# Inverse map of a one to one function

1. Aug 9, 2010

### xaos

I'm trying to show an inverse map composed with its noninverse results in the identity
in terms of the set map f:X-->Y between topological spaces when f is one to one function. If I define the inverse map of a set as the disjoint union of the inverse map of each point in the set in Y, then let that set be the image of a subset U in X. My problem is the change in indexing under the union operator. It looks that I have to go from "y is in f(U)" to "x is in U" when each inverse y is the singleton x (using one to one), but this appears to assume one to one correspondence between U and f(U), but isn't that what I'm trying to show?, so my thinking is running in circles. Maybe its technically illegal to index a union of points in a set by the points in the set. I'm sure there's a trivial solution to this, but I'm failing to see it.

2. Aug 9, 2010

### owlpride

You should go from y is in f(U) to f^(-1) (y) is in f^(-1)(f (U)).

This only makes sense if f is onto. In general there might be points in Y that are not hit by points in X. If your only assumption is that f is 1-1, you can only get a "partial inverse" on the subset f(X) in Y.

3. Aug 10, 2010

### xaos

Thankyou for the response. But I think I need to restate my problem more clearly:
Define f^(-1)(S)=DisjointUnion{f^(-1)(y); for all y in S}.

So let f^(-1) o f(U)=DisjointUnion{f^(-1)(y); for all y in f(U)}
When f is 1-1, show this is equal to: DisjointUnion{ {x}, for all x in U} which is just =U

1. if you just replace the indexing set "y is in f(U)" with f^(-1)(y) is in f^(-1) o f(U)" this doesn't resolve my problem since my purpose is already trying to show this is "x is in U"

2. f^(-1) o f(U) is nonempty for all nonempty subsets U in X when f is 1-1. I want to show this can be no bigger than U.

Attempt#1: assume there is a point in x in f^(-1) o f(U) not in U. then f(x) is in f(U). but f^(-1) o f(x) must contain points in U, otherwise f(x) may not be in f(U). since f is 1-1, there is only one such point in U. therefore x is in U. contradiction.

Okay, I think I can fix this up, but I'm still not sure about the indexed set part...

Last edited: Aug 10, 2010
4. Aug 10, 2010

### owlpride

Can you state clearly how you define your inverse map? And what you are trying to show?

Given only the assumption that f is 1-1, your only shot at defining an inverse is as follows: let f^(-1) be the unique function that satisfies f^(-1) (f(x)) = x for all x in X. (f^(-1) exists thanks to f being 1-1. This is something you need to check though.)

Now are you trying to show that f(f^(-1) (y)) = y for all y in the image of f? I would do this as follows: if y is in the image of f, then y = f(x) for some x. So f(f^(-1)(y)) = f( f^(-1)(f(x))) = f(x) = y.

Last edited: Aug 10, 2010
5. Aug 10, 2010

### xaos

Let f:X-->Y be a function.
then f^(-1)(y) is defined as "the set of all x in X such that f(x)=y"
similarly f^(-1)(U) is "the set of all x in X such that f(x) is in U"
I need to show f^(-1) o f(U) is contained in U (and then equal to U) when f is 1-1.

6. Aug 10, 2010

### owlpride

Thanks for clarifying. I was confused because I was not reading your earlier posts careful enough. My apologies.

The obvious solution to your indexing problem is to note that f^(-1) o f (x) = {x} when f is 1-1. Then replace y with f(x) in the line you already had:

f^(-1) o f(U) = DisjointUnion{f^(-1)(y); for all y in f(U)}
... = DisjointUnion{f^(-1)(f(x)); for all x in U}
... = DisjointUnion{{x}; for all x in U}
... = U

f^(-1) o f(U) is always nonempty when U is nonempty, regardless of what f does. In fact, it's usually bigger than U unless f is 1-1.

7. Aug 10, 2010

### xaos

"f^(-1) o f(U) = DisjointUnion{f^(-1)(y); for all y in f(U)}
... = DisjointUnion{f^(-1)(f(x)); for all x in U}
... = DisjointUnion{{x}; for all x in U}
... = U"

I am assuming that your are getting from "y in f(U)" to "x in U" by 1-1 correspondence? Again, isn't that what we're trying to show? I beginning to think that this DisjointUnion definition only works after showing already 1-1 correspondence...

"f^(-1) o f(U) is always nonempty when U is nonempty, regardless of what f does. In fact, it's usually bigger than U unless f is 1-1."

yes, I was thinking "f(x)=f(y)={} but x isn't y". but this isn't allowed anyway.

8. Aug 11, 2010

### owlpride

First off, all the DisjointUnions were meant to be regular unions (copied and pasted without thinking) but you probably figured that out yourself. There's no need for disjoint unions based on your definition of f^(-1).

No, that's just by the definition of f(U). Instead of indexing over y in f(U), you can list all f(x) for x in U. Any redundancy would be absorbed by the union operation. I am using the 1-1 correspondance when I am moving from f^(-1) o f (x) to {x}. The existence of this 1-1 correspondance comes straight from the definition of injectivity. You can write a little proof by contradiction, as you did above, if you want to be really careful.

It seems like you are asking if set notation lets you prove your statement without using f^(-1) o f (x) = {x} in any way. I don't think so. Set notation gives you a compact way to work with information about the individual elements of a set - but you do need that information to start with.

9. Aug 12, 2010

### xaos

If we have (definition#1)"the set of all f^(-1) o f(x) such that f(x) is in f(U)", f(U) itself cannot be directly replaced by "x is in U" since they are not in general the same size sets.
The workaround I see is to redefine the original statement (definition#2):
f^(-1) o f(U) = "the set of all f^(-1) o f(x) such that x is in U" and =U follows immediately with 1-1. Which is the trivial solution I thought existed.
However, I do not see any resolve with definition #1.

10. Aug 12, 2010

### owlpride

This definition of f^(-1) is usually called a "pre-image". That's the definition I was working with in my last post. It also agrees with your non-standard earlier definition

Let S = f(U). Both definitions allow me to say that f^(-1) (S) = Union[ f^(-1) (f(x)), for all x in U]. Here's an explicit example to convince you of this fact.

Let f: {1,2,3,4}->{a,b}, with f(1) = a, f(2) = a, f(3) = a, f(3) = b. I am choosing a map that fails to be 1-1 on purpose. Now let U = {1,2} and S=f(U)={a}.

Then f^(-1) (S) = f^(-1) ({a}) = {1,2,3}.
Union[ f^(-1) (f(x)), for all x in U] = Union[f^(-1) (a), f^(-1)(a)] = Union[{1,2,3}, {1,2,3}] = {1,2,3}.

11. Aug 12, 2010

### xaos

Thankyou for your patience. What you're saying is that its set equivalence rather than index equivalence which allows us to go from different indexes...
f^(-1) o f(U) = Union {of all f^(-1) o f(x) s.t. x is in U} (by composition on a set)
f^(-1) o f(U) = Union {of all f^(-1) o f(x) s.t. x is in f^(-1) o f(U)} (trivially)
and these are equal as sets...