External Direct Product of Projection with Canonical Injection/General Result

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Theorem

Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$ be algebraic structures with identities $e_1, e_2, \dotsc, e_j, \dotsc, e_n$ respectively.

Let $\ds \struct {S, \circ} = \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}$ be the external direct product of $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$.

Let $\pr_j: \struct {S, \circ} \to \struct {S_j, \circ_j}$ be the $j$th projection from $\struct {S, \circ}$ to $\struct {S_j, \circ_j}$.

Let $\inj_j: \struct {S_j, \circ_j} \to \struct {S, \circ}$ be the canonical injection from $\struct {S_j, \circ_j}$ to $\struct {S, \circ}$.


Then:

$\pr_j \circ \inj_j = I_{S_j}$

where $I_{S_j}$ is the identity mapping from $S_j$ to $S_j$.


Proof

Let $\ds \tuple {s_1, s_2, \dotsc, s_{j - 1}, s_j, s_{j + 1}, \dotsc, s_n} \in \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}$.

So:

$s_j \in S_j$

From the definition of the canonical injection:

$\map {\inj_j} {s_j} = \tuple {e_1, e_2, \dotsc, e_{j - 1}, s_j, e_{j + 1}, \dotsc, e_n}$

So from the definition of the $j$th projection:

$\map {\pr_j} {e_1, e_2, \dotsc, e_{j - 1}, s_j, e_{j + 1}, \dotsc, e_n} = s_j$


Thus:

$\map {\pr_j \circ \inj_j} {s_j} = s_j$

and the result follows from the definition of the identity mapping.

$\blacksquare$


Sources

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