Projection is Injection iff Factor is Singleton/Family of Sets/Necessary Condition

Theorem
Let $\family {S_i}_{i \mathop \in I}$ be a non-empty family of non-empty sets where $I$ is an arbitrary index 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 $S$.

Let $\pr_j$ be an injection.

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

Proof
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 non-empty.

Let $z \in S$.

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

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

Let $z' \in S$ be defined by:
 * $\map {z'} i = \begin{cases}

\map z i & : i \in I \setminus \set {k} \\ x_k & : i = k \end{cases}$ 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.