Look Moe can you just write down the actual bijection and ill show you what part of it I don't understand. Swear to jebus we'll be here forever if you want me to discover it myself.
I've as much as given it to you.
To finish it off, first I'll introduce a notation:
Let 'r' stand for restriction. I.e., for sets X and Y, we have
XrY = {<p q> | <p q> in X and q in Y}.
In other words, XrY is the set of all the ordered pairs <p q> in X such that p is in Y.
Each f in {f | f is a function from BuC into A} is of this form:
frB u frC.
That's just another way of saying what I mentioned in my last post.
So for f in {f | f is a function from BuC into A} let's map f to
<frB frC>.
So let J be this mapping:
For f in {f | f is a function from BuC into A},
let J(f) = <frB frC>.
In other words, for f in {f | f is a function from BuC into A}, I'm mapping f to some g and some h such that
<g h> is in
{g | g is a function from B into A} X {h | h is a function from C into A}, and
g and h are specifically frB and frC.
Now, all that remains is to prove that this J as just defined IS a bijection from {f | f is a function from BuC into A} onto
{g | g is a function from B into A} X {h | h is a function from C into A}.
So, I'd like you now to do that.
No. This is taking too much of my time up. The proofs not as basic as I thought if would be and I have waaaay too much other stuff to be doing. 9 exams and this probably wont show up on the one exam it could be in. Not worth the time. I thank you for your effort.
Thread marked as solved.