Solvable radical is normal-homomorph-containing

Statement
Suppose $$H$$ is the solvable radical of a group $$G$$, i.e., the unique largest solvable normal subgroup of $$G$$. (Note: The solvable radical may not exist for every group. It does, however, exist for finite groups, virtually solvable groups, and slender groups).

Then, $$H$$ is a normal-homomorph-containing subgroup of $$G$$. In other words, for any homomorphism $$f:H \to G$$ such that $$f(H)$$ is a normal subgroup of $$G$$, $$f(H) \le H$$.