Canonical Injection is Injection

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.

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 injections.

Proof
Let $x, x' \in S_1$.

Suppose that $\operatorname{in}_1 \left({x}\right) = \operatorname{in}_1 \left({x'}\right)$.

That is, suppose $\left({x, e_2}\right) = \left({x', e_2}\right)$.

By Equality of Ordered Pairs, $x = x'$.

Similarly, let $x, x' \in S_2$.

Suppose that $\operatorname{in}_2 \left({x}\right) = \operatorname{in}_2 \left({x'}\right)$.

That is, suppose $\left({e_1, x}\right) = \left({e_1, x'}\right)$.

Again by Equality of Ordered Pairs, $x = x'$.

Hence $\operatorname{in}_1$ and $\operatorname{in}_2$ are injections.