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Composition of Inverse Functions 
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#1
Sep1108, 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. 


#2
Sep1208, 03:06 PM

HW Helper
P: 1,372

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
Sep1208, 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) 


#4
Sep1408, 12:05 AM

P: 308

Composition of Inverse Functions
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
Sep1408, 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 counterexample is. 


#6
Sep1408, 08:34 PM

P: 813




#7
Sep1508, 01:31 AM

P: 308




#8
Sep1508, 12:47 PM

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#9
Sep1508, 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 


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