Canonical Injection is Monomorphism

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

Then the canonical injections:


 * $$\operatorname{in}_1: \left({S_1, \circ_1}\right) \to \left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right): \forall x \in S_1: \operatorname{in}_1 \left({x}\right) = \left({x, e_2}\right)$$


 * $$\operatorname{in}_2: \left({S_2, \circ_2}\right) \to \left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right): \forall x \in S_2: \operatorname{in}_2 \left({x}\right) = \left({e_1, x}\right)$$

are monomorphisms.

Generalized Version
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
 * $$\operatorname{in}_j: \left({S_j, \circ_j}\right) \to \prod_{i=1}^n \left({S_i, \circ_i}\right)$$

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