Canonical Injection is Monomorphism/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.

Then the canonical injection
 * $\displaystyle \operatorname{in}_j: \left({S_j, \circ_j}\right) \to \prod_{i=1}^n \left({S_i, \circ_i}\right)$

defined as:


 * $\operatorname{in}_j \left({x}\right) = \left({e_1, e_2, \ldots, e_{j-1}, x, e_{j+1}, \ldots, e_n}\right)$

is a monomorphism.

Proof
First it needs to be established that the canonical injections are in fact injective.

Suppose $x, y \in S_j: \operatorname{in}_j \left({x}\right) = \operatorname{in}_j \left({y}\right)$.

Then $\left({e_1, e_2, \ldots, e_{j-1}, x, e_{j+1}, \ldots, e_n}\right) = \left({e_1, e_2, \ldots, e_{j-1}, y, e_{j+1}, \ldots, e_n}\right)$.

By the definition of equality of ordered $n$-tuples, it follows directly that $x = y$.

Thus the canonical injections are injective.

Now to prove the morphism property.

Let $x, y \in \left({S_j, \circ_j}\right)$.

Then:

and the morphism property has been demonstrated to hold.

Thus $\displaystyle \operatorname{in}_j: \left({S_j, \circ_j}\right) \to \prod_{i=1}^n \left({S_i, \circ_i}\right)$ has been shown to be an injective homomorphism and therefore a monomorphism.