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Let Y be path-connected, locally path-connected, and simply connected. Let $G_1$ and $G_2$ be subgroups of Homeo(Y) defining covering space actions of Y. (this means that each point y has a neighborhood U such that for $g_1, g_2 \in G_1, g_1(U) \cap g_2(U) \neq \emptyset\ \text {implies}\ g_1 = g_2$) Show that the orbit spaces $Y/G_1$ and $Y/G_2$ are homeomorphic iff $G_1$ and $G_2$ are conjugate subgroups of Homeo(Y).

Left to right is easy since we have that $G_1$ is isomorphic to $\pi _1(Y/G_1)$ because Y is simply connected. Same with $G_2$. (proposition 1.40 in Allen Hatcher's algebraic topology) It's right to left that I'm having trouble with. I was thinking of just building a map from $Y/G_1$ to $Y/G_2$ by directly sending one orbit to another, but I wasn't sure how to show it was continuous. Is there a way to use the quotient topology to do this? Or should I be looking at this problem another way entirely?

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Say $G_2 = h G_1 h^{-1}$. Then the map $f: Y \to Y/G_2$ that is given by the action of $h$ followed by the quotient map, i.e., $y \mapsto \overline{h \cdot y}$

  • is continuous, since $h$ acts via a homeomorphism and the quotient map is continuous, and
  • collapses orbits of $G_1$: if $y' = g_1 \cdot y$ for some $g_1 \in G_1$, then $$f(y') = \overline{hg_1 \cdot y} = \overline{(h g_1^{-1} h^{-1}) \cdot hg_1 \cdot y} = \overline{h \cdot y} = f(y).$$

Therefore, by the universal property of quotient maps, there exists a continuous map $\bar{f}: Y/G_1 \to Y/G_2$ extending $f$.

By exchanging the roles of $G_1$ and $G_2$, you can construct another continuous map $Y/G_2 \to Y/G_1$ which you can check is inverse to $\bar{f}$. Hence $Y/G_1 \cong Y/G_2$.

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  • $\begingroup$ Thanks a bunch mate! $\endgroup$
    – fosterc4
    Mar 5 at 15:59

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