Thread: Is my work correct? Prove the following is bijective.

The problem says:

Let A and B be sets. Show that there is a bijective map f: A x B --> B x A.

My professor helped me out and said that the function should be given by f((a,b)) = (b,a) (where a exists in A and b exists in B).

Here is the work I have so far.

Proof. (Injection) Let f((a1,b1)) = f((a2,b2)), where (a1,b1) , (a2,b2) exist in A x B. Then, (b1,a1) = (b2,a2), which implies f((b1,a1)) = f((b2,a2)), which implies (a1,b1) = (a2,b2). So, f is injective.

(Surjection) Fix (b,a) which exists in B x A. Put x = (a,b) and observe that f(x) = f((a,b)) = (b,a). So, f is surjective.

Since f is both injective and surjective, then f is bijective.

I feel like my Injection proof is wrong. Do any of you see anything wrong? Thanks in advance

Some times subscripts are troublesome.
Suppose that then

That means so that

Thanks. That makes sense, I just wasn't sure if it was really that simple or not.

Originally Posted by Lprdgecko

The problem says:

Let A and B be sets. Show that there is a bijective map f: A x B --> B x A.

My professor helped me out and said that the function should be given by f((a,b)) = (b,a) (where a exists in A and b exists in B).

Here is the work I have so far.

Proof. (Injection) Let f((a1,b1)) = f((a2,b2)), where (a1,b1) , (a2,b2) exist in A x B. Then, (b1,a1) = (b2,a2), which implies b1=b2 and a1=a2,

not {f((b1,a1)) = f((b2,a2))} Actually, this is also implied, but it's not relevant.

which implies (a1,b1) = (a2,b2). So, f is injective. (In other words: f(p)=f(q) implies p=q.)

(Surjection) For any (b,a) which exists in B x A, there exists (a,b) in AxB such that f((a,b)) = (b,a). So, f is surjective.

Since f is both injective and surjective, then f is bijective.

I feel like my Injection proof is wrong. Do any of you see anything wrong? Thanks in advance

I made some notes above, in red.

In general the proof is pretty good - just needs that one significant change.

Often, when you, the student, define a function, the professor likes you to prove that the function is "well defined" - i.e, p=q, implies f(p)=f(q). In this case, you would show: if (a1,b1)=(a2,b2) in AxB, then f(a1,b1)=f(a2,b2) in BxA. This is often done "indirectly" by showing that implies that .

[QUOTE=SammyS;589482]

Often, when you, the student, define a function, the professor likes you to prove that the function is "well defined" - i.e, p=q, implies f(p)=f(q).
I don't think that's the case, unless you're doing something with equivalence classes like