Projection is Epimorphism

Theorem
Let $\left({S, \circ}\right)$ be the external direct product of the algebraic structures $\left({S_1, \circ_1}\right)$ and $\left({S_2, \circ_2}\right)$.


 * $\operatorname{pr}_1$ is an epimorphism from $\left({S, \circ}\right)$ to $\left({S_1, \circ_1}\right)$;
 * $\operatorname{pr}_2$ is an epimorphism from $\left({S, \circ}\right)$ to $\left({S_2, \circ_2}\right)$.

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

Generalized Version
Let $\left({S, \circ}\right)$ be the external direct product of the algebraic structures $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_k, \circ_k}\right), \ldots, \left({S_n, \circ_n}\right)$.

Then for each $j \in \left[{1 \,. \, . \, n}\right]$, $\operatorname{pr}_j$ is an epimorphism from $\left({S, \circ}\right)$ to $\left({S_j, \circ_j}\right)$

where $\operatorname{pr}_j: \left({S, \circ}\right) \to \left({S_j, \circ_j}\right)$ is the $j$th projection from $\left({S, \circ}\right)$ to $\left({S_j, \circ_j}\right)$.

Proof
From Projections are Surjections, $\operatorname{pr}_j$ is a surjection for all $j$.

We now need to show it is a homomorphism.

Let $s, t \in \left({S, \circ}\right)$ where $s = \left({s_1, s_2, \ldots, s_j, \ldots, s_n}\right)$ and $t = \left({t_1, t_2, \ldots, t_j, \ldots, t_n}\right)$.

Then:

... and thus the morphism property is demonstrated.