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Let $A$ a set. Define $C$ to be the collection of all functions $f:\{0,1\}\to A$. Construct a bijection $F:A\times A\to C$. For any $a,b\in A$, define $g_{ab}:\{0,1\}\to A$ by $g(0)=a$ and $g(1)=b$. Now define $F:A\times A\to C$ as $$F(a,b)=g_{ab}$$. Injective: Suppose $F(a,b)=F(c,d)$, then $g_{ab}=g_{cd}\implies \{(0,a),(1,b)\}=\{(0,c),(1,d)\}$ which means $a=c$ and $b=d$. Therefore $F$ is injective. Surjective: Pick any $g\in C$, then let $a=g(0)$ and $b=g(1)$. Then $g_{ab}=g$ because they have the same domain and codomain, and $g_a(0)=g(0)=a$ and $g_b(1)=g(1)=b$. So $F(a,b)=g$.
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Tue, 26 Nov 2019 18:57 GMT