Isomorphism of External Direct Products/General Result

Theorem
Let:


 * $(1): \quad \displaystyle \left({S, \circ}\right) = \prod_{k \mathop = 1}^n S_k = \left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right) \times \cdots \times \left({S_n, \circ_n}\right)$
 * $(2): \quad \displaystyle \left({T, \ast}\right) = \prod_{k \mathop = 1}^n T_k = \left({T_1, \ast_1}\right) \times \left({T_2, \ast_2}\right) \times \cdots \times \left({T_n, \ast_n}\right)$

be external direct products of algebraic structures.

Let $\phi_k: \left({S_k, \circ_k}\right) \to \left({T_k, \ast_k}\right)$ be 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
By definition of isomorphism, each of $\phi_k$ is a homomorphism which is a bijection.

From Cartesian Product of Bijections is Bijection: General Result, $\phi$ is a bijection.

From Homomorphism of External Direct Products: General Result, $\phi$ is a homomorphism.

Hence the result.