External Direct Product of Projection with Canonical Injection

Theorem
Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be algebraic structures with identity elements $e_1$ and $e_2$ respectively.

Let $\struct {S_1 \times S_2, \circ}$ be the external direct product of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$

Let:
 * $\pr_1: \struct {S_1 \times S_2, \circ} \to \struct {S_1, \circ_1}$ be the first projection from $\struct {S_1 \times S_2, \circ}$ to $\struct {S_1, \circ_1}$
 * $\pr_2: \struct {S_1 \times S_2, \circ} \to \struct {S_2, \circ_2}$ be the second projection from $\struct {S_1 \times S_2, \circ}$ to $\struct {S_2, \circ_2}$.

Let:
 * $\inj_1: \struct {S_1, \circ_1} \to \struct {S_1 \times S_2, \circ}$ be the canonical injection from $\struct {S_1, \circ_1}$ to $\struct {S_1 \times S_2, \circ}$


 * $\inj_2: \struct {S_2, \circ_2} \to \struct {S_1 \times S_2, \circ}$ be the canonical injection from $\struct {S_2, \circ_2}$ to $\struct {S_1 \times S_2, \circ}$.

Then:
 * $(1): \quad \pr_1 \circ \inj_1 = I_{S_1}$
 * $(2): \quad \pr_2 \circ \inj_2 = I_{S_2}$

where $I_{S_1}$ and $I_{S_2}$ are the identity mappings on $S_1$ and $S_2$ respectively.

Proof
Let $\tuple {s_1, s_2} \in S_1 \times S_2$.

So, $s_1 \in S_1$ and $s_2 \in S_2$.

From the definition of the canonical injection, we have:
 * $\map {\inj_1} {s_1} = \tuple {s_1, e_2}$
 * $\map {\inj_2} {s_2} = \tuple {e_1, s_2}$

From the definition of the first projection:
 * $\map {\pr_1} {s_1, e_2} = s_1$

and similarly from the definition of the second projection:
 * $\map {\pr_2} {e_1, s_2} = s_2$

Thus:
 * $\map {\pr_1 \circ \inj_1} {s_1} = s_1$
 * $\map {\pr_2 \circ \inj_2} {s_2} = s_2$

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