I need to prove the following:
1) Let f ∈ F(A,B). Prove that if f is bijective, so is f-¹
2) Let f ∈ F(A,B) and g ∈ F(B,C). prove that if f and g are invertible so is gf and that (gf)-¹ = f-¹g-¹
3) Let A,B, and C be nonempty sets, and let f ∈ F(A,B) and g, h ∈ F(B,C)
a) prove that if f is surjective and gf=hf, then g=h.
b) Give an example in which gf=hf, but g ≠ h
I'm completely lost on those 3. Any help would be appreciated.
Let f ∈ F(A,B). Prove that if f is bijective, so is f-¹
Your notation looks a bit weird to me. Is that exactly from a question? What normal notation would be:
Let f:A⇒B. Prove that if f is bijective, so is f-¹
Well, we know that f is a function. Since f is a map:
For every element a∈A, there must exist a b∈B such that f(a) = b.
Applying an inverse to both sides, we get:
f-¹(f(a)) = f-¹(b)
a = f-¹(b)
For all a and some b. Thus, f-¹ must be onto.
Now let f-¹(b1) = f-¹(b2) for any b1, b2 in B. Since f is 1-1, f(f-¹(b1)) = f(f-¹(b2)). But f(f-¹(b1)) = b1 and f(f-¹(b2)) = b2. So b1 = b2. Thus, f-¹ is 1-1.
Let f ∈ F(A,B) and g ∈ F(B,C). prove that if f and g are invertible so is gf and that (gf)-¹ = f-¹g-¹
Notice how you can go from A->B and then from B->C. But remember, that f and g are invertible, so you can go from B->A and from C->B. Use this, and show that if an element is in C, then the element must correspond to some element in A.
Try this, and if you can get it, the thrid should become much easier.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
f ∈ F(A,B)
I think that this notatin is very weird too.
I accept that it may mean
and it's the logicest.
But it may mean that f is 2-argument function with first argument from A and second from B.
Kazy must reply.
IPBLE: Increasing Performance By Lowering Expectations.