Set Products on Same Set are Equivalent

Theorem

Let $S$ and $T$ be sets.

Let $\struct {P, \phi_1, \phi_2}$ and $\struct {Q, \psi_1, \psi_2}$ be products of $S$ and $T$.

Then there exists a unique bijection $\chi: Q \to P$ such that:

$\phi_1 \circ \chi = \psi_1$

$\phi_2 \circ \chi = \psi_2$

We have that $\struct {P, \phi_1, \phi_2}$ is a set product.

From the definition of set product $\chi: Q \to P$ is the unique mapping such that:

$\phi_1 \circ \chi = \psi_1$

$\phi_2 \circ \chi = \psi_2$

Similarly, we have that $\struct {Q, \psi_1, \psi_2}$ is a set product.

So from the definition of set product there exists a unique mapping $\xi: P \to Q$ such that:

$\psi_1 \circ \xi = \phi_1$

$\psi_2 \circ \xi = \phi_2$

It can be seen that $\chi \circ \xi: P \to P$ is the unique mapping such that:

$\phi_1 \circ \chi \circ \xi = \phi_1$

$\phi_2 \circ \chi \circ \xi = \phi_2$

and similarly that $\xi \circ \chi: Q \to Q$ is the unique mapping such that:

$\psi_1 \circ \xi \circ \chi = \psi_1$

$\psi_2 \circ \xi \circ \chi = \psi_2$

So it appears that:

$\chi \circ \xi = I_P$

$\xi \circ \chi = I_Q$

where $I_P$ and $I_Q$ are the identity mappings on $P$ and $Q$ respectively.

From the corollary to Bijection iff Left and Right Inverse it follows that $\chi$ and $\xi$ are bijections such that $\chi = \xi^{-1}$.

$\blacksquare$

1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.13$