Projection is Epimorphism/General Result

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Theorem

Let $\struct {\SS, \circ}$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_k, \circ_k}, \ldots, \struct {S_n, \circ_n}$.

Then:

for each $j \in \closedint 1 n$, $\pr_j$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_j, \circ_j}$

where $\pr_j: \struct {\SS, \circ} \to \struct {S_j, \circ_j}$ is the $j$th projection from $\struct {\SS, \circ}$ to $\struct {S_j, \circ_j}$.


Proof

From Projection is Surjection, $\pr_j$ is a surjection for all $j$.

We now need to show it is a homomorphism.

Let $\mathbf s, \mathbf t \in \struct {\SS, \circ}$ where $\mathbf s = \tuple {s_1, s_2, \ldots, s_j, \ldots, s_n}$ and $\mathbf t = \tuple {t_1, t_2, \ldots, t_j, \ldots, t_n}$.

Then:

\(\ds \map {\pr_j} {\mathbf s \circ \mathbf t}\) \(=\) \(\ds \map {\pr_j} {\tuple {s_1, s_2, \ldots, s_j, \ldots, s_n} \circ \tuple {t_1, t_2, \ldots, t_j, \ldots, t_n} }\) Definition of $\mathbf s$ and $\mathbf t$
\(\ds \) \(=\) \(\ds \map {\pr_j} {\tuple {s_1 \circ_1 t_1, s_2 \circ_2 t_2, \ldots, s_j \circ_j t_j, \ldots, s_n \circ_n t_n} }\)
\(\ds \) \(=\) \(\ds s_j \circ_j t_j\)
\(\ds \) \(=\) \(\ds \map {\pr_j} {\mathbf s} \circ_j \map {\pr_j} {\mathbf t}\)

and thus the morphism property is demonstrated.

$\blacksquare$


Sources

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