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I need to prove that $|(A^C) × (B^C)| = |(A×B)^C| $ .

I've tried to find a bijection but I'm stuck: we need $f: A^C × B^C \to (A×B)^C$. Let $l:C \to A$ and $k: C \to B$. $f$ inputs a pair of functions $(l,k)$ and outputs a function $f(l,k): C \to A×B$, which outputs a pair $(a,b) \in A×B$. I'm confused, I don't know how to proceed.

Thanks.

Lstoi
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2 Answers2

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Let $f_{l,k}(x)=(l(x), k(x))$.

Mario Carneiro
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  • Is my reasoning wrong? – Lstoi Apr 29 '14 at 04:35
  • Your setup is correct, but you haven't completed the proof yet. That's a suggestion for how to define your function $f$. – Mario Carneiro Apr 29 '14 at 04:36
  • Thank you for your suggestion. However, I'm having trouble proving that $f$ is surjective: let $(l(x),k(x)) \in (A × B)^C$, so there exists $f(l,k) : C \to A × B$ and, by definition of $f$, there exists $x \in C$ such that $f(l,k)(x) = (l(x),g(x))$. Is this correct? – Lstoi Apr 29 '14 at 05:47
  • @user132349 In order to show that $f$ is surjective, you need each $g\in(A\times B)^C$ to have some $l,k$ such that $f_{l,k}=g$. Choose $l=g_1$ and $k=g_2$. Then for all $x$, $f_{g_1,g_2}(x)=(g_1(x),g_2(x))=g(x)$, so $f_{g_1,g_2}=g$. (For functions to be equal, you need their values to be equal for all $x$, not for some $x$.) – Mario Carneiro Apr 29 '14 at 12:38
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And, on the other direction

$$ g: (A \times B)^C \longrightarrow A^C \times B^C \ , $$

given a map $\alpha : C \longrightarrow A\times B$ you just have to use the projections $\pi_A : A\times B \longrightarrow A$ and $\pi_B : A\times B \longrightarrow B$ to define the inverse $g$ to your $f$ as

$$ g(\alpha ) = (\pi_A \circ \alpha , \pi_B \circ \alpha) \ . $$

You can check indeed that $g \circ f = \mathrm{id}$ and $f\circ g = \mathrm{id}$. So you have bijections $f$ and $g$ inverse one to another.

Agustí Roig
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