External Direct Product of Projection with Canonical Injection/General Result

Theorem
Let $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_j, \circ_j}\right), \ldots, \left({S_n, \circ_n}\right)$ be algebraic structures with identities $e_1, e_2, \ldots, e_j, \ldots, e_n$ respectively.

Let $\displaystyle \left({S, \circ}\right) = \prod_{i=1}^n \left({S_i, \circ_i}\right)$ be the external direct product of $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_j, \circ_j}\right), \ldots, \left({S_n, \circ_n}\right)$.

Let $\operatorname{pr}_j: \left({S, \circ}\right) \to \left({S_j, \circ_j}\right)$ be the $j$th projection from $\left({S, \circ}\right)$ to $\left({S_j, \circ_j}\right)$.

Let $\operatorname{in}_j: \left({S_j, \circ_j}\right) \to \left({S, \circ}\right)$ be the canonical injection from $\left({S_j, \circ_j}\right)$ to $\left({S, \circ}\right)$.

Then:
 * $\operatorname{pr}_j \circ \operatorname{in}_j = I_{S_j}$

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

Proof
Let $\displaystyle \left({s_1, s_2, \ldots, s_{j-1}, s_j, s_{j+1}, \ldots, s_n}\right) \in \prod_{i=1}^n \left({S_i, \circ_i}\right)$.

So, $s_j \in S_j$.

From the definition of the canonical injection, we have $\operatorname{in}_j \left({s_j}\right) = \left({e_1, e_2, \ldots, e_{j-1}, s_j, e_{j+1}, \ldots, e_n}\right)$.

So from the definition of the $j$th projection, we have $\operatorname{pr}_j \left({\left({e_1, e_2, \ldots, e_{j-1}, s_j, e_{j+1}, \ldots, e_n}\right)}\right) = s_j$.

Thus $\operatorname{pr}_j \circ \operatorname{in}_j \left({s_j}\right) = s_j$ and the result follows from the definition of the identity mapping.