Openness of a map of $G$-spaces This is question 10 from chapter 1 of Bredon, Introduction to Compact Transformation Groups.
Let $G$ be a compact group, $X$ and $Y$ be $G$-spaces (Hausdorff spaces with a continuous $G$-action), and $f: X \to Y$ a $G$-equivariant map. Suppose that $f$ restricts to a homeomorphism of each orbit $f: G(x) \overset{\cong} \to G(f(x))$, and that the induced map $f': X/G \to Y/G$ is open. Prove that $f$ is open.
This intuitively feels true, but I haven't been able to come up with a formal argument. Given $U \subset X$ open, the projection $f(U) \to Y \to Y/G$ has open image, and the fibers $f(U) \cap G(y)$ are relatively open in the orbits, but that alone doesn't get you there. I believe I could do it if there were local sections of $Y \to Y/G$, but those don't necessarily exist. Anyone point me in the right direction?
 A: $$\require{AMScd} \begin{CD}
X  @>f>> Y \\
@Vp_XVV @VVp_YV \\
X/G @>>f'>  Y/G
\end{CD}$$
Let $U \subset X$ be open, $y$ be a limit point of $f(U)$, and $x$ be any preimage of $y$. Let $\mathcal{A}_y = \{V_\alpha\}_{\alpha \in A}$ and $\mathcal{B}_x = \{W_\beta\}_{\beta \in B}$ be the collections of all open sets in $Y$ and $X$ containing $y$ and $x$ respectively. Give $A$ and $B$ orderings by reverse inclusion, e.g. $\alpha \ge \alpha'$ iff $V_\alpha \subset V_{\alpha'}$. Then put the Cartesian ordering on $A \times B$ and define nets in $Y \setminus f(U)$ and $X/G$ by choosing for each $(\alpha, \beta)$ points $y_{\alpha,\beta} \in V_\alpha \cap Gf(W_\beta) \setminus f(U)$ and $\sigma_{\alpha, \beta} \in p_X(W_\beta)$ such that $f'(\sigma_{\alpha,\beta}) = Gy_{\alpha,\beta}$. (We used here that $Gf(W_\beta) = p_Y^{-1} \circ f' \circ p_x(W_\beta)$ is open.) Then let $x_{\alpha, \beta} = (f |_{\sigma_{\alpha,\beta}})^{-1}(y_{\alpha,\beta}) $, and choose $g_{\alpha,\beta} \in G$ such that $g_{\alpha,\beta}x_{\alpha,\beta} \in W_\beta$. By construction, these nets converge $y_{\alpha,\beta} \to y$ and $\sigma_{\alpha,\beta} \to Gx$ and $g_{\alpha,\beta}x_{\alpha,\beta} \to x$.
Since $G$ is compact, we may assume after passing to a subnet that $g_{\alpha,\beta}$ converges to some $g$. Since these spaces are Hausdorff, limits are unique, so
$$y = f(x) = \lim g_{\alpha,\beta}f(x_{\alpha,\beta}) = \lim g_{\alpha,\beta}y_{\alpha,\beta} = gy.$$
Thus $g$ is in the stabilizer $G_y$ of $y$, and subsequently $g \in G_x$ as well, since $f|_{Gx}: Gx \to Gy$ is a homeomorphism and $G$ is compact. Then we have convergence
$$x_{\alpha,\beta} = g_{\alpha,\beta}^{-1}(g_{\alpha,\beta}x_{\alpha,\beta}) \to g^{-1} x = x.$$
Since $x_{\alpha,\beta} \in X \setminus U$, we conclude $x \in X \setminus U$, and since $x \in f^{-1}(y)$ was arbitrary, $y \in Y \setminus f(U)$.
