Normal-homomorph-containing implies strictly characteristic

Statement with symbols
Suppose $$N$$ is a normal-homomorph-containing subgroup of a group $$G$$: for any homomorphism $$\varphi:N \to G$$ such that $$\varphi(N)$$ is normal in $$G$$, we have $$\varphi(N) \le N$$.

Then, $$N$$ is a strictly characteristic subgroup of $$G$$: for any surjective endomorphism $$\alpha$$ of $$G$$, $$\alpha(N) \le N$$.

Facts used

 * 1) uses::Normality satisfies image condition: The image of a normal subgroup under a surjective homomorphism is a normal subgroup of the image.

Proof
Given: A normal subgroup $$N$$ of a group $$G$$ such that whenever $$\varphi:N \to G$$ is a homomorphism such that $$\varphi(N)$$ is normal in $$G$$, we have $$\varphi(N) \le N$$. A surjective endomorphism $$\alpha$$ of $$G$$.

To prove: $$\alpha(N) \le N$$.

Proof:


 * 1) (Given data used: $$N$$ is normal in $$G$$, $$\alpha$$ is a surjective endomorphism): $$\alpha(N)$$ is a normal subgroup of $$G$$: By fact (1), the image $$\alpha(N)$$ is a normal subgroup of $$\alpha(G)$$. By surjectivity, we have $$\alpha(G) = G$$, so $$\alpha(N)$$ is normal in $$G$$.
 * 2) (Given data used: $$N$$ is normal-homomorph-containing): $$\alpha(N) \le N$$: Let $$\varphi:N \to G$$ be the restriction of $$\alpha$$ to $$N$$. Then $$\varphi(N) = \alpha(N)$$ by definition, and by step (1), $$\varphi(N)$$ is normal in $$G$$. Since $$N$$ is normal-homomorph-containing, we get $$\varphi(N) \le N$$, so $$\alpha(N) \le N$$.