# Composition of Inverse Functions

by John Creighto
Tags: composition, functions, inverse
 P: 813 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.
 HW Helper P: 1,344 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.
 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)
P: 308

## Composition of Inverse Functions

 Quote by 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.
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 ;

$$(g\circ f)^{-1}(A) = f^{-1}(g^{-1}(A))$$
 P: 367 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.
P: 813
 Quote by 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.
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.
P: 308
 Quote by Moo Of Doom If $g(f(x)) \in A$, then certainly $f(x) \in g^{-1}(A)$ by definition. We already know that $f(x) \in T$. .