[proofplan] We must show the quotient map $\pi: X \to X/G$ is open, i.e., for every open $U \subset X$, the image $\pi(U)$ is open in $X/G$. By definition of the quotient topology, $\pi(U)$ is open if and only if $\pi^{-1}(\pi(U))$ is open in $X$. We show that $\pi^{-1}(\pi(U))$ is the saturation $G \cdot U = \bigcup_{g \in G} g \cdot U$, which is a union of open sets (since each $g \cdot U$ is open, as $x \mapsto g \cdot x$ is a homeomorphism). [/proofplan]

[step:Identify the preimage $\pi^{-1}(\pi(U))$ as the saturation $G \cdot U$]Let $U \subset X$ be open. The image $\pi(U) \subset X/G$ consists of all orbits $G \cdot x$ with $x \in U$. The preimage $\pi^{-1}(\pi(U))$ consists of all points in $X$ whose orbit intersects $U$: \begin{align*} \pi^{-1}(\pi(U)) = \{x \in X : G \cdot x \cap U \neq \varnothing\} = \{x \in X : \exists\, g \in G,\; g^{-1} \cdot x \in U\} = \bigcup_{g \in G} g \cdot U \end{align*} where $g \cdot U := \{g \cdot u : u \in U\}$. To verify the last equality: if $x \in \pi^{-1}(\pi(U))$, then there exists $u \in U$ with $\pi(x) = \pi(u)$, meaning $x = g \cdot u$ for some $g \in G$, so $x \in g \cdot U$. Conversely, if $x = g \cdot u$ for some $g \in G$ and $u \in U$, then $\pi(x) = \pi(u) \in \pi(U)$, so $x \in \pi^{-1}(\pi(U))$.[/step]

[guided]The set $\pi^{-1}(\pi(U))$ is the **saturation** of $U$ — the smallest saturated (i.e., union of complete orbits) subset of $X$ containing $U$. It consists of every point of $X$ that belongs to the same orbit as some element of $U$. The identity $\pi^{-1}(\pi(U)) = \bigcup_{g \in G} g \cdot U$ expresses this saturation as the orbit of $U$ under the group action: we "spread" $U$ across all group translates.[/guided]

[step:Show each translate $g \cdot U$ is open using the homeomorphism hypothesis]For each $g \in G$, define the orbit map \begin{align*} \lambda_g: X &\to X \\ x &\mapsto g \cdot x. \end{align*} By hypothesis, $\lambda_g$ is a homeomorphism for each $g \in G$. In particular, $\lambda_g$ is an open map: it sends open sets to open sets. Since $U$ is open, $g \cdot U = \lambda_g(U)$ is open in $X$.[/step]

[guided]The hypothesis that $G$ acts by homeomorphisms is essential here. It guarantees that each group element $g$ acts as a bijection $\lambda_g: X \to X$ with both $\lambda_g$ and $\lambda_g^{-1} = \lambda_{g^{-1}}$ continuous. In particular, $\lambda_g$ maps open sets to open sets. If $G$ merely acted by continuous maps (not homeomorphisms), the translates $g \cdot U$ could fail to be open, and the saturation $\bigcup_{g \in G} g \cdot U$ could fail to be open. The homeomorphism condition ensures that each individual translate preserves the open set structure.[/guided]

[step:Conclude that $\pi(U)$ is open via the quotient topology]The saturation $\pi^{-1}(\pi(U)) = \bigcup_{g \in G} g \cdot U$ is a union of open sets, hence open in $X$. By definition of the quotient topology on $X/G$, a subset $W \subset X/G$ is open if and only if $\pi^{-1}(W)$ is open in $X$. Since $\pi^{-1}(\pi(U))$ is open, $\pi(U)$ is open in $X/G$. Since $U$ was an arbitrary open subset of $X$, the map $\pi$ sends open sets to open sets: $\pi$ is an open map.[/step]

[guided]This step applies the defining property of the quotient topology: openness of $\pi(U) \subset X/G$ is *equivalent* to openness of $\pi^{-1}(\pi(U)) \subset X$. We have shown that $\pi^{-1}(\pi(U)) = \bigcup_{g \in G} g \cdot U$ is a union of open sets (each $g \cdot U$ is open by the homeomorphism property), so $\pi^{-1}(\pi(U))$ is open, and therefore $\pi(U)$ is open. Note that the continuity of the action map $G \times X \to X$ is not directly used in this argument — only the fact that each $\lambda_g$ is a homeomorphism. The joint continuity of the action is a hypothesis of the theorem that ensures the action is "well-behaved" in ways relevant to other applications (e.g., the existence of continuous local sections), but for the openness of $\pi$, the individual homeomorphism condition suffices. The conclusion that $\pi$ is open has an important consequence: since $\pi$ is a continuous, surjective, open map, it is a quotient map by [Open and Closed Surjections Are Quotient Maps](/theorems/1030), and the product map $\pi \times \pi: X \times X \to (X/G) \times (X/G)$ is also an open surjection, hence a quotient map. This makes the [Hausdorff Criterion for Quotient Spaces](/theorems/1032) directly applicable to orbit spaces.[/guided]