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$\forall set\; A,B,C\;\;$Let $a=card(A),\;\; b=card(B),\;\; c=card(C)$

Claim

prove $(ab)^c = a^cb^c$

We need to show that $(A\times B)^C \approx A^c \times B^c$.

Thus let $\pi_A:A\times B\rightarrow A\;$ $\pi_B: A\times B \rightarrow B$ be the projections.

then the function: $$\phi:(A\times B)^C \rightarrow A^C \times B^C$$ prescribed by: $$h \mapsto (\pi_A \circ h, \pi_B \circ h)\tag{3.19}$$ is bijective.

Question

my textbook requires to find the inverse function of (3.19).

Is it sufficient for me to denote the inverse of (3.19) as follow?

$(\pi_A \circ f, \pi_B \circ g) \mapsto f : A^C \times B^C \rightarrow (A\times B)^C $ where $f: C \rightarrow A \times B$

drhab
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Daschin
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  • https://math.stackexchange.com/questions/1751288/equinumerousity-of-operations-on-cardinal-numbers – Asaf Karagila Jun 05 '17 at 07:59
  • please check my edit. – drhab Jun 05 '17 at 08:19
  • this site is full of people who are so much benevolent.. I decided to start getting a bachelor degree in mathematics from the help of you guys.. thx – Daschin Jun 05 '17 at 08:36
  • @drhab What's the difference between function and projections? is function including the projection? – Daschin Jun 05 '17 at 10:17
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    A projection is a special function. In this context $\pi_A:A\times B\to A$ is a the function prescribed by $\langle a,b\rangle\mapsto a$. Writing it as a set of ordered pairs we have $\pi_A:={\langle\langle a,b\rangle, a\rangle\mid a\in A\wedge b\in B}$. – drhab Jun 05 '17 at 10:22
  • Consequently if $h:C\to A\times B$ is prescribed by $c\mapsto\langle h_1(c),h_2(c)\rangle$ then $\pi_A\circ h=h_1$. – drhab Jun 05 '17 at 10:33

2 Answers2

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Just writing that doesn't show the function is invertible; one could just as easily say that the inverse of the function sending $x$ to $x^2$ is the inverse of the function sending $x^2$ to $x$, even though $x\mapsto x^2$ (defined on $\mathbb R$) is not invertible.

Here's the inverse: let $g:A^C\times B^C\to (A\times B)^C$ be defined such that given inputs $\alpha: C\to A$ and $\beta: C\to B$, and given $c\in C$, we have (recalling that $g((\alpha,\beta))$ should be a function taking elements of $C$ as inputs and elements of $A\times B$ as outputs) $$g((\alpha, \beta))(c) = (\alpha(c), \beta(c))$$ Can you show that $g$ is the inverse of $h$?

florence
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2

The prescription for the inverse is:$$\langle f,g\rangle\mapsto\{\langle c,\langle f(c),g(c)\rangle\rangle\mid c\in C\}$$

Written as a set of ordered pairs it is:$$\psi:=\{\langle\langle f,g\rangle,\{\langle c,\langle f(c),g(c)\rangle\rangle\mid c\in C\}\rangle\mid f\in A^C\wedge g\in B^C\}\subseteq (A^C\times B^C)\times (A\times B)^C\tag1$$

It is enough to verify that $\phi\circ\psi$ and $\psi\circ\phi$ are identities, where: $$\phi:=\{\langle h,\langle \pi_A\circ h,\pi_B\circ h\rangle\rangle\mid h\in (A\times B)^C\}\subseteq (A\times B)^C\times(A^C\times B^C)\tag2$$

Endnote: $$\phi\left(\psi\left(\left\langle f,g\right\rangle \right)\right)=\phi\left(\left\{ \langle c,\left\langle f\left(c\right),g\left(c\right)\right\rangle \mid c\in C\right\} \right)=$$$$\left\langle \pi_{A}\circ\left\{ \langle c,\left\langle f\left(c\right),g\left(c\right)\right\rangle \mid c\in C\right\} ,\pi_{B}\circ\left\{ \langle c,\left\langle f\left(c\right),g\left(c\right)\right\rangle \mid c\in C\right\} \right\rangle =\left\langle f,g\right\rangle $$ and

$$\psi\left(\phi\left(h\right)\right)=\psi\left(\left\langle \pi_{A}\circ h,\pi_{B}\circ h\right\rangle \right)=\left\{ \langle c,\left\langle \pi_{A}\circ h\left(c\right),\pi_{B}\circ h\left(c\right)\right\rangle \mid c\in C\right\} =h$$

drhab
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