Is this continuous bijection a homeomorphism?

Question

Let $G$ be a topological group acting on a topological space $X$ and let $$\pi:X\to X/G$$ be the quotient map. Suppose that $Y$ is a subspace of $X$ such that the restriction $$\pi|_Y:Y\to X/G$$ is bijective. Is $\pi|_Y$ necessarily a homeomorphism?

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Attempts. Since $\pi|_Y$ is a continuous bijection, it is a homeomorphism if and only if it is an open map. Hence, it suffices to show that if $U\subseteq Y$ is open, then $$G\cdot U=\{g\cdot u:g\in G,u\in U\}$$ is open in $X$. Is that true? Also, we know that $\pi$ itself is an open map (since if $V\subseteq X$ is open then $\pi^{-1}(\pi(V))=\bigcup_{g\in G}g\cdot V$ is a union of open sets). Does that help?

Answer 1

This is false; consider, for instance, $\mathbb Z$ acting on $\mathbb R$ by translation, so that $X/G = S^1$, and your question would give $[0,1) \cong S^1$. This is even a relatively nice choice for $Y$; note that we simply have to pick a representative for each coset $x + \mathbb Z \in \mathbb R/\mathbb Z$ (as a group) which can be really ugly sets. In general, you should expect this to be true only in the most trivial cases, and even possible (i.e. for exactly the right $Y$) in the nicest of cases.