Prehomomorph-contained implies strictly characteristic

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., prehomomorph-contained subgroup) must also satisfy the second subgroup property (i.e., strictly characteristic subgroup)
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Statement

Suppose $H$ is a prehomomorph-contained subgroup of a group $G$. In other words, for any surjective homomorphism of groups $\alpha :K\to H$ with $K\le G$, $H$ is contained in $K$. Then, $H$ is a strictly characteristic subgroup of $G$.

Proof

Given: A prehomomorph-contained subgroup $H$ of a group $G$. A surjective endomorphism $\rho$ of $G$.

To prove: $\rho (H)\le H$.

Proof: Let $K={\rho }^{-1}(H)$ and let $\alpha$ be the restriction of $\rho$ to $K$. Since $\rho$ is surjective, $\rho (K)=H$, so $\alpha :K\to H$ is surjective. By the assumption that $H$ is prehomomorph-contained in $G$, we get $H\le K$. Thus, $H\le {\rho }^{-1}(H)$, so $\rho (H)\le H$.