Weird proof of a cardinal numbers identity

[Darya]

In discrete maths class we started on cardinal numbers and one of the identities we proved was [MATH]p^q*p^r=p^{q+r}[/MATH]. It was done this way:
All parts of the proof to me seem reasonable (hope I translated it correctly). I just don't see why we had to show bijection 2 times. What for did we have to show this weird transformation from [f2,f2] to (f1 ∪ f2) to [f1 ∪ f2\B, f1 ∪ f2\A] to [f2,f2] and then from g and all the way back to g? Why isn't it enough to show the bijection between [f2,f2] and (f1 ∪ f2)? Those are elements from LHD and RHD of the equation we're proving.. Thanks!

 

[Steven G]

You write f₁: B->A which is a function f₁ from A to B by definition.
However you write f₁: B->A is an element from A^B. What does that mean exactly?

 

[Darya]

You write f₁: B->A which is a function f₁ from A to B by definition.
However you write f₁: B->A is an element from A^B. What does that mean exactly?

f_1: B ->A is a function from B to A, at least in terms of notation I'm learning. A^B expresses all functions from B to A. I'm not sure if I can explain it well but to me it seems reasonable when you look at it that way: if card(A)=p, card(B)=q, the number of all functions from B to A is p^q which is card(A)^card(B) or |A|^ |B|. Is it ok now?

 

[Dr.Peterson]

this weird transformation from [f2,f2] to (f1 ∪ f2) to [f1 ∪ f2\B, f1 ∪ f2\A] to [f2,f2]

I think you meant "from [f₁,f₂] to (f₁ ∪ f₂) to [f₁ ∪ f₂\B, f₁ ∪ f₂\A] to [f₁,f₂]" And I don't think "\" is the right notation, but I can live with it. (Do you understand what the notation means?)

Why isn't it enough to show the bijection between [f1,f2] and (f1 ∪ f2)?

I think that's what they're doing. How would you do it (that is, show that what they have defined is a bijection)?

If you think only one part is needed, explain in words how that one part proves the bijection. We want to see if anything is missing. (The whole proof needs more words of explanation -- were there any in the original?)