Composition of Inverse Functions


by John Creighto
Tags: composition, functions, inverse
John Creighto
John Creighto is offline
#1
Sep11-08, 01:55 PM
P: 813
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.
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
statdad
statdad is offline
#2
Sep12-08, 03:06 PM
HW Helper
P: 1,344
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.
evagelos
evagelos is offline
#3
Sep12-08, 06:45 PM
P: 308
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)

evagelos
evagelos is offline
#4
Sep14-08, 12:05 AM
P: 308

Composition of Inverse Functions


Quote Quote by John Creighto View Post
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]
Moo Of Doom
Moo Of Doom is offline
#5
Sep14-08, 06:36 AM
P: 367
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.
John Creighto
John Creighto is offline
#6
Sep14-08, 08:34 PM
P: 813
Quote Quote by Moo Of Doom View Post
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.
evagelos
evagelos is offline
#7
Sep15-08, 01:31 AM
P: 308
Quote Quote by Moo Of Doom View Post


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
HallsofIvy
HallsofIvy is offline
#8
Sep15-08, 12:47 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,894
Quote Quote by evagelos View Post
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).
evagelos
evagelos is offline
#9
Sep15-08, 07:49 PM
P: 308
write a proof where you justify each of your steps ,if you wish.

The above proof is not very clear


Register to reply

Related Discussions
Composition of functions Calculus & Beyond Homework 3
Inverse functions Introductory Physics Homework 1
Composition of 2 functions Set Theory, Logic, Probability, Statistics 6
Composition and Identity of functions Calculus 3
Inverse Functions General Math 7