there is some ambiguity regarding the terminology used (inverse image vs. preimage) and the implications of mapping elements.

Contextual Notes

There is a reliance on the definitions of continuity and the properties of mappings, which may not be fully resolved in the discussion. The example provided assumes familiarity with the concept of preimages and their relation to continuity.

JasonRox
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I have a question regarding this.

I wish I were home right now so I can give the exact words.

Anyways, the book is talking about continuous maps from one space to another. This is basically what it says...

Let f be a function with domain D in R. Then the following statements are equivalent:
f is continuous
If D is open, then the inverse image of every open set under f is again open.
If D is closed, then the inverse image of every closed set under f is again closed.

There are others, but that's not important.

I just want to clarify that f does not need to be one-to-one, correct?
 
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No, f does not need to be one to one. Instead of 'inverse image' try using the phrase 'preimage' of a set as someone here once pointed out. Can't remember who to credit it with.
 
matt grime said:
No, f does not need to be one to one. Instead of 'inverse image' try using the phrase 'preimage' of a set as someone here once pointed out. Can't remember who to credit it with.

So, using the "preimage" we can "map" to two elements?

I used quotes for map because it doesn't really satisfy the definition.
 
Last edited:
I'm not sure what you mean by "map" two elements but here is an example:

If f(x)= x2 (not one-to-one) and D is the interval (-1, 4) then f-1(D) = (-2, 2), the set of numbers whose square is between -1 and 4. Since (-1, 4) is open and f is continuous, that inverse image (preimage if you prefer) is open.
 
HallsofIvy said:
I'm not sure what you mean by "map" two elements but here is an example:

If f(x)= x2 (not one-to-one) and D is the interval (-1, 4) then f-1(D) = (-2, 2), the set of numbers whose square is between -1 and 4. Since (-1, 4) is open and f is continuous, that inverse image (preimage if you prefer) is open.

That's exactly what I wanted to confirm. Thanks.
 
My most embarassing moment: My first semester in graduate school, I was called on to do a proof in Topology class that involved f-1(A) where A is a set. Without thinking, I did the proof assuming that f had an inverse! (Hey, it said f-1!)
 
HallsofIvy said:
My most embarassing moment: My first semester in graduate school, I was called on to do a proof in Topology class that involved f-1(A) where A is a set. Without thinking, I did the proof assuming that f had an inverse! (Hey, it said f-1!)

I'll remember this whenever I get embarrased. :wink:

Were you lucky enough to redo your proof?
 
I did later submit a proof on paper to the professor but that day I just shrank into a small lump in my chair. The next time I was called on to present a proof in class I did well and I actually passed the course!
 

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