[proofplan]
It suffices to show that $\pi_\beta$ maps every basis element of the [product topology](/page/Product%20Topology) to an [open set](/page/Open%20Set) in $X_\beta$, since every open set in the product is a union of basis elements and $\pi_\beta$ preserves unions. A basis element of the product [topology](/page/Topology) is a product $\prod_{\alpha \in A} V_\alpha$ where each $V_\alpha$ is open in $X_\alpha$ and $V_\alpha = X_\alpha$ for all but finitely many $\alpha$. Its image under $\pi_\beta$ is $V_\beta$, which is open in $X_\beta$.
[/proofplan]
[step:Reduce to basis elements]
Let $W \subset \prod_{\alpha \in A} X_\alpha$ be open in the [product topology](/page/Product%20Topology). Then $W = \bigcup_{i \in I} B_i$ for some collection of basis elements $B_i$ of the product [topology](/page/Topology). Since the projection $\pi_\beta$ preserves arbitrary unions,
\begin{align*}
\pi_\beta(W) = \pi_\beta\Bigl(\bigcup_{i \in I} B_i\Bigr) = \bigcup_{i \in I} \pi_\beta(B_i).
\end{align*}
A union of [open sets](/page/Open%20Set) is open, so it suffices to show that $\pi_\beta(B)$ is open in $X_\beta$ for every basis element $B$ of the product topology.
[guided]
To prove that $\pi_\beta$ is an open map, we must show that the image of every open set in $\prod_{\alpha \in A} X_\alpha$ is open in $X_\beta$. Rather than working with arbitrary open sets, we reduce to basis elements.
Every open set $W$ in the product topology is a union of basis elements: $W = \bigcup_{i \in I} B_i$. Since image commutes with union for any function — $\pi_\beta(\bigcup_{i \in I} B_i) = \bigcup_{i \in I} \pi_\beta(B_i)$ — the image $\pi_\beta(W)$ is a union of the sets $\pi_\beta(B_i)$. If each $\pi_\beta(B_i)$ is open in $X_\beta$, then $\pi_\beta(W)$ is open as a union of open sets.
This reduces the problem to a single computation: determine the image of a basis element under $\pi_\beta$.
[/guided]
[/step]
[step:Compute the image of a basis element under $\pi_\beta$]
A basis element of the [product topology](/page/Product%20Topology) on $\prod_{\alpha \in A} X_\alpha$ has the form
\begin{align*}
B = \prod_{\alpha \in A} V_\alpha,
\end{align*}
where each $V_\alpha$ is open in $X_\alpha$ and $V_\alpha = X_\alpha$ for all but finitely many indices $\alpha$. (This is the standard basis consisting of products of [open sets](/page/Open%20Set) with all but finitely many factors equal to the full space.)
The image under $\pi_\beta$ is
\begin{align*}
\pi_\beta(B) = \pi_\beta\Bigl(\prod_{\alpha \in A} V_\alpha\Bigr) = V_\beta.
\end{align*}
To verify this equality: if $y \in V_\beta$, then for each $\alpha \neq \beta$ choose any $x_\alpha \in V_\alpha$ (each $V_\alpha$ is nonempty since it is a nonempty open subset of $X_\alpha$; in particular, $V_\alpha = X_\alpha$ for all but finitely many $\alpha$, and the finitely many remaining $V_\alpha$ are nonempty open sets). Define the element $x \in \prod_{\alpha \in A} X_\alpha$ by setting $x_\beta = y$ and $x_\alpha \in V_\alpha$ for $\alpha \neq \beta$. Then $x \in B$ and $\pi_\beta(x) = y$, so $V_\beta \subset \pi_\beta(B)$. Conversely, if $x \in B$ then $x_\alpha \in V_\alpha$ for every $\alpha$, so $\pi_\beta(x) = x_\beta \in V_\beta$, giving $\pi_\beta(B) \subset V_\beta$.
Since $V_\beta$ is open in $X_\beta$ by construction, $\pi_\beta(B)$ is open in $X_\beta$.
[guided]
The basis for the product [topology](/page/Topology) consists of all sets of the form $B = \prod_{\alpha \in A} V_\alpha$ where:
- Each $V_\alpha$ is open in $(X_\alpha, \tau_\alpha)$, and
- $V_\alpha = X_\alpha$ for all but finitely many indices $\alpha \in A$.
The projection $\pi_\beta: \prod_{\alpha \in A} X_\alpha \to X_\beta$ extracts the $\beta$-th coordinate. We claim $\pi_\beta(B) = V_\beta$.
**Containment $\pi_\beta(B) \subset V_\beta$:** If $x = (x_\alpha)_{\alpha \in A} \in B$, then $x_\alpha \in V_\alpha$ for every $\alpha$. In particular, $\pi_\beta(x) = x_\beta \in V_\beta$.
**Containment $V_\beta \subset \pi_\beta(B)$:** Fix any $y \in V_\beta$. We need to exhibit a point $x \in B$ with $\pi_\beta(x) = y$. For each $\alpha \neq \beta$, the set $V_\alpha$ is nonempty: for the finitely many $\alpha$ where $V_\alpha \neq X_\alpha$, the set $V_\alpha$ is a nonempty open subset of $X_\alpha$ (a basis element of a topology on a nonempty space must be nonempty); for all remaining $\alpha$, $V_\alpha = X_\alpha$ which is nonempty. Choose any element $x_\alpha \in V_\alpha$ for each $\alpha \neq \beta$, and set $x_\beta = y$. Then $x = (x_\alpha)_{\alpha \in A}$ satisfies $x_\alpha \in V_\alpha$ for every $\alpha$, so $x \in B$, and $\pi_\beta(x) = y$.
Since $V_\beta$ is open in $X_\beta$ (it is one of the open sets in the product), $\pi_\beta(B) = V_\beta$ is open. This completes the proof.
[/guided]
[/step]
