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:
- $\phi_1 \circ \chi = \psi_1$
- $\phi_2 \circ \chi = \psi_2$
Proof
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 $\chi = \xi^{-1}$.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.13$