Projection Mappings of Finite Topological Products

# Projection Mappings of Finite Topological Products

Recall from the Finite Topological Products of Topological Spaces page that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces then the resulting topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ has the topology whose basis is:

(1)\begin{align} \quad \mathcal B = \left \{ \prod_{i=1}^{n} U_i : U_i \: \mathrm{is \: open \: in \:} X_i, \: \forall i \in \{ 1, 2, ..., n \} \right \} \end{align}

We claimed that the product topology is the initial topology induced by the projection maps $\{ p_1, p_2, ..., p_n \}$. We prove that statement below.

Theorem 1: Let $\{ X_1, X_2, ..., X_n \}$ be a finite collection of topological spaces and let $\displaystyle{\prod_{i=1}^{n} X_i}$ denote the corresponding topological product. For each $j \in \{1, 2, ..., n \}$ and for all $\displaystyle{\mathbf{x} = (x_1, x_2, ..., x_n) \in \prod_{i=1}^{n} X_i}$ define the map $\displaystyle{p_j : \prod_{i=1}^{n} X_i \to X_j}$ by $p_j(\mathbf{x}) = x_j$. Then:a) $p_j$ is surjective, open, and continuous for all $i \in \{1, 2, ..., n \}$.b) The product topology on $\displaystyle{\prod_{i=1}^{n} X_i}$ is the coarsest topology which makes all of the maps $\{ p_1, p_2, ..., p_n \}$ continuous. |

**Proof of a):**We begin by showing that each $p_j$ is surjective. Let $j \in \{ 1, 2, ..., n \}$ and consider the map $\displaystyle{p_j : \prod_{i=1}^{n} \to X_j}$. Let $b \in X_j$. Then if $\mathbf{a} = (x_1, x_2, ..., x_{j-1}, b, x_{j+1}, ..., x_n)$ then:

\begin{align} \quad p_j(\mathbf{a}) = b \end{align}

- So each $p_j$ is surjective.

- We now show that each $p_j$ is surjective by showing that the images of basis sets are open. Let $\displaystyle{U = \prod_{i=1}^{n} U_i}$ be any basis set in $\displaystyle{\prod_{i=1}^{n} X_i}$. Then $U_i$ is open in $X_i$ for all $i \in \{1, 2, ..., n \}$, and:

\begin{align} \quad p_j(U) = U_j \end{align}

- So $p_j(U)$ is open in $X_j$. So each $p_j$ is open.

- Lastly, we show that each $p_j$ is continuous. Let $U$ be any open set in $X_j$. Then we have that:

\begin{align} \quad p_j^{-1}(U_j) = X_1 \times X_2 \times ... \times X_{j-1} \times U_j \times X_{j+1} \times ... \times X_n \end{align}

- But each $X_i$ is open in itself, and $U_j$ is open in $X_j$, so $p^{-1}(U_j)$ is open in $\displaystyle{\prod_{i=1}^{n} X_i}$ which shows that $p_j$ is continuous. $\blacksquare$

**Proof of b)**Let $\tau$ denote the product topology on the finite product $\displaystyle{\prod_{i=1}^{n} X_i}$ and let $\tau'$ denote any other topology which makes all of the projection maps $\{ p_1, p_2, ..., p_n \}$ continuous. To show that the product topology $\tau$ is the one which makes all of these maps continuous, we must show that $\tau \subseteq \tau'$.

- Let $U_j$ be an open set in $X_j$ for all $j \in \{1, 2, ..., n \}$. Then $\displaystyle{\prod_{i=1}^{n} U_i}$ is contained in $\tau$. Now the inverse image of $U_j$ under $p_j$ is:

\begin{align} \quad p_j^{-1}(U_j) = X_1 \times X_2 \times ... \times X_{j-1} \times U_j \times X_{j+1} \times ... \times X_n \end{align}

- If $\tau'$ is to make each of the projection maps continuous, then $p_j^{-1}(U_j)$ must be open in $\displaystyle{\prod_{i=1}^{n} X_i}$. Moreover, arbitrary (in this case, finite) intersections of sets must be open in $\displaystyle{\prod_{i=1}^{n} X_i}$, and so the following set must be contained in $\tau'$.

\begin{align} \quad \bigcap_{i=1}^{n} p_i^{-1}(U_i) = \prod_{i=1}^{n} U_i \end{align}

- Therefore $\tau \subseteq \tau'$ which shows that the product topology on $\displaystyle{\prod_{i=1}^{n} X_i}$ is the coarsest topology which makes the projection maps $\{ p_1, p_2, ..., p_n \}$ continuous. $\blacksquare$