Isomorphism is Equivalence Relation

Theorem
Isomorphism is an equivalence on the set of algebraic structures.

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


 * Reflexive:

Using Identity Mapping is a Bijection, define $$I_S: S \to S: \forall s \in S: I_S \left({s}\right) = s$$.

This is bijective, and is trivially a homomorphism, so $$S \cong S$$, and isomorphism is seen to be Reflexive.


 * Symmetric:

If $$\left({S, \circ}\right) \cong \left({T, *}\right)$$, then $$\exists \phi: \left({S, \circ}\right) \to \left({T, *}\right)$$ such that $$\phi$$ is an Isomorphism.

Thus, from Inverse Isomorphism, $$\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$$ is also an isomorphism, and $$T \cong S$$.

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


 * Transitive:

Let $$\left({S_1, \circ_1}\right) \cong \left({S_2, \circ_2}\right)$$, and $$\left({S_2, \circ_2}\right) \cong \left({S_3, \circ_3}\right)$$.

We can define these two isomorphisms:


 * $$\phi: \left({S_1, \circ_1}\right) \to \left({S_2, \circ_2}\right)$$
 * $$\psi: \left({S_2, \circ_2}\right) \to \left({S_3, \circ_3}\right)$$

So as to minimise notational confusion, we temporarily denote composition of mappings $$\psi \bullet \phi$$.

We want to show that $$\psi \bullet \phi$$ is an isomorphism between $$S_1$$ and $$S_3$$.

As $$\phi$$ and $$\psi$$ are both bijective, then the composition $$\psi \bullet \phi$$ is likewise a bijection.

From Composition of Homomorphisms, we know that the composite of homorphisms is itself a homomorphism.

So we have established that a composite of isomorphisms is a bijective homomorphism, and thus that $$\psi \bullet \phi$$ is in fact an isomorphism, thus demonstrating that $$S_1 \cong S_3$$.

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


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