Isomorphism is Equivalence Relation

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Theorem

Isomorphism is an equivalence on a set of magmas.


This result applies to all magmas: rings, groups, R-algebraic structures etc.


Proof

To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive.

So, checking in turn each of the criteria for equivalence:


Reflexive

From Identity Mapping is Automorphism, the identity mapping $I_S: S \to S$ is an automorphism, which is an isomorphism from a magma onto itself.

So $S \cong S$, and isomorphism is seen to be reflexive.

$\Box$


Symmetric

If $S \cong T$, then $\exists \phi: S \to T$ such that $\phi$ is an isomorphism.

Thus, from Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi^{-1}: T \to S$ is also an isomorphism, and $T \cong S$.

Thus we have shown that $S \cong T \implies T \cong S$, so $\cong$ is symmetric.

$\Box$


Transitive

Let $S_1 \cong S_2$, and $S_2 \cong S_3$.

We can define these two isomorphisms:

$\phi: S_1 \to S_2$
$\psi: S_2 \to S_3$


From Composite of Isomorphisms in Algebraic Structure is Isomorphism, we have that $\psi \circ \phi$ is an isomorphism between $S_1$ and $S_3$.


Thus we have shown that $\cong$ is transitive.

$\Box$


Thus isomorphism is reflexive, symmetric and transitive, and therefore an equivalence.

$\blacksquare$


Sources