The cartesian product of a family $(A_i\mid i\in I)$ is defined as $$\prod_{i\in I}A_i=\{f:I\to\bigcup A_i\mid f(i)\in A_i\}$$
I'm trying to prove the below theorem. While I'm able to construct a desired mapping and intuitively found that it is a bijection, I'm unable to prove that this mapping is actually bijective in a formal proof. I also found that it's easier to use the definition of cartesian product as n-array, but I would like to use the definition of cartesian product as a function.
Let $I_{n+1} = \{i\in\mathbb N\mid 1\leq i\leq n+1\}$ and $(A_i\mid i\in I_{n+1})$ be a family of finite sets. Then $|A_{n+1}\times\prod_{i\in I_n}A_i|=|\prod_{i\in I_{n+1}}A_i|$ where $|X|$ is the cardinality of $X$.
First, we generate a family of indexed sets $(B_i\mid i\in I_2)$ as follows: $B_1=A_{n+1}$ and $B_2=\prod_{i\in I_n}A_i$. Then we construct a mapping $f:B_1\times B_2\to\prod_{i\in I_{n+1}}A_i$ such that $\{(0,b_1),(1,b_2)\}\mapsto \{(n+1,b_1)\cup b_2\}$ where $b_1\in B_1=A_{n+1},b_2\in B_2=\prod_{i\in I_n}A_i$.
How can I proceed to prove that $f$ is bijective?
Best Answer
Consider
$$\begin {array}{l|rcl} f : & A_{n+1}\times\prod\limits_{i\in I_{n}}A_i & \longrightarrow & \prod\limits_{i\in I_{n+1}}A_i \\ & \phi & \longmapsto & f(\phi) \end{array}$$
where $\phi: \{1,2\} \to A_{n+1} \bigcup \prod\limits_{i\in I_{n}}A_i$ with $\phi(1) \in A_{n+1}$ and $\phi(2) \in \prod\limits_{i\in I_{n}}A_i$.
$f(\phi)$ is defined by $$f(\phi)(i)=\begin{cases}\phi(1) & \text{ if } i=n+1\\ \phi(2)(i) & \text{ if } i \in I_n \end {cases}$$
$f$ is surjective
For $\psi \in \prod\limits_{i\in I_{n+1}}A_i$ define $\phi \in A_{n+1}\times\prod\limits_{i\in I_{n}}A_i$ by $$\phi(i)=\begin{cases}\psi(n+1) & \text{ if } i=1\\ \phi(2) & \text{ if } i =2 \end {cases}$$ where $\phi(2)(j)=\psi(j)$ for $j \in I_n$. $\phi(2)$ is a map belonging to $\prod\limits_{i\in I_{n}}A_i$. You’ll verify that $f(\phi)=\psi$ proving that $f$ is surjective.
$f$ is injective
This is an easy verification that $\phi_1=\phi_2$ if $f(\phi_1)=f(\phi_2)$.