Projection on Group Direct Product is Epimorphism/Proof 1

Theorem
Let $\left({G_1, \circ_1}\right)$ and $\left({G_2, \circ_2}\right)$ be groups.

Let $\left({G, \circ}\right)$ be the group direct product of $\left({G_1, \circ_1}\right)$ and $\left({G_2, \circ_2}\right)$.

Then:
 * $\operatorname{pr}_1$ is an epimorphism from $\left({G, \circ}\right)$ to $\left({G_1, \circ_1}\right)$
 * $\operatorname{pr}_2$ is an epimorphism from $\left({G, \circ}\right)$ to $\left({G_2, \circ_2}\right)$

where $\operatorname{pr}_1$ and $\operatorname{pr}_2$ are the first and second projection respectively of $\left({G, \circ}\right)$.

Proof
From Projections are Surjections, $\operatorname{pr}_1$ and $\operatorname{pr}_2$ are surjections.

We now need to show they are homomorphisms.

Let $g, h \in \left({G, \circ}\right)$ where $g = \left({g_1, g_2}\right)$ and $h = \left({h_1, h_2}\right)$.

Then:

... and thus the morphism property is demonstrated for both $\operatorname{pr}_1$ and $\operatorname{pr}_2$.