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.
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.
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)
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]
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.
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).