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 $$\left({T_1 \times T_2, *}\right)$$.

Generalized Result
Let:


 * $$\left({S, \circ}\right) = \prod_{k=1}^n S_k = \left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right) \times \cdots \times \left({S_n, \circ_n}\right)$$
 * $$\left({T, \ast}\right) = \prod_{k=1}^n T_k = \left({T_1, \ast_1}\right) \times \left({T_2, \ast_2}\right) \times \cdots \times \left({T_n, \ast_n}\right)$$

be external direct products of algebraic structures.

If $$\phi_k: \left({S_k, \circ_k}\right) \to \left({T_k, \ast_k}\right)$$ is an isomorphism for each $$k \in \left[{1 \,. \, . \, n}\right]$$, then:
 * $$\phi: \left({s_1, \ldots, s_n}\right) \to \left({\phi_1 \left({s_1}\right), \ldots, \phi_n \left({s_n}\right)}\right)$$

is an isomorphism from $$\left({S, \circ}\right)$$ to $$\left({T, \ast}\right)$$.

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.