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g: B->A such that g(f(a))=a for all aEA
h:B->A such that f(h(b)=b for all bEB
show that g(b)=h(b) for all bEB
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Set f(a) = b. Substitute to get g(b) = g(f(a)), which we know from the definition of g is equal to a. Thus, g(b) = a.
Do the same for h to get h(f(a)). We know from the definition of h that f(h(f(a))) = f(a), which tells us that h(f(a)) = a. Substitute back to get h(b) = a = g(b).
Wrap it in bacon
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No more copying our solutions for your homework without understanding what youre supposed to learn. So tell us honestly first: do you understand the solutions weve been giving you?
EDIT: Grr at TheDude for posting before me. Also, I see that TheDudes proof is almost (but not quite) complete: TheDude is assuming that f is both surjective and injective. This is in fact true but it still has to be stated why.
So, @ xXxEmZyxXx: If you copy TheDudes solution blindly, you will NOT get full marks for your homework, haha! Now concentrate on understanding rather than copying.
Last edited by JaneFairfax (2007-10-31 02:22:20)
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A much better method is to let c = h(b), where b ∈ B. But I will not tell xXxEmZyxXx how to proceed. xXxEmZyxXx must by now have learnt how to do some things by himself/herself.
Well, all right, heres a hint to help xXxEmZyxXx: f(c) = b. (Why?) ∴ g(b) = g(f(c)) = ?
Last edited by JaneFairfax (2007-10-31 02:20:37)
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Heh, your method is a lot easier.
Wrap it in bacon
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