Projection is Injection iff Factor is Singleton/Family of Sets

Theorem
Let $I$ be an indexing set.

Let $\family {S_i}_{i \mathop \in I}$ be a non-empty family of non-empty sets where $I$ is the indexing set.

Let $S = \displaystyle \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.

Let $\pr_j: S \to S_j$ be the $j$th projection on $S$.

Then $\pr_j$ is an injection $S_i$ is a singleton for all $i \in I \setminus \set {j}$.

Sufficient Condition
Let $S_i = \set {s_i}$ for all $i \in I \setminus \set {j}$.

Let $\map {\pr_j} x = \map {\pr_j} y = s_j$ for $x, y \in S$.

By definition of $j$th projection:
 * $\map x j = \map {\pr_j} x = s_j$
 * $\map y j = \map {\pr_j} y = s_j$

and so $\map x j = \map y j$.

By the definition of Cartesian product, for all $i \in I \setminus \set {j}$:
 * $\map x i, \map y i \in S_i = \set {s_i}$

and so $\map x i = \map y i$ for all $i \in I \setminus \set {j}$.

Thus $x = y$.

Thus $\pr_j$ is an injection by definition.

Necessary Condition
Let $\pr_j$ be an injection.

Then:
 * $\forall x, y \in S: \map {\pr_j} x = \map {\pr_j} y \implies x = y$

Since $\family {S_i}_{i \mathop \in I}$ is a non-empty family of non-empty sets, by the axiom of choice $S$ is not empty.

Let $z \in S$.

By the definition of Cartesian product $S$:
 * $\forall i \in I: \map z i \in S_i$.

It remains to show that:
 * $\forall k \in I \setminus \set j: S_k = \set {z_k}$

Let $x_k \in S_k$ for some $k \in I \setminus \set j$

Let $z' \in S$ be defined by:
 * $\map {z'} i = \map z i$ if $i \in I \setminus \set {k}$
 * $\map {z'} k = \map z k$

Since $j \neq k$ then:

Thus $z' = z$.

In particular:
 * $x_k = \map {z'} k = \map z k = z_k$

It follows that $S_k = \set {z_k}$.

Since $k$ was arbitrary then:
 * $\forall k \in I \setminus \set j: S_k = \set {z_k}$

The result follows.