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

1. ## 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

2. Some times subscripts are troublesome.
Suppose that $f((p,q))=f((r,s))$ then $(q,p)=(s,r).$

That means $q=s~\&~p=r$ so that $(p,q)=(r,s).$

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

4. 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 $f(a_1,b_1)\neq f(a_2,b_2)$ implies that $(a_1,b_1)\neq (a_2,b_2)$.

5. [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 $f:\mathbb{Q}\to\mathbb{Q}:\frac{p}{q}\mapsto pq$

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### show that there exist one-one mapping from AxB to BxA

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