Bijection of a Generalised Cartesian Product

I’m halfway through a proof and I’m kind of stuck. The exercise is:

Let ${A_i}_{iin {0, 1}}$ be an indexed system of sets. Show that the map $$mathbf f: prod{{A_i}_{iin {0, 1} }} rightarrow A_0 times A_1$$
given by
$$f(g) =(g(0), g(1))$$
is a Bijection.

First we would like to show that $f$ is Injective: Consider two instances of input function $g$, e.g. $x$
and $y$ hence:
$$therefore x(0) = y(0) land x(1)=y(1) $$
$$mathbf {dom}(x) = mathbf{dom}(y) = {0, 1}$$
$$therefore x(i) = y(i) forall i in {0, 1}$$
$therefore f(g)$ is Injective. Is my reasoning correct so far?

Now to showing $f$ is Surjective, we consider the the domain and codomain of the mapping, namely, the Generalised Cartesian Product of an Indexed Family of Sets:
$$prod{{A_i}_{iin {0, 1} }}={{(0,x_0 in A_0),(1, x_1 in A_1)}}$$
and the usual Cartesian Product:
$$A_0 times A_1 =(x_0 in A_0, x_1 in A_1)$$

It is trivial to see that the mapping given by $f$ is Surjective, as the two ordered pair elements of $f=(g(0),g(1))$ span over all $A_0$ and $A_1$ respectively and considering that $g$ could take any arbitrary function of $mathbf {domain}$ ${0, 1}$by definition of the Generalised Cartesian Product of an Indexed Family of Sets. But how do I write this formally?

Mathematics Asked on November 12, 2021

1 Answers

One Answer

Your proof of injectivity is correct, though I would add one more line concluding that $x=y$. For the proof of surjectivity, let $langle a_0,a_1ranglein A_0times A_1$ be arbitrary; then $a_0in A_0$ and $a_1in A_1$. Define

$$x:{0,1}to A_0cup A_1:begin{cases} 0mapsto a_0\ 1mapsto a_1;; end{cases}$$

then $f(x)=langle x(0),x(1)rangle=langle a_0,a_1rangle$, so $f$ is surjective.

Answered by Brian M. Scott on November 12, 2021

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