Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Surjection
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Theorem
Let $\family {X_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the Cartesian product of $\family {X_i}_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $p_i = \pr_i {\restriction_{Y_i} }$, where $\pr_i$ is the projection from $X$ to $X_i$.
Then:
- $p_i$ is a surjection.
Proof
Note that by definitions of a restriction and a projection then:
- $\forall y \in Y_i: \map {p_i} y = y_i$
Let $x \in X_i$.
Let $y \in Y_i$ be defined by:
- $\forall j \in I: y_j = \begin{cases}
z_j & j \ne i \\ x & j = i \end{cases}$
Then:
- $\map {p_i} y = y_i = x$
It follows that $p_i$ is an surjection by definition.
$\blacksquare$