Composition of Inverse Functions

In summary: Let x \in (g \circ f)^{-1}(A). By definition, this means that x \in S and g(f(x)) \in A. Since g(f(x)) \in A, we know that f(x) \in g^{-1}(A) by definition of g^{-1}(A). Therefore, x \in f^{-1}(g^{-1}(A)), which means that (g \circ f)^{-1}(A) \subseteq f^{-1}(g^{-1}(A)).Now, let x \in f^{-1}(g^{-1}(A)). By definition, this means that x \in S and f(x) \in g^{-
  • #1
John Creighto
495
2
In Micheal C. Gemignani, "Elementary Topology" in section 1.1 there is the following exercise

2)
i)
If [tex]f:S \rightarrow T[/tex] and [tex]G: T \rightarrow W [/tex], then [tex](g \circ f)^{-1}(A) = f^{-1}(g^{-1}(A))[/tex] for any [tex]A \subset W [/tex].

I think the above is only true if A is in the image of g yet the book says to prove the above. I have what I believe is a counter example. Any comments? I will give people two days to prove the above or post a counter example. After this time I'll post my counter example for further comment.
 
Physics news on Phys.org
  • #2
How is [tex] \mathbf{W} [/tex] used here - does [tex] g [/tex] map to all of [tex] \mathbf{W} [/tex] or only into [tex] \mathbf{W} [/tex]? That could explain the possible confusion.
 
  • #3
The way the problem has been written you can only prove that:

...f^-1[g^-1(A)] IS a subset of (g*f)^-1(A) i.e the right hand side of the above is a subset of the left hand side

FOR the above to be equal we must have that: f(S)={ f(x): xεS } MUST be a subset of

g^-1(A)
 
  • #4
John Creighto said:
In Micheal C. Gemignani, "Elementary Topology" in section 1.1 there is the following exercise

2)
i)
If [tex]f:S \rightarrow T[/tex] and [tex]G: T \rightarrow W [/tex], then [tex](g \circ f)^{-1}(A) = f^{-1}(g^{-1}(A))[/tex] for any [tex]A \subset W [/tex].

I think the above is only true if A is in the image of g yet the book says to prove the above. I have what I believe is a counter example. Any comments? I will give people two days to prove the above or post a counter example. After this time I'll post my counter example for further comment.

xε[tex]f^{-1}(g^{-1}(A))[/tex]<====> xεS & f(x)ε[tex]g^{-1}(A)[/tex]====> xεS & g(f(x))εA <====> xε[tex](g\circ f)^{-1}(A)[/tex]

since [tex]g^{-1}(A)[/tex] = { y: yεT & g(y)εΑ}

ΙΝ the above proof all arrows are double excep one which is single and for that arrow to become double we must have :

........f(S)[tex]\subseteq g^{-1}(A)[/tex]........

and then we will have ;

[tex](g\circ f)^{-1}(A) = f^{-1}(g^{-1}(A))[/tex]
 
  • #5
Let [itex]x \in (g \circ f)^{-1}(A)[/itex]. Then [itex]x \in S[/itex] with [itex]g(f(x)) \in A[/itex]. This means [itex]f(x) \in g^{-1}(A)[/itex] and thus [itex]x \in f^{-1}(g^{-1}(A))[/itex].

The other direction has been shown.

How are those not all double arrows, evagelos? If [itex]g(f(x)) \in A[/itex], then certainly [itex]f(x) \in g^{-1}(A)[/itex] by definition. We already know that [itex]f(x) \in T[/itex].

I'm curious as to what this supposed counter-example is.
 
  • #6
Moo Of Doom said:
Let [itex]x \in (g \circ f)^{-1}(A)[/itex]. Then [itex]x \in S[/itex] with [itex]g(f(x)) \in A[/itex]. This means [itex]f(x) \in g^{-1}(A)[/itex] and thus [itex]x \in f^{-1}(g^{-1}(A))[/itex].

The other direction has been shown.

How are those not all double arrows, evagelos? If [itex]g(f(x)) \in A[/itex], then certainly [itex]f(x) \in g^{-1}(A)[/itex] by definition. We already know that [itex]f(x) \in T[/itex].

I'm curious as to what this supposed counter-example is.

Your proof looks correct. There appears to be a mistake in my counter example. I'll spend a few futile minutes anyway trying to think up a counterexample anyway.
 
  • #7
Moo Of Doom said:
If [itex]g(f(x)) \in A[/itex], then certainly [itex]f(x) \in g^{-1}(A)[/itex] by definition. We already know that [itex]f(x) \in T[/itex].

.

What definition,write down please
 
  • #8
evagelos said:
What definition,write down please
g-1(A) is defined as the set of all x such that g(x) is in A. If g(f(x)) is in A, then, by that definition, f(x) is in g-1(A).
 
  • #9
write a proof where you justify each of your steps ,if you wish.

The above proof is not very clear
 

What is the definition of inverse functions?

Inverse functions are functions that "undo" each other. In other words, if a function f(x) maps an input x to an output y, then the inverse function g(y) will map y back to x. This means that when you compose the two functions, f(g(y)) or g(f(x)), you will end up with the original input value.

How do you find the inverse of a function?

To find the inverse of a function, you can follow these steps:

  1. Replace f(x) with y.
  2. Swap the x and y variables.
  3. Solve for y.
  4. Replace y with inverse notation, g(x).
If the resulting function is valid, then it is the inverse of the original function.

What is the domain and range of inverse functions?

The domain of an inverse function is the range of the original function, and vice versa. This means that the input values for the inverse function will be the output values of the original function, and the output values for the inverse function will be the input values of the original function. In other words, the domains and ranges of inverse functions are swapped.

Can every function have an inverse?

No, not every function has an inverse. A function must be one-to-one (have a unique output for every input) in order to have an inverse. This means that if two different inputs have the same output, the function does not have an inverse. For example, the function f(x) = x² does not have an inverse because f(2) = 4 and f(-2) = 4.

How do you use the composition of inverse functions?

The composition of inverse functions is used to verify that two functions are inverses of each other. To do this, you can compose the two functions, f(g(x)) and g(f(x)), and if the result is x, then the functions are inverses. This can also be used to simplify complex functions by breaking them down into smaller, simpler functions.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
847
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
746
  • Set Theory, Logic, Probability, Statistics
Replies
7
Views
365
  • Set Theory, Logic, Probability, Statistics
2
Replies
54
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
2K
  • Precalculus Mathematics Homework Help
Replies
7
Views
272
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
Back
Top