Composite of Isomorphisms is Isomorphism

Theorem
Let: be algebraic structures.
 * $\left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right)$
 * $\left({S_2, *_1, *_2, \ldots, *_n}\right)$
 * $\left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$

Let: be isomorphisms.
 * $\phi: \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_2, *_1, *_2, \ldots, *_n}\right)$
 * $\psi: \left({S_2, *_1, *_2, \ldots, *_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$

Then the composite of $\phi$ and $\psi$ is also a isomorphism.

R-Algebraic Structures
Let: be $R$-algebraic structures with the same number of operations.
 * $\left({S_1, \ast_1}\right)_R$
 * $\left({S_2, \ast_2}\right)_R$
 * $\left({S_3, \ast_3}\right)_R$

be isomorphisms.
 * $\phi: \left({S_1, \ast_1}\right)_R \to \left({S_2, \ast_2}\right)_R$
 * $\psi: \left({S_2, \ast_2}\right)_R \to \left({S_3, \ast_3}\right)_R$

Then the composite of $\phi$ and $\psi$ is also a isomorphism.

Proof
If $\phi$ and $\psi$ are both isomorphisms, then they are by definition:
 * homomorphisms;
 * bijections.

So:


 * From Composition of Homomorphisms we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both homomorphisms;
 * From Composite of Bijections we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both bijections;

... and hence by definition also isomorphisms.