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I started with proving f is one to one.

I considered $(a_{1},b_{1}), (a_{2},b_{2}) \in A \times B$ such that $a_{1}=a_{2}$ and $b_{1}=b_{2}$.

I have to prove $(a_{1},b_{1})=(a_{2},b_{2})$. I am given that $f(a_{1},b_{1})=f(a_{2},b_{2})$. This implies that $(b_{1},a_{1})=(b_{2},a_{2})$. But $(b_{1},a_{1})= (a_{1},b_{1})$ iff $a_{1}=b_{1}$. How to conclude $a_{1}=b_{1}$?

Thanks in advance...!!!

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    It doesn't make sense to ask that $a_1=b_1$. they are elements of different sets. – 2017-01-02
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    If you are trying to prove injectivity you have to start with two points with $f(a_1,b_1)=f(a_2,b_2)$ and show that this implies $a_1=a_2,b_1=b_2$. – 2017-01-02
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    Yeah that's what I was also thinking. I'm struck after that step. Is there any other way to prove this? @lulu – 2017-01-02
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    No, you have to prove injectivity and surjectivity, and lulu's comment shows the most direct way to prove injectivity. – 2017-01-02
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    Follow the steps I sketched in my last comment. Suppose $f(a_1,b_1)=f(a_2,b_2)$. this means that $(b_1,a_1)=(b_2,a_2)$. What does that mean? – 2017-01-02
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    You could, alternatively, try to construct an inverse function $g:B\times A \to A\times B$, and show that for the specific $g$ you created, $g(f(a, b)) = (a, b)$ and $f(g(b, a)) = (b, a)$ for all $a\in A, b\in B$. – 2017-01-02
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    @lulu it means $b_{1}=b_{2}$ and $a_{1}=a_{2}$ right? – 2017-01-02
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    @Arthur I think that's a long way – 2017-01-02
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    Yes, that's correct. As practice, you should also follow the steps outlined by @Arthur . Constructing inverses, or trying to, is a good way to understand functions. – 2017-01-02
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    And don't forget you still need to show surjectivity. My process only addresses injectivity. – 2017-01-02
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    Sorry @lulu I still have confusion in one more thing. Here in order to prove that f is one to one it is enough if I prove that if $f(a_{1},b_{1}) = f(a_{2},b_{2})$, then $(a_{1},b_{1})=(a_{2},b_{2})$. – 2017-01-02
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    But according to our discussion here I have proved $(b_{1},a_{1}) = (b_{2},a_{2})$. We already know that Cartesian product is not commutative – 2017-01-02
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    The Cartesian product is defined so that $(x,y)=(w,v)\iff x=w,y=v$. – 2017-01-02
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    @lulu agreed but how does it conclude $(a_{1},b_{1})=(a_{2},b_{2})$ – 2017-01-02
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    If $(b_1,a_1)=(b_2,a_2)$ then $b_1=b_2,a_1=a_2$. That's how the Cartesian product works. – 2017-01-02

1 Answers 1

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Injectivity: Suppose $f(a,b)=f(c,d)$. Then $(b, a)=(d,c)$, so $b=d$ and $a=c$. Thus $(a,b)=(c, d)$.

Surjectivity: Suppose $(b,a)\in B\times A$. Then $b\in B$ and $a\in A$, so $(a,b)\in A\times B$, and $f(a,b)=(b,a)$

So $f$ is bijective.

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    Surjectivity part is very clear to me thank you @MPW – 2017-01-02
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    can you please explain how last sentence in your injectivity proof is true? – 2017-01-02
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    how $b=d$ and $a=c $implies that $(a,b)=(c,d)$? – 2017-01-02
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    @Sahana By definition of the direct product $A\times B$. Two elements in the direct product are equal iff each of their components are equal. – 2017-01-02
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    @Arthur I understand that. I understood how $b=d$ and $a=c$ came. My question is how does it conclude $(a,b)=(c,d)$? Sorry if I am troubling. But I seriously don't get it – 2017-01-02
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    Ordered pairs are equal if and only if the corresponding components are equal. Think about complex numbers: $z=w$ if and only if $Re(z)=Re(w)$ and $Im(z)=Im(w)$. In the proof, the second sentence concludes that the components are equal, and the third sentence uses what we just said to conclude that the ordered pairs are equal. – 2017-01-02
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    @Sahana That's my point, though. When you, in the direct product $A\times B$, say that "$(a, b) = (c, d)$", that is _entirely equivalent_ to saying "$a = c$ and $b = d$". This convention is part of the _definition_ of what $A\times B$ is. It's not something you argue is true because of this and that, and you may or may not understand the argument. It just _is_. – 2017-01-02