Functions real analysis

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Let f and g be functions such that (g[itex]\circ[/itex]f)(x)=x for all x [itex]\epsilon[/itex]D(f) and (f[itex]\circ[/itex]g)(y)=y for all y [itex]\epsilon[/itex]D(g). Prove that a g = f^-1

Pf/

How would you go about starting this besides saying

Let f and g be functions such that (g[itex]\circ[/itex]f)(x)=x for all x [itex]\epsilon[/itex]D(f) and (f[itex]\circ[/itex]g)(y)=y for all y [itex]\epsilon[/itex]D(g).

isn't obvious that the functions would have to be the inverse of each other, How else could you get the identity? So how do you prove it, can I just say clearly it is.
 
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  • #2
benorin
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What does g = f^-1 mean? My calc text gives this definition:

g = f^-1 if, and only if, f(x) is a one-to-one function with domain D(f) and range D(g) such that

g(y)=x <=> f(x)=y for all y in D(g).

Start with the given and prove this definition must be true.
 
  • #3
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g= f^-1 means that g is the inverse of f
 
  • #4
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How do you prove a definition true. If its a definition shouldn't it always be true?
 
  • #5
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This is what I have been able to get to

(g[itex]\circ[/itex]f)(x)=x for all x ϵD(f) and (f[itex]\circ[/itex]g)(y)=y for all y ϵD(g).


So choose
g(y)=x and f(x)=y

so

x=g(y)=f^-1(y)

ans since f and g are bijections we can conclude g=f^-1


does that work
 

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