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.