External Direct Product of Projection with Canonical Injection

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

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

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

Then $$pr_j \circ in_j = I_{S_j}$$, where $$I_{S_j}$$ is the identity mapping from $$S_j$$ to $$S_j$$.

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

So, $$s_j \in S_j$$.

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

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

Thus $$pr_j \circ in_j \left({s_j}\right) = s_j$$ and the result follows from the definition of the identity mapping.