Prehomomorph-contained implies strictly characteristic

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 $$\rho(H) \le H$$.