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  • Definition.$1$
    $$X\times_1 Y=\{ (x,y)=\{\{x\},\{x,y\}\}: x\in X, y\in Y\}$$
  • Definition.$2$
    $$X\times_2 Y=\left\{ f\in Fonk\left( \left\{ 0,1\right\},X\cup Y\right) : f(0)\in X, f(1)\in Y\right\}$$

Show there is a bijection between Definition $1$ of Cartesian Product and Definition $2$ of Cartesian Product.

My proof trying:
Let $T$ be a function from $X\times_1 Y$ to $X\times_2 Y$. We will show that $T$ is bijection.

Case 1:
One-to-one. Let $a_1,a_2$ be any elements of domain of $T$. Then, $a_1$ of the form is $a_1:=(x_1,y_1)$, and $a_2$ of the form is $a_2:=(x_2,y_2)$. We need to show $T(a_1)=T(a_2)$.
I couldn't continue my proof, can you help?

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  • $\begingroup$ I assume $Fonk(A,B)$ is the set of functoins $A\to B$ (more often written $B^A$)? $\endgroup$ – Hagen von Eitzen Nov 1 '18 at 11:52
  • $\begingroup$ @HagenvonEitzen yes. $\endgroup$ – NewMoon Nov 1 '18 at 11:53
  • $\begingroup$ Edited definitions. $\endgroup$ – NewMoon Nov 1 '18 at 11:55
  • 3
    $\begingroup$ "Let $T$ be a function ... We will show that $T$ is a bijection" -- Taken literally, this cannot work. You better define a specific function $T$ instead of trying to show that any arbitrary function is a bijection $\endgroup$ – Hagen von Eitzen Nov 1 '18 at 11:56
  • $\begingroup$ "Fonk" as functions? $\endgroup$ – Ennar Nov 1 '18 at 11:56
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We can define $$\begin{align}T\colon X\times_2 Y&\to X\times_1 Y\\f&\mapsto(f(0),f(1))\end{align}$$ (because $f\in X\times_2 Y$ implies $f(0)\in X$ and $f(1)\in Y$ as required).

Showing that $T$ is injective and surjective is straightforward. Alternatively, also define $$\begin{align}U\colon X\times_1 Y&\to X\times_2 Y\\(x,y)&\mapsto t\mapsto \begin{cases}x&t=0\\y&t=1\end{cases}\end{align}$$ and show that $U\circ T$ and $T\circ U$ are the identities on the two versions of product.

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  • $\begingroup$ thanks for answer and comments $\endgroup$ – NewMoon Nov 2 '18 at 12:47

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