Isomorphism of External Direct Products

Theorem
Let:
 * $\left({S_1 \times S_2, \circ}\right)$ be the external direct product of two algebraic structures $\left({S_1, \circ_1}\right)$ and $\left({S_2, \circ_2}\right)$


 * $\left({T_1 \times T_2, *}\right)$ be the external direct product of two algebraic structures $\left({T_1, *_1}\right)$ and $\left({T_2, *_2}\right)$


 * $\phi_1$ be an isomorphism from $\left({S_1, \circ_1}\right)$ onto $\left({T_1, *_1}\right)$


 * $\phi_2$ be an isomorphism from $\left({S_2, \circ_2}\right)$ onto $\left({T_2, *_2}\right)$.

Then the mapping $\phi_1 \times \phi_2: \left({S_1 \times S_2, \circ}\right) \to \left({T_1 \times T_2, *}\right)$ defined as:
 * $\left({\phi_1 \times \phi_2}\right) \left({\left({x, y}\right)}\right) = \left({\phi_1 \left({x}\right), \phi_2 \left({y}\right)}\right)$

is an isomorphism from $\left({S_1 \times S_2, \circ}\right)$ to $\left({T_1 \times T_2, *}\right)$.

Proof
From Homomorphism of External Direct Products we have that $\phi_1 \times \phi_2: \left({S_1 \times S_2, \circ}\right) \to \left({T_1 \times T_2, *}\right)$ is a homomorphism.

The fact that it is an isomorphism follows by the fact that the composite of bijections is a bijection.