Continous Mappings (Functions)

[JasonRox]

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?

 

[matt grime]

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.

 

[JasonRox]

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.

 

[HallsofIvy]

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.

 

[JasonRox]

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.

 

[HallsofIvy]

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!)

 

[JasonRox]

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.

Were you lucky enough to redo your proof?

 

[HallsofIvy]

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!