0
$\begingroup$

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

Let $\{A_i\}_{i\in \{0, 1\}}$ be an indexed system of sets. Show that the map $$\mathbf f: \prod{\{A_i\}_{i\in \{0, 1\} }} \rightarrow A_0 \times A_1$$ 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: $$f(x)=f(y)$$ $$\therefore(x(0),x(1))=(y(0),y(1))$$ $$\therefore x(0) = y(0) \land x(1)=y(1) $$ $$\mathbf {dom}(x) = \mathbf{dom}(y) = \{0, 1\}$$ $$\therefore x(i) = y(i) \forall i \in \{0, 1\}$$ $\therefore f(g)$ 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: $$\prod{\{A_i\}_{i\in \{0, 1\} }}=\{\{(0,x_0 \in A_0),(1, x_1 \in A_1)\}\}$$ and the usual Cartesian Product: $$A_0 \times A_1 =(x_0 \in A_0, x_1 \in A_1)$$ respectively.

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?

$\endgroup$
1
$\begingroup$

Your proof of injectivity is correct, though I would add one more line concluding that $x=y$. For the proof of surjectivity, let $\langle a_0,a_1\rangle\in A_0\times A_1$ be arbitrary; then $a_0\in A_0$ and $a_1\in A_1$. Define

$$x:\{0,1\}\to A_0\cup A_1:\begin{cases} 0\mapsto a_0\\ 1\mapsto a_1\;; \end{cases}$$

then $f(x)=\langle x(0),x(1)\rangle=\langle a_0,a_1\rangle$, so $f$ is surjective.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.