Correct statement about composite and inverse functions

[songoku]

Homework Statement

Given that X and Y are nonempty subsets of ℝ and let f : X → Y be a function and f^(-1) (y) = {x ∈ X | f(x) = y}. If A ⊆ X and B ⊆ Y, which statement must be true?
a. f^(-1) (f(A)) ⊆ f(f^(-1) (B))
b. A ⊆ f^(-1) (f(A))
c. f(f^(-1) (B)) = f^(1-) (B)
d. B ⊆ f(f^(-1) (B))
e. f^(-1) (f(A)) ⊆ A

Relevant Equations

Inverse function
Composite function

f_-1(f(A)) = A and f_-1(f(B)) = B so options (a) and (c) are wrong.

For (b), I get A ⊆ A
For (d), I get B ⊆ B
For (e), I get A ⊆ A

So there are three correct statements? Thanks

 

[mjc123]

f-1(f(A)) = A and f-1(f(B)) = B

Is this correct? Consider for example f(x) = x², and A = {1,2,3}. What is f^-1(f(A))?

 

[Delta2]

$f^{-1}(f(A))=A$ and $f(f^{-1}(B))=B$ hold only if f is 1-1. if f is not 1-1 then you can prove that $A\subseteq f^{-1}(f(A))$ but you can't prove that $f^{-1}(f(A))\subseteq A$. Try to prove the latter and you ll see that you ll need f to be 1-1.

 

[songoku]

Is this correct? Consider for example f(x) = x², and A = {1,2,3}. What is f^-1(f(A))?

f(x) = x² so f(A) = A² with domain A = {1,2,3}. Then the range of f(A) will be {1, 4, 9}

From there, {1, 4, 9} will become domain of f^-1(A) and {1,2,3} will be range of f^-1(A)

f^-1(A) maps {1, 4, 9} to {1,2,3} then continuing this process to get f^-1(f(A)), f(A) will map {1,2,3} to {1, 4, 9} so we end up with same numbers as we start which is {1, 4, 9}

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

I am not sure whether this is what you mean.

$f^{-1}(f(A))=A$ and $f(f^{-1}(B))=B$ hold only if f is 1-1. if f is not 1-1 then you can prove that $A\subseteq f^{-1}(f(A))$ but you can't prove that $f^{-1}(f(A))\subseteq A$. Try to prove the latter and you ll see that you ll need f to be 1-1.

Sorry I don't understand.

I want to clarify several things:
1. from f^-1 (y) = {x ∈ X | f(x) = y} can we say that f(x) is one - one function? In my opinion, yes I think f(x) is one - one function and f(x) has inverse

2. for A ⊆ X and B ⊆ Y, can we use same f(x) to define relation between A and B? I don't think we can. My example is like this:
Let f(x) = x + 5 , X = {1, 2, 3, 4, 5} and Y = {6, 7, 8, 9, 10}. Then I take A = {1, 2, 3} as subset of X and B = {6, 9, 10} so this means that no relation between A and B

 

[Delta2]

Sorry I don't understand.

I want to clarify several things:
1. from f^-1 (y) = {x ∈ X | f(x) = y} can we say that f(x) is one - one function? In my opinion, yes I think f(x) is one - one function and f(x) has inverse

No you can't say that f(x) is 1-1 just because you have defined $f^{-1}$ the way you defined it. In order for f to be 1-1, $f^{-1}(y)$ must contain only one element for any $y$. Take for example $f(x)=x^2$and $y=1$. Then $f^{-1}(y)=\left \{-1,1 \right\}$ (because $f(-1)=f(1)=y=1$)hence $f$ is not 1-1

2. for A ⊆ X and B ⊆ Y, can we use same f(x) to define relation between A and B? I don't think we can. My example is like this:
Let f(x) = x + 5 , X = {1, 2, 3, 4, 5} and Y = {6, 7, 8, 9, 10}. Then I take A = {1, 2, 3} as subset of X and B = {6, 9, 10} so this means that no relation between A and B

I don't really understand what you trying to say here. I just said that $f^{-1}(f(A))=A$ and $f(f^{-1}(B))=B$ only if f is 1-1, i did not say that there is any relation between A and B.

 

[FactChecker]

Regarding part e.:
The preimage set, $f^{-1}(x)$, can be "scattered" all over the domain. A lot of points scattered all over the place can be mapped by the function $f$ to the same value. The preimage does not have to all be inside any proper subset, A, of the preimage.
Consider the constant function $f(x)=0, \forall x \epsilon \mathbb{R}$ as an extreme, but simple, example.

 

[mathwonk]

Your use of language may be confusing you. "Inverse functions" are not relevant here because in your problem f^(-1) is not the inverse function of f. Your f^(-1) is defined for all functions, whether or not they are injective or surjective, but in the general case, f^(-1) does not have the nice properties of an inverse function. In your problem you are looking at the behavior of f and f^(-1) on subsets rather than on elements. In this setting they are not fully inverse to each other. I.e. by definition of your f^(-1), all you know is that:

1) A is contained in f^(-1)(f(A)), and
2) f(f^(-1)(B)) is contained in B.

Indeed these are tautological, i.e. immediate from the definition of f^(-1).
In general however f(f^(-1)(B)) may be smaller than B, and f^(-1)(f(A)) may be larger than A.

The key properties that do hold are that:

3) f preserves unions and (therefore) inclusions, (but not strict inclusions, and not intersections);
4) f^(-1) preserves unions, intersections, and inclusions (but again not strict inclusions).

In general, one thing to come away with is that f^(-1) is better behaved than f, on subsets.

Of course if f were a genuine inverse function, (only possible when f is bijective), so that fof^(-1) and f^(-1)of are both identity functions, then both f and f^(-1) are equally well behaved, and completely symmetrical.

 

[songoku]

Regarding part e.:
The preimage set, $f^{-1}(x)$, can be "scattered" all over the domain. A lot of points scattered all over the place can be mapped by the function $f$ to the same value. The preimage does not have to all be inside any proper subset, A, of the preimage.
Consider the constant function $f(x)=0, \forall x \epsilon \mathbb{R}$ as an extreme, but simple, example.

Your use of language may be confusing you. "Inverse functions" are not relevant here because in your problem f^(-1) is not the inverse function of f. Your f^(-1) is defined for all functions, whether or not they are injective or surjective, but in the general case, f^(-1) does not have the nice properties of an inverse function.

Yes I think this is the source of my mistake. I interpret " f^-1 (y) = {x ∈ X | f(x) = y} " as a claim by the question that the inverse function f^-1 (x) exists so it means that f(x) is one - one function. Apparently it is wrong because I think it is only to say that "if f(x) = y then y = f^-1 (x)" and no claim whatsoever that the inverse function exists or f(x) is one - one function.

In your problem you are looking at the behavior of f and f^(-1) on subsets rather than on elements. In this setting they are not fully inverse to each other. I.e. by definition of your f^(-1), all you know is that:

1) A is contained in f^(-1)(f(A)), and
2) f(f^(-1)(B)) is contained in B.

Indeed these are tautological, i.e. immediate from the definition of f^(-1).
In general however f(f^(-1)(B)) may be smaller than B, and f^(-1)(f(A)) may be larger than A.

The key properties that do hold are that:

3) f preserves unions and (therefore) inclusions, (but not strict inclusions, and not intersections);
4) f^(-1) preserves unions, intersections, and inclusions (but again not strict inclusions).

In general, one thing to come away with is that f^(-1) is better behaved than f, on subsets.

Of course if f were a genuine inverse function, (only possible when f is bijective), so that fof^(-1) and f^(-1)of are both identity functions, then both f and f^(-1) are equally well behaved, and completely symmetrical.

Sorry I am really not familiar with all these terms. What is the meaning of "f^-1 is better behaved than f" ?

$f^{-1}(f(A))=A$ and $f(f^{-1}(B))=B$ hold only if f is 1-1. if f is not 1-1 then you can prove that $A\subseteq f^{-1}(f(A))$ but you can't prove that $f^{-1}(f(A))\subseteq A$. Try to prove the latter and you ll see that you ll need f to be 1-1.

Maybe I got something from all the hints given. Let me try:
Suppose $f(x) = y = x^2$ where $x = {1, 2, 3}$ then $f^{-1}(y) = x =\pm \sqrt{y}$ so $f^{-1}(f(x))$ for $x = {1, 2, 3}$ will give result ${\pm1, \pm 2, \pm 3}$
Hence the statement $f^{-1}(f(A)) \subseteq A$ is wrong

Is that correct up until this part? Thanks

 

[Delta2]

Yes you are correct. As others have said, $f^{-1}(y)$ is not the inverse function its a whole set (even if for a single y). It becomes a function if that set contains only one element for any y, and this condition is met iff f is 1-1.

 

[songoku]

I tried the same thing for option (d) but got confused.

Suppose $f(x) = y = x^2$ where $x = \{1, 2, 3\}$ then $f^{-1}(y) = x =\pm \sqrt{y}$ so $f(f^{-1}(x))$ for:
$x=1 \rightarrow f(f^{-1}(1)) = f(\pm \sqrt{1}) = f(\pm 1) = 1$

Doing it again for $x=2$ and $x=3$ give the same result so I get $f(f^{-1}(x))=x$

I am not sure that is correct. Thanks

 

[Delta2]

I think you are right here , $f(f^{-1}(B))=B$ regardless if f is 1-1. But we need f to be onto(surjective) for this.

 

[mathwonk]

In saying that f^(-1) is "better behaved" than f on subsets, I meant that it preserves intersections, as well as unions and inclusions. I.e. it is a "boolean homomorphism". That means if you take any collection, even an infinite collection, of subsets of Y, and connect them up by any combination of intersection and union signs and parentheses, you can put f^(-1) around the whole expresion, or you can put f^(-1) on every subset, and the result will be the same subset of X.

The statement that f(f^(-1)(B)) = B is true when f is surjective, and more generally is true if and only if B is contained in f(X); but if f is the constant function f:R-->R taking every element x to 0 say, then for every subset B which contains 0, f(f^(-1)(B)) = {0}, while f(f^(-1)(B)) is empty if B does not contain 0. This example of factchecker is very illuminating. In particular it shows that f(f^(-1)(B)) can be smaller than B.

The statement that f^(-1)(f(A)) = A is true if f is injective, and more generally is true if and only if f(p) ≠ f(q) whenever p is in A and q is not; but again the constant function gives us f^(-1)(f(A)) = R if A is any non empty subset of R. This shows that f^(-1)(f(A)) can be larger than A.

Remark: I have edited this answer to remove an erroneous statement, proving I am not at all immune to them myself. But I think what remains is correct. As often holds, the best way forward is to examine a good example, like factchecker's. Notice factchecker has chosen an example that is as far away as possible from being injective, and also from being surjective. I urge you to check carefully the truth of the statements in the two paragraphs just above.

 

[mathwonk]

One useful way to use the concept of f^(-1) is to apply it to one point subsets of Y, i.e. for each element y of Y, to look at all those elements x of X, such that f(x) = y. This set, f^(-1)(y), is called the "fiber (of f) over y". We visualize X as lying over Y, with f being the projection sending each point of X down to the point of Y under it. Then the fiber of f over y is the part of X lying directly over the point y.

E.g. the projection of the real x,y plane down onto the x axis, sending the point (x,y) to x, has as fiber over the point a, just the vertical line perpendicular to the x-axis at (a,0), i.e. it consists of all points of form (a,y) for all y.

If we take any subset A of the plane, and project it down to the x axis, and call the restricted projection f:A-->R, then the fiber of f over the point a, is the intersection of the set A with the vertical line perpendicular to the x-axis at (a,0).

With the concept of fiber, we can say that a function f:X-->Y is injective if and only if the fiber over every point of Y contains at most one element, and we can say f is surjective if and only if the fiber over every point of Y contains at least one element.

 

[songoku]

In saying that f^(-1) is "better behaved" than f on subsets, I meant that it preserves intersections, as well as unions and inclusions. I.e. it is a "boolean homomorphism". That means if you take any collection, even an infinite collection, of subsets of Y, and connect them up by any combination of intersection and union signs and parentheses, you can put f^(-1) around the whole expresion, or you can put f^(-1) on every subset, and the result will be the same subset of X.

The statement that f(f^(-1)(B)) = B is true when f is surjective, and more generally is true if and only if B is contained in f(X); but if f is the constant function f:R-->R taking every element x to 0 say, then for every subset B which contains 0, f(f^(-1)(B)) = {0}, while f(f^(-1)(B)) is empty if B does not contain 0. This example of factchecker is very illuminating. In particular it shows that f(f^(-1)(B)) can be smaller than B.

The statement that f^(-1)(f(A)) = A is true if f is injective, and more generally is true if and only if f(p) ≠ f(q) whenever p is in A and q is not; but again the constant function gives us f^(-1)(f(A)) = R if A is any non empty subset of R. This shows that f^(-1)(f(A)) can be larger than A.

Remark: I have edited this answer to remove an erroneous statement, proving I am not at all immune to them myself. But I think what remains is correct. As often holds, the best way forward is to examine a good example, like factchecker's. Notice factchecker has chosen an example that is as far away as possible from being injective, and also from being surjective. I urge you to check carefully the truth of the statements in the two paragraphs just above.

One useful way to use the concept of f^(-1) is to apply it to one point subsets of Y, i.e. for each element y of Y, to look at all those elements x of X, such that f(x) = y. This set, f^(-1)(y), is called the "fiber (of f) over y". We visualize X as lying over Y, with f being the projection sending each point of X down to the point of Y under it. Then the fiber of f over y is the part of X lying directly over the point y.

E.g. the projection of the real x,y plane down onto the x axis, sending the point (x,y) to x, has as fiber over the point a, just the vertical line perpendicular to the x-axis at (a,0), i.e. it consists of all points of form (a,y) for all y.

If we take any subset A of the plane, and project it down to the x axis, and call the restricted projection f:A-->R, then the fiber of f over the point a, is the intersection of the set A with the vertical line perpendicular to the x-axis at (a,0).

With the concept of fiber, we can say that a function f:X-->Y is injective if and only if the fiber over every point of Y contains at most one element, and we can say f is surjective if and only if the fiber over every point of Y contains at least one element.

That's a lot of information for me to digest. I will re-read it many time to understand it.

Thank you very much for all the help mjc123 , Delta2, FactChecker , mathwonk

 

[mathwonk]

yes rereading is useful. working out statements in view of fact checkers example may be even more useful. remember: examples teach best. good luck!