Composite of Isomorphisms is Isomorphism/Algebraic Structure

Theorem

Let:

$\struct {S_1, \odot_1, \odot_2, \ldots, \odot_n}$

$\struct {S_2, *_1, *_2, \ldots, *_n}$

$\struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$

be algebraic structures.

Let:

$\phi: \struct {S_1, \odot_1, \odot_2, \ldots, \odot_n} \to \struct {S_2, *_1, *_2, \ldots, *_n}$

$\psi: \struct {S_2, *_1, *_2, \ldots, *_n} \to \struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$

be isomorphisms.

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

Proof

If $\phi$ and $\psi$ are both isomorphisms, then they are by definition:

homomorphisms

bijections.

From Composite of Homomorphisms on Algebraic Structure is Homomorphism:

$\phi \circ \psi$ and $\psi \circ \phi$ are both homomorphisms.

From Composite of Bijections is Bijection:

$\phi \circ \psi$ and $\psi \circ \phi$ are both bijections.

Hence by definition both are also isomorphisms.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Theorem $6.1: \ 3^\circ$
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.2$
• 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\S 1.2$