External Direct Product of Projection with Canonical Injection

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

Let $\left({S_1 \times S_2, \circ}\right)$ be the external direct product of $\left({S_1, \circ_1}\right)$ and $\left({S_2, \circ_2}\right)$

Let:
 * $\operatorname{pr}_1: \left({S_1 \times S_2, \circ}\right) \to \left({S_1, \circ_1}\right)$ be the first projection from $\left({S_1 \times S_2, \circ}\right)$ to $\left({S_1, \circ_1}\right)$
 * $\operatorname{pr}_2: \left({S_1 \times S_2, \circ}\right) \to \left({S_2, \circ_2}\right)$ be the second projection from $\left({S_1 \times S_2, \circ}\right)$ to $\left({S_2, \circ_2}\right)$.

Let:
 * $\operatorname{in}_1: \left({S_1, \circ_1}\right) \to \left({S_1 \times S_2, \circ}\right)$ be the canonical injection from $\left({S_1, \circ_1}\right)$ to $\left({S_1 \times S_2, \circ}\right)$


 * $\operatorname{in}_2: \left({S_2, \circ_2}\right) \to \left({S_1 \times S_2, \circ}\right)$ be the canonical injection from $\left({S_2, \circ_2}\right)$ to $\left({S_1 \times S_2, \circ}\right)$.

Then:
 * $(1): \quad \operatorname{pr}_1 \circ \operatorname{in}_1 = I_{S_1}$
 * $(2): \quad \operatorname{pr}_2 \circ \operatorname{in}_2 = I_{S_2}$

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

Proof
Let $\displaystyle \left({s_1, s_2}\right) \in S_1 \times S_2$.

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

From the definition of the canonical injection, we have:
 * $\operatorname{in}_1 \left({s_1}\right) = \left({s_1, e_2}\right)$
 * $\operatorname{in}_2 \left({s_2}\right) = \left({e_1, s_2}\right)$

So from the definition of the first projection:
 * $\operatorname{pr}_1 \left({\left({s_1, e_2}\right)}\right) = s_1$

and similarly from the definition of the second projection:
 * $\operatorname{pr}_2 \left({\left({e_1, s_2}\right)}\right) = s_2$

Thus:
 * $\operatorname{pr}_1 \circ \operatorname{in}_1 \left({s_1}\right) = s_1$
 * $\operatorname{pr}_2 \circ \operatorname{in}_2 \left({s_2}\right) = s_2$

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