# Continuous mappings in topology.

1. Sep 20, 2012

### DeadOriginal

I am trying to understand the theorem:

Let f:S->T be a transformation of the space S into the space T. A necessary and sufficient condition that f be continuous is that if O is any open subset of T, then its inverse image $f^{-1}(O)$ is open in S.

First off, I don't really understand what the inverse image is. Is it just the inverse function? The proof of this theorem deals with the inverse image a lot so I believe that not understanding what an inverse image is, is limiting my ability to understand the proof.

Thanks to anyone who can offer any advice.

2. Sep 20, 2012

### lavinia

The inverse image of a set is the points in the domain of the function that are mapped into the set.

Often the definition of continuous is that the inverse image of open sets is open. So what definition are you starting with?

Last edited: Sep 20, 2012
3. Sep 20, 2012

### DeadOriginal

I think that I would start with the definition of continuity and try to deduce that the inverse image is open.

4. Sep 20, 2012

### SteveL27

What's the definition of continuity in a topological space? The theorem you're trying to prove is usually given as the definition. Therefore there must be some other definition given in your book or class. What is it?

5. Sep 20, 2012

### DeadOriginal

My book states that a transformation f:S->T is continuous provided that if p is a limit point of X a subset of S then f(p) is a point or limit point of f(X).

6. Sep 20, 2012

### lavinia

use this definition to deduce that inverse images of open sets is open. This is a good exercise.

Here is a suggestion. If the inverse image were not open then it would contain a point so that every open ball around it no matter how small its radius, would contain a point outside of the inverse image.

Last edited: Sep 20, 2012
7. Sep 20, 2012

### DeadOriginal

Thanks! I will get on it. My main problem was understanding what an inverse image was but now that it has been clarified I should be able to do this.

8. Sep 20, 2012

### mathwonk

a function f:X-->Y takes each point of X to a single point of Y.

If we try to define an inverse of f, from Y to X, by sending each point of Y to the point of X that maps to it via f, we may not get a function because some points of Y may be the image of more than one point of X, and some may not be images of any points of X.

But we can always define an inverse of f from Y to subsets of X, by sending each point q of Y to the subset of all those points of X that map to q by f. This set may have more than one point or may have no points, but it does give a good definition of a map Y-->subsets of X.

The image of a point q of Y under this map is called the "inverse image" of q under f.

i.e. it is the image of q under the (set theoretic) inverse of f. Since it always exists for any f, and is the closest thing we have to an inverse function of f, it is very important.

oh yes, and it also exists as a map subsets of Y-->subsets of X, as in the current situation, sending any subset W of Y to the subset of all elements of X that map by f to any element of W.

9. Sep 20, 2012

### DeadOriginal

This helps a lot! Thanks. Why couldn't the book just say so.....

10. Sep 20, 2012

### mathwonk

well most book authors don't really think about what they are writing, but just repeat what they themselves memorized.

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