Projection from Product Topology is Open and Continuous/General Result

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Theorem

Let $\family {T_i}_{i \mathop \in I} = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.

Let $\ds S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.

Let $\tau$ denote the product topology on $S$.

Let $\pr_i: S \to S_i$ be the corresponding projection from $S$ onto $S_i$.


Then $\pr_i$ is open and continuous for all $i \in I$.


Proof

Projection is Continuous

By definition of the product topology on $S$:

$\tau$ is the initial topology on $S$ with respect to $\family {\pr_i}_{i \mathop \in I}$

By definition of the Initial Topoplogy:Definition 2:

$\tau$ is the coarsest topology on $S$ such that each $\pr_i: S \to S_i$ is a $\struct{\tau, \tau_i}$-continuous.

$\Box$

Projection is Open

Let $U \in \tau$.

It follows from the definition of product topology that $U$ can be expressed as:

$\ds U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} }$

where:

$J$ is an arbitrary index set
$n_j \in \N$
$i_{k, j} \in I$
$U_{k, j} \in \tau_{i_{k, j} }$.


For all $i' \in I$, define $V_{i', k, j} \in \tau_{i'}$ by:

$V_{i', k, j} = \begin {cases} U_{k, j} & : i' = i_{k, j} \\ S_{i'} & : i' \ne i_{k, j} \end {cases}$


For all $i \in I$, we have:

\(\ds \map {\pr_i} U\) \(=\) \(\ds \bigcup_{j \mathop \in J} \map {\pr_i} {\bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1} } { U_{k,j} } }\) Image of Union under Relation: Family of Sets
\(\ds \) \(=\) \(\ds \bigcup_{j \mathop \in J} \map {\pr_i} {\bigcap_{k \mathop = 1}^{n_j} \prod_{i' \mathop \in I} V_{i', k, j} }\) Definition of Projection
\(\ds \) \(=\) \(\ds \bigcup_{j \mathop \in J} \map {\pr_i} {\prod_{i' \mathop \in I} \bigcap_{k \mathop = 1}^{n_j} V_{i', k, j} }\) Cartesian Product of Intersections: General Case
\(\ds \) \(=\) \(\ds \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i,k,j}\) Definition of Projection


As:

$\ds \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i, k, j} \in \tau_i$

it follows that $\pr_i$ is open.

$\blacksquare$