# Thread: inverse function

1. Hello!

I got something which I already proved, but I am not sure whether it is correct.

f:X -> Y, A subset of X, B subset of Y

I need to prove that A is subset of f^-1 ( f ( A ) )

I proved it like this.

From definition of inverse function, there must be true the composition:

f^-1(f(x))=x

f(f^-1(x))=x

If x is in A then x is in X.

Because of this

f^-1(f(A))=A

So it follows that A is subset of f^-1(f(A)) and f^-1(f(A)) is subset of A.

Is my proof correct?

Thanks in advance.

Regards.

2.

3. Originally Posted by scientist91
Hello!

I got something which I already proved, but I am not sure whether it is correct.

f:X -> Y, A subset of X, B subset of Y

I need to prove that A is subset of f^-1 ( f ( A ) )

I proved it like this.

From definition of inverse function, there must be true the composition:

f^-1(f(x))=x

f(f^-1(x))=x

If x is in A then x is in X.

Because of this

f^-1(f(A))=A

So it follows that A is subset of f^-1(f(A)) and f^-1(f(A)) is subset of A.

Is my proof correct?

Thanks in advance.

Regards.
No, your proof is not correct. The theorem does not require that f be an invertible function. f^-1(B) is defnined (the inverse image of B) even if f is not invertible as a function.

4. Originally Posted by scientist91
Hello!

I got something which I already proved, but I am not sure whether it is correct.

f:X -> Y, A subset of X, B subset of Y

I need to prove that A is subset of f^-1 ( f ( A ) )

I proved it like this.

From definition of inverse function, there must be true the composition:

f^-1(f(x))=x

f(f^-1(x))=x

If x is in A then x is in X.

Because of this

f^-1(f(A))=A

So it follows that A is subset of f^-1(f(A)) and f^-1(f(A)) is subset of A.

Is my proof correct?

Thanks in advance.

Regards.
Not quite true.
f^-1(f(x))=x.
is not necessarily so.

Simple example. Let X and Y be reals, let A be positive reals. Let f(x)=x^2.
Then f^-1(f(x))=x or -x

5. Originally Posted by mathman
Not quite true.
f^-1(f(x))=x.
is not necessarily so.

Simple example. Let X and Y be reals, let A be positive reals. Let f(x)=x^2.
Then f^-1(f(x))=x or -x
In fact often not meaningful, but the theorem remains true.

, for then

6. f^-1(f(x))=x
f(f^-1(x))=x

is true under the condition that f is injection.

If f is injection then f^(-1) is injection.

if both f and g are injection then the composition fg(x) is injection.

In this case f(f^(-1))(x) and f^-1(f(x)) is injection.

When f is injection my proof is correct (I believe).

But when f is not injection, then f^-1 is also not injection.

I believe the statement:
A is subset of f^-1 ( f ( A ) )
will not work in this case.

Example.

f: R->R

Let A=Z and B=Z (the set of integers)

f(x)=x^2

We need to divide f(x) on two parts.

f(x)=x^2, x>=0 , f:Z^+ (positive integers) --> Z^+

f(x)=x^2, x<0, f:Z^- (negative integers) ---> Z^+

y=x^2

Sqrt[y]=Abs[x]

For x>=0,

Sqrt[y]=x

f^(-1)(x)=Sqrt[x], f^(-1): Z^+ ---> Z^+

and

f^(-1)(x)=-Sqrt[x], f^(-1): Z^+ ---> Z^-

Now because A=Z lets check it:

f^(-1)(f(A))=Z^+

and

f^(-1)(f(A))=Z^-

So A=Z is not subset of Z^+ and A=Z is not subset of Z^-

Or maybe it refers to A=Z is subset of Z^+ U Z^- = Z ????

7. Originally Posted by scientist91
.

I believe the statement:
A is subset of f^-1 ( f ( A ) )
will not work in this case.
You believe incorrectly.

Not only is the statement true, it is trivial.

8. Originally Posted by DrRocket
Originally Posted by scientist91
.

I believe the statement:
A is subset of f^-1 ( f ( A ) )
will not work in this case.
You believe incorrectly.

Not only is the statement true, it is trivial.
Ok. Did I do the example correctly? Then how will I prove it when it is not injection?

9. scientist: I strongly recommend you look more carefully at the help you have been given so far. No, injections and surjections are irrelevant here, unless you want define a bijection (in which case you may assume that are mutual inverses - don't do that!)

I also suggest you work first on elements rather than sets to get your thinking straight..

As to your example: if then, one has that . That is whose preimage is simply . Therefore

But it is certainly true that .

In the above, your function is surjective, but in the general case one must assume that is "smaller than or equal to" , otherwise one would have that there exist elements in which have more than a single image in . This is not allowed, by the definition of a function.

So what if, say, ? Is it possible that ? Is this a subset of ? If so, how would you describe this function? (Granted, it's an extreme case - but not unusual, think)

PS Please use the Tex capability kindly provided by the team here

10. Originally Posted by Guitarist
scientist: I strongly recommend you look more carefully at the help you have been given so far. No, injections and surjections are irrelevant here, unless you want define a bijection (in which case you may assume that are mutual inverses - don't do that!)

I also suggest you work first on elements rather than sets to get your thinking straight..

As to your example: if then, one has that . That is whose preimage is simply . Therefore

But it is certainly true that .

In the above, your function is surjective, but in the general case one must assume that is "smaller than or equal to" , otherwise one would have that there exist elements in which have more than a single image in . This is not allowed, by the definition of a function.

So what if, say, ? Is it possible that ? Is this a subset of ? If so, how would you describe this function? (Granted, it's an extreme case - but not unusual, think)

PS Please use the Tex capability kindly provided by the team here
Guitarist thank you.

In my example, the function is not surjective, because the range of f(x) is not same as the codomain Z. The range of f(x) is:

,

f(x) is not also injective.

does not actually exist because f(x) is not injective.

We should not consider this case.

Injection is very important here, because it is the necessary condition for the inverse function.

If then it is possible that , because it is clearly that A doesn't necessarily be the domain of the function.

For ex. and . In this case let the subset A={-1,-2,-3} and then

Clearly the domain of f(x) would be .

Yes, the empty set is subset of Y.

Still, do not understand what do you want to tell me.

11. Originally Posted by scientist91
does not actually exist because f(x) is not injective.
No! First this makes little sense, as I thought we might be assuming that .

ALWAYS exists, it is called the preimage of y under f ; it is always a subset of the domain.

Injection is very important here because it is the necessary condition for the inverse function.
It is, but it is not sufficient - you need your function to be surjective also

If then it is possible that , because it is clearly that A doesn't necessarily be the domain of the function.
Why not? Functions are not entitled to "ignore" elements in their domain. So if A subset X, then it must have an image in Y. If the image set is empty, so what? (this is called an injection BTW)

Clearly the domain of f(x) would be .
No! is an element (in the codomain). It is NOT a function, and therefore cannot have a domain.

Once again, until you are quite clear about functions on sets (and what properties these have) I suggest you stick to elements.

I also suggest you get very clear the difference between preimages and inverses.

PS I grant you, the notation here is a bit naughty - for a surjection, for example, the preimage , so that if is bijective, then which is called the singleton set in X.

Everybody, but everybody, forgets the difference between the singleton and the element and writes, for this bijection - purists might say they shouldn't, perhaps, but they do

12. Ok, I agree that I was wrong about some things, but I was partially right.

Here are some definitions that clear the thing up:

Definition of domain.

This means that the range doesn't necessarily is equal to the codomain of the function.

Here are one more very important definition:

Definition of inverse function

The conclusion:

"a function is bijective if and only if it has an inverse function".

So if an inverse function exists then the function is bijective.

Here is another important definition, that I stated in the proof:

Composition of inverse and a function

Now I see that can't be true.

My conclusion is that, for the prove to be valid we need to assume that exists that is if and only if it is both one-to-one and onto.

Am I right?
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13. Originally Posted by scientist91
Ok, I agree that I was wrong about some things, but I was partially right.

Here are some definitions that clear the thing up:

Definition of domain.

This means that the range doesn't necessarily is equal to the codomain of the function.

Here are one more very important definition:

Definition of inverse function

The conclusion:

"a function is bijective if and only if it has an inverse function".

So if an inverse function exists then the function is bijective.

Here is another important definition, that I stated in the proof:

Composition of inverse and a function

Now I see that can't be true.

My conclusion is that, for the prove to be valid we need to assume that exists that is if and only if it is both one-to-one and onto.

Am I right?
[/url]
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No. You are totally wrong.

If is a function then for any subsetset of the range of exists and is defined wherther or not f is bijective. This is an important point, and not a quibble.

14. I don't know why you think that I am totally wrong, when I showed you links where are the correct definitions.

I will give you example again.

is not injective neither surjective so it is not bijective.

The inverse of it should be .

But there is a problem. Because f(x) is not surjective, not all elements of will be mapped into Z, so is not a function.

Because f(x) is not injective, for ex. could not be determined whether it is 1 or -1.

We can also confirm this by the test with vertical lines.

Conclusion: must be bijection.

15. Originally Posted by scientist91
I don't know why you think that I am totally wrong,
here's why

is not injective neither surjective so it is not bijective.
Take another look at the definitions of injection and surjection

not all elements of will be mapped into Z, so is not a function.
Your second statement is certainly true, are you completely sure about the first?

16. Originally Posted by Guitarist
Originally Posted by scientist91
I don't know why you think that I am totally wrong,
here's why

is not injective neither surjective so it is not bijective.
Take another look at the definitions of injection and surjection

not all elements of will be mapped into Z, so is not a function.
Your second statement is certainly true, are you completely sure about the first?
Yes. Here is why:

Def. of injection:

If f(x)=f(y) then x=y

in this case f(-2)=f(2) but x is not equal to y, hence true implies not true, and that's false.

is not injection.

Now. def. of surjection:

In mathematics, a function is said to be surjective or onto if its range is equal to its codomain.

Here: f(-n)=n^2,.....,f(-2)=4, f(-1)=1, f(0)=0, f(1)=1, f(2)=4, f(3)=9.....f(n)=n^2

What about the elements 2,3,5,6,7,8,10,11,12,13,14,15,17.... of the codomain ??

Clearly the range {0,1,4,9,16....} is not equal to the codomain {0,1,2,3,4,5,6....}

Hence the function is not surjective.

f is bijective if it is a one-to-one correspondence between those sets; i.e., both one-to-one (injective) and onto (surjective).

Hence, the function is not bijection.

What is the problem here?

17. Originally Posted by scientist91
I don't know why you think that I am totally wrong, when I showed you links where are the correct definitions.

I will give you example again.

is not injective neither surjective so it is not bijective.

The inverse of it should be .

But there is a problem. Because f(x) is not surjective, not all elements of will be mapped into Z, so is not a function.

Because f(x) is not injective, for ex. could not be determined whether it is 1 or -1.

We can also confirm this by the test with vertical lines.

Conclusion: must be bijection.
Wrong, and I don't care what links you have.

In your case

The main poit is that is a set and that set is defined whether or not f is injective, surjective or bijective. It makes no difference.

18. Ok, I agree that it is set.

But in that case the formula will only work only if f is bijection.

For example:

Yes, it is true that {0,1,4,9....} is subset of Z, but how will I prove it?

Here is f(x):

If y=f(x) then

Now lets suppose as in the task.

f(A) would be defined as:

And

I am stuck here. I know that A=f^{-1}(B) is not always true, because of f(x) is not injection always.

How will I show that A is a subset of f^{-1}(B)?

19. Originally Posted by scientist91
Ok, I agree that it is set.

But in that case the formula will only work only if f is bijection.

For example:

Yes, it is true that {0,1,4,9....} is subset of Z, but how will I prove it?

Here is f(x):

If y=f(x) then

Now lets suppose as in the task.

f(A) would be defined as:

And

I am stuck here. I know that A=f^{-1}(B) is not always true, because of f(x) is not injection always.

How will I show that A is a subset of f^{-1}(B)?

Let then and therefore from whence QED

Note that this applies to any function f, which need not be injective, surjective or bijective. Note also that this is completely trivial and follows from the definition. Such a statement would rarely be called a theorem or be demonstrated in detail; rather it would be simply be a remark which should be obvious to the reader.

20. Thank you.

I believe you did a little mistake but it does not matter.

It should be:

Let implies which implies which equals to

So then

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