Composition of Inverse Functions

[John Creighto]

In Micheal C. Gemignani, "Elementary Topology" in section 1.1 there is the following exercise

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

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.

 

[statdad]

How is \mathbf{W} used here - does g map to all of \mathbf{W} or only into \mathbf{W}? That could explain the possible confusion.

 

[evagelos]

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]

In Micheal C. Gemignani, "Elementary Topology" in section 1.1 there is the following exercise

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

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.

xεf^{-1}(g^{-1}(A))<====> xεS & f(x)εg^{-1}(A)====> xεS & g(f(x))εA <====> xε(g\circ f)^{-1}(A)

since g^{-1}(A) = { 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)\subseteq g^{-1}(A)........

and then we will have ;

(g\circ f)^{-1}(A) = f^{-1}(g^{-1}(A))

 

[Moo Of Doom]

Let x \in (g \circ f)^{-1}(A). Then x \in S with g(f(x)) \in A. This means f(x) \in g^{-1}(A) and thus x \in f^{-1}(g^{-1}(A)).

The other direction has been shown.

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

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

 

[John Creighto]

Let x \in (g \circ f)^{-1}(A). Then x \in S with g(f(x)) \in A. This means f(x) \in g^{-1}(A) and thus x \in f^{-1}(g^{-1}(A)).

The other direction has been shown.

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

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]

If g(f(x)) \in A, then certainly f(x) \in g^{-1}(A) by definition. We already know that f(x) \in T.

.

What definition,write down please

 

[HallsofIvy]

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]

write a proof where you justify each of your steps ,if you wish.

The above proof is not very clear