Composite of Isomorphisms is Isomorphism/Algebraic Structure
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Theorem
Let:
- $\struct {S_1, \odot_1, \odot_2, \ldots, \odot_n}$
- $\struct {S_2, *_1, *_2, \ldots, *_n}$
- $\struct {S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}$
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:
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): $\S 1.2$