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.