Projection is Injection iff Factor is Singleton

Theorem
Let $S_1, S_2, \ldots, S_n$ be non-empty sets.

Let $\displaystyle S = \prod_{i \mathop = 1}^n S_i$ be the cartesian product of $S_1, S_2, \ldots, S_n$.

Let $\operatorname{pr}_j: S \to S_j$ be the $j$th projection on $S$.

Then $\operatorname{pr}_j$ is an injection $S_k$ is a singleton for all $k \in \left\{{1, 2, \ldots, n}\right\}$ where $k \ne j$.

Sufficient Condition
Suppose $S_k = \left\{{s_k}\right\}$ for all $k \in \left\{{1, 2, n}\right\}$ where $k \ne j$.

Let $\operatorname{pr}_j \left({x}\right) = \operatorname{pr}_j \left({y}\right) = z$ for $x, y \in S$.

Then by definition of $j$th projection, $x, y \in S$ are given by:
 * $x = \left({s_1, s_2, \ldots, s_{j-1}, z, s_{j+1}, \ldots, s_n}\right)$
 * $y = \left({s_1, s_2, \ldots, s_{j-1}, z, s_{j+1}, \ldots, s_n}\right)$

and so $x = y$.

Thus $\operatorname{pr}_j$ is an injection by definition.

Necessary Condition
Let $\operatorname{pr}_j$ be an injection.

Thus:
 * $\forall x, y \in S: \operatorname{pr}_j \left({x}\right) = \operatorname{pr}_j \left({y}\right) \implies x = y$

Let $x, y \in S$ be such that:
 * $x = \left({x_1, x_2, \ldots x_{j-1}, z, x_{j+1}, \ldots, x_n}\right)$
 * $y = \left({y_1, y_2, \ldots y_{j-1}, z, y_{j+1}, \ldots, y_n}\right)$

Then:
 * $\operatorname{pr}_j \left({x}\right) = \operatorname{pr}_j \left({y}\right) = z$

For $x = y$ it is necessary that:
 * $x_1 = y_1, x_2 = y_2, \ldots, x_{j-1} = y_{j-1}, x_{j+1} = y_{j+1}, \ldots, x_n = y_n$

That is:
 * $\forall k \in \left\{{1, 2, \ldots, n}\right\}, k \ne j: x_k = y_k$

That is, that $S_k$ is a singleton for all $k \in \left\{{1, 2, \ldots, n}\right\}$ where $k \ne j$.