Composite of Isomorphisms is Isomorphism/R-Algebraic Structure
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Theorem
Let:
- $\struct {S_1, \ast_1}_R$
- $\struct {S_2, \ast_2}_R$
- $\struct {S_3, \ast_3}_R$
be $R$-algebraic structures with the same number of operations.
Let:
- $\phi: \struct {S_1, \ast_1}_R \to \struct {S_2, \ast_2}_R$
- $\psi: \struct {S_2, \ast_2}_R \to \struct {S_3, \ast_3}_R$
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:
So:
- From Composite of Homomorphisms for R-Algebraic Structures is Homomorphism we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both homomorphisms
- From Composite of Bijections is Bijection we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both bijections;
and hence by definition also isomorphisms.
$\blacksquare$