# Bijections on ordered pairs formed from other bijections?

How could you go about proving the following:

Let $f_1$ and $f_2$ denote bijections where $f_1:A_1 \rightarrow B_2$ and $f_2:A_2 \rightarrow B_2$. If $g:A_1 \times A_2 \rightarrow B_1 \times B_2$ where $g(x, y) = (f_1(x), f_2(y))$, then $g$ is a bijection.

This is a sample exam problem. I'm not really sure how to go about proving this, as it seems intuitively correct to me, so I'm not really sure where to start.

I'm not so much interested in a proof as some sort of starting point or hint that could tell me what I need to prove or what I might be able to use.

EDIT: Just to be clear, we're using what I presume is the standard definition for a bijection, a function that is both an injection (one-to-one) and a surjection (onto).

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A good starting point would be your definition of "bijection". –  Chris Eagle Dec 6 '11 at 23:28

Have you tried showing that $g$ is injective / surjective? What I will say will not really be hints, but more of "what you should've thought of to solve this problem" :

$\circ$ To show that $g$ is injective, you need to show that $g(x_a,y_a) = g(x_b,y_b)$ implies that $x_a=x_b$ and $y_a=y_b$. But $g(x,y) = (f_1(x),f_2(x))$. What do you know about $f_1$ and $f_2$ that could help you sort this out?

$\circ$ To show that $g$ is surjective, you need to show that for each element $(z_a, z_b) \in B_1 \times B_2$, there is an element $(x_a, x_b)$ such that $g(x_a, x_b) = (f_1(x_a),f_2(x_b)) = (z_a,z_b)$. What do you know about $f_1$ and $f_2$ that could give you such $x_a$'s and $x_b$'s?

Hope that helps,

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Thanks for the helpful tips. If I'm correct, it's clear that you could find a $z_a \in B_1$ and a $z_b \in B_2$ in the range of $f_1$ and $f_2$ since they are both surjections. –  Rafe Kettler Dec 7 '11 at 0:13
Yes, you're correct. In fact, you can be more general, and you can show that $f_1$ and $f_2$ are injections $\Longleftrightarrow$ $g$ is an injection, and you can show that $f_1$ and $f_2$ are surjections $\Longleftrightarrow$ $g$ is a surjection. Just read the parts of your proof where you consider injectivity to get the first result, and read the part where you consider surjectivity to get the second one. –  Patrick Da Silva Dec 7 '11 at 0:18

In a nutshell, you need to show that the function defined is a bijection. What does that mean? It is 1-1 and onto.

How does one show a function is 1-1? Take $(a_1,a_2),(b_1,b_2)$ from the domain with the same image. It would mean that $f_i(a_i)=f_i(b_i)$. What can you deduce from that?

Similarly to show $g$ is onto.

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Prove it in two steps: first, show that it's an injection, and then show that it's a surjection.

To show that it's an injection, suppose $g(x,y)=g(x',y')$. Use the definition of $g$ to show that this implies $x=x', y=y'$, and hence $(x,y)=(x',y')$.

To show that it's a surjection, take any $(b_1, b_2) \in B_1 \times B_2$. Can you find $(a_1, a_2) \in A_1 \times A_2$ such that $g(a_1, a_2)=(b_1, b_2)$?

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