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.

xε$$f^{-1}(g^{-1}(A))$$<====> xεS & f(x)ε$$g^{-1}(A)$$====> 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