Isomorphism Preserves Identity/Proof 1

Theorem
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an isomorphism.

Then $\circ$ has an identity $e_S$ iff $\phi \left({e_S}\right)$ is the identity for $*$.

Proof
Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ has an identity $e_S$.

Then $\forall x \in S: x \circ e_S = x = e_S \circ x$.

The result follows directly from the morphism property of $\circ$ under $\phi$:

As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.

Thus the result for $\phi \left({e_S}\right)$ can be applied to $\phi^{-1} \left({\phi \left({e_S}\right)}\right)$.