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I have the following question

Let $G$ be a locally connected topological group and $H < G$ a closed locally connected subgroup, show that if $H$ is not connected then $G/H$ is not simply connected.

However the book does not provide the definition of simply connectedness in the context of topological groups, i know the the definition that is being path connected and with paths that can shrink to a point (something like that).

The hint is to use the identity connected component, so let $H_0$ be this component.

I know that $H_0$ is open since $H$ is loc connected, also $H_0$ is closed and proper.

My intuition tells me there is something to do with the action that takes $g \in G$ and goes to $gH_0 \in G/H_0$. But as i don't know what i should be looking to prove here i don't know how to procede. If anyone could enlighten me i would apreciate.

Edit: Apparently the way is showing that $G/H_0$ is homeomorphic to $G/H$ when $G/H$ is simply connected, once i can do this i will post the result.

Edit 2: $G$ needs to be connected

Daniel Moraes
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    "Simply connected" is a term that makes sense to use for any topological space. It means that the space is path connected and that all continuous maps from $S^1$ to the space are nullhomotopic. This might not help much but I hope it helps you start down the right path! – diracdeltafunk Mar 23 '21 at 18:25
  • @diracdeltafunk i know i didn't write the proper definition, but even when comparing to it i feel lost – Daniel Moraes Mar 23 '21 at 18:27
  • "The action that takes $g \in G$ and goes to $gH_0 \in G/H_0$" -- this describes a map $G \to G/H_0$, which is not an action. – diracdeltafunk Mar 24 '21 at 05:20
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    Wait what if $G = \mathbb{R} \times \mathbb{Z}/2$ with $H = {0} \times \mathbb{Z}/2$. Then $H$ is closed, locally connected, and a subgroup, and $H$ is not connected. However, $G/H \cong \mathbb{R}$ is simply connected. – diracdeltafunk Mar 24 '21 at 05:39
  • @diracdeltafunk as for this part i may be very wrong also, i thought it described an action – Daniel Moraes Mar 24 '21 at 10:12
  • I will also bring this example to my teacher, i guess it makes sense to me at least – Daniel Moraes Mar 24 '21 at 10:13
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    An action of a group $G$ on a set $X$ is a function $G \times X \to X$ satisfying certain axioms. You just gave a map $G \to X$. – diracdeltafunk Mar 24 '21 at 15:56
  • @diracdeltafunk in your case is $G$ loc connected also? i'm thinking not – Daniel Moraes Mar 26 '21 at 10:55
  • actually would be a collection of straight lines so it should be loc connected right? – Daniel Moraes Mar 26 '21 at 11:27
  • Correct, $\mathbb{R} \times \mathbb{Z}/2$ is locally connected. – diracdeltafunk Mar 26 '21 at 19:29
  • Well, this is a result in San Martin's Lie group book, i'm waiting for a meeting with my professor to check it, when i know better i will post here as an answer – Daniel Moraes Mar 26 '21 at 19:33
  • @diracdeltafunk we need $G$ to be connected also, there was a typo in the book, with that i could prove, i will wait and if it is correct i can post my solution. – Daniel Moraes Apr 01 '21 at 20:27

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