Monomorphism Image is Isomorphic to Domain

Theorem
The image of a monomorphism is isomorphic to its domain.

That is, if $\phi: S_1 \to S_2$ is a monomorphism, then:
 * $\phi: S_1 \to \operatorname{Im} \left({\phi}\right)$

is an isomorphism.

Proof
Let $\left({S_1, \circ_1}\right)$ and $\left({S_2, \circ_2}\right)$ be closed algebraic structures.

Let $\phi$ be a monomorphism from $\left({S_1, \circ_1}\right)$ to $\left({S_2, \circ_2}\right)$.

Let $T = \operatorname{Im} \left({\phi}\right)$ be the image of $\phi$.

By Morphism Property Preserves Closure, $\left({T, \circ_2}\right)$ is closed.

As $\phi$ is a monomorphism, it is by definition an injective homomorphism.

From Restriction of Injection is Injection, $\phi: S_1 \to \operatorname{Im} \left({\phi}\right)$ is an injection.

From Surjection iff Image equals Codomain, $\phi: S_1 \to \operatorname{Im} \left({\phi}\right)$ is a surjection.

Thus $\phi \to \operatorname{Im} \left({\phi}\right)$ is by definition a bijection.

Thus $\phi: S_1 \to \operatorname{Im} \left({\phi}\right)$ is a bijective homomorphism.

Hence, by definition, $\phi: S_1 \to \operatorname{Im} \left({\phi}\right)$ is an isomorphism.