Perfect core is homomorph-containing

Statement
Suppose $$G$$ is a group and $$H$$ is the perfect core of $$G$$: it is the unique largest perfect subgroup of $$G$$. Then, $$H$$ is a homomorph-containing subgroup of $$G$$: given any homomorphism $$\varphi:H \to G$$, we have $$\varphi(H) \le H$$.

Related facts

 * Fitting subgroup is normal-isomorph-free in finite
 * Solvable core is normal-isomorph-free in finite

Facts used

 * 1) uses::Perfectness is quotient-closed: The image of a perfect group under a surjective homomorphism is perfect.

Proof
Given: A group $$G$$ with perfect core $$H$$ and a homomorphism $$\varphi:H \to G$$.

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

Proof: $$\varphi(H)$$ is perfect by fact (1). By the definition of perfect core, any perfect subgroup of $$G$$ is contained in $$H$$. Thus, $$\varphi(H) \le H$$.