Projection is Epimorphism

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Theorem

Let $\struct {\SS, \circ}$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.


Then:

$\pr_1$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_1, \circ_1}$
$\pr_2$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_2, \circ_2}$

where $\pr_1$ and $\pr_2$ are the first and second projection respectively of $\struct {\SS, \circ}$.


General Result

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, $\pr_1$ and $\pr_2$ are surjections.

We now need to show they are homomorphisms.

Let $s, t \in \struct {\SS, \circ}$ where $s = \tuple {s_1, s_2}$ and $t = \tuple {t_1, t_2$.

Then:

\(\ds \map {\pr_1} {s \circ t}\) \(=\) \(\ds \map {\pr_1} {\tuple {s_1, s_2} \circ \tuple {t_1, t_2} }\)
\(\ds \) \(=\) \(\ds \map {\pr_1} {\tuple {s_1 \circ_1 t_1, s_2 \circ_2 t_2} }\)
\(\ds \) \(=\) \(\ds s_1 \circ_j t_1\)
\(\ds \) \(=\) \(\ds \map {\pr_1} s \circ_1 \map {\pr_1} t\)


\(\ds \map {\pr_2} {s \circ t}\) \(=\) \(\ds \map {\pr_2} {\tuple {s_1, s_2} \circ \tuple {t_1, t_2} }\)
\(\ds \) \(=\) \(\ds \map {\pr_2} {\tuple {s_1 \circ_1 t_1, s_2 \circ_2 t_2} }\)
\(\ds \) \(=\) \(\ds s_2 \circ_j t_2\)
\(\ds \) \(=\) \(\ds \map {\pr_2} s \circ_2 \map {\pr_2} t\)

Thus the morphism property is demonstrated for both $\pr_1$ and $\pr_2$.

$\blacksquare$


Sources

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