Subhomomorph-containment is transitive

Statement
Suppose $$H \le K \le G$$. Suppose $$K$$ is a subhomomorph-containing subgroup of $$G$$ and $$H$$ is a subhomomorph-containing group of $$K$$. Then, $$H$$ is a subhomomorph-containing subgroup of $$G$$.

Related facts

 * Homomorph-containment is not transitive
 * Subhomomorph-containing implies right-transitively homomorph-containing
 * Full invariance is transitive

Proof
Given: Groups $$H \le K \le G$$. A homomorphism $$\varphi: A \to G$$ for a subgroup $$A$$ of $$H$$.

To prove: $$\varphi(A)$$ is a subgroup of $$H$$.

Proof: Since $$A$$ is a subgroup of $$H$$, $$A$$ is a subgroup of $$K$$. Thus, $$\varphi:A \to G$$ is a homomorphism from a subgroup of $$K$$. Since $$K$$ is subhomomorph-containing in $$G$$, $$\varphi(A)$$ is contained in $$K$$. Thus, $$\varphi$$ can be viewed as a map from $$A$$ to $$K$$.

Thus, $$\varphi:A \to K$$ is a map from a subgroup $$A$$ of $$H$$. Since $$H$$ is subhomomorph-containing in $$K$$, $$\varphi(A) \le H$$, and we are done.