Image under Epimorphism of Center is Subset of Center

Theorem
Let $G$ and $H$ be groups.

Let $\theta: G \to H$ be an epimorphism.

Let $\map Z G$ denote the center of $G$.

Then:
 * $\theta \sqbrk {\map Z G} \subseteq \map Z H$

Proof
Let $y \in \theta \sqbrk {\map Z G}$.

Let $t \in H$.

We have that:
 * $y = \map \theta z$

for some $z \in \map Z G$

As $\theta$ is an epimorphism, it is by definition surjective.

Then:
 * $t = \map \theta s$

for some $s \in G$.

Hence:

As $t$ is arbitrary, it follows that $y \in \map Z H$.

The result follows.