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Inverse composite proof (wording of the proof)

  • Thread starter psycho2499
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Homework Statement


Let f:A->B and g:B->C be invertible mappings. Show (g o f)^-1 = f^-1 o g^-1.


Homework Equations


A mapping is invertible iff it is bijective


The Attempt at a Solution


I understand why these are equivalent statements; however, I can't figure out the wording of the proof.

The best I can think of is:

Suppose f and g are invertible mappings defined f:A->B and g:B->C. Let a, b, and c be elements of sets A, B, and C respectively such that f(a)=b and g(b)=c. Since f and g are bijective, f^-1(b)=a and g^-1(c)=b. So (g o f)^-1(c) = a = f^-1(g^-1(c)).
 

Answers and Replies

  • #2
Office_Shredder
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(g o f)^-1(c) = a
You haven't actually proven this to be the case.


We want to show for all [itex]c\in C[/itex] that [itex](g\circ f)^{-1}(c) = f^{-1}(g^{-1}(c))[/itex]. Let [itex]b\in B[/itex] be the unique element of B such that [itex] g^{-1}(c)=b[/itex] and [itex]a\in A[/itex] the unique element of A such that [itex]f^{-1}(b) = a[/itex]. These both exist because f and g are invertible.

Clearly [itex]f^{-1}(g^{-1}(c)) = a[/itex] just by how a and b were defined. Since c is arbitrary, it suffices to prove that [itex](g\circ f)^{-1}(c) = a[/itex] as well. How can you do that?
 
  • #3
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"Since c is arbitrary, it suffices to prove that (g∘f)−1(c)=a as well. How can you do that?"

I don't quiet understand your question, or the statement before it.
 
  • #4
Deveno
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what other statement involving a and c can you use, here?

where did a and c come from?

can we get c from a, somehow using f and g? how?

our assumptions have been as follows:

g-1:c-->b

f-1:b-->a

what is f(a)?
what is g(b)?
 
  • #5
Office_Shredder
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"Since c is arbitrary, it suffices to prove that (g∘f)−1(c)=a as well. How can you do that?"

I don't quiet understand your question, or the statement before it.

To prove that two functions a(x) and b(x) are equal, you can prove that given any possible input x0, a(x0)=b(x0) (this shouldn't be particularly surprising) So to prove that [itex] (g\circ f)^{-1} = f^{-1}\circ g^{-1}[/itex], you can prove that given any [itex]c\in C[/itex], [itex] (g\circ f)^{-1}(c) = f^{-1}(g^{-1}(c))[/itex]. We know that the right hand side of this last equation is a by how we defined b and a in my post... how can you prove that the left hand side is equal to a as well?
 
  • #6
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This is the part that confuses me. Should I be saying something to the effect:
For any a[itex]\in[/itex]A there exists a unique b[itex]\in[/itex]B and a unique c[itex]\in[/itex]C such that f(a)=b and g(b)=c. It follows that g(f(a))=c. Thus (g o f)^(-1)(c)=a
 

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