For Lie Groups $ H < G$, where H is embedded. The projection $\pi :G \rightarrow G/H$ maps components onto components Let $G$ be a Lie-group and $H < G$ an embedded Lie-subgroup. Show that:
The natural projection $\pi:G \rightarrow G/H$ maps connected components onto connected components.
We already showed in the lecture that for an embedded Lie-subgroup $H < G$ the projection is a smooth submersion. The hint was to use that fact. But can't we just argue that since the projection $\pi$ is continuous the images of connected components are again connected ? And we can "reach" every component in $G/H$ since $\pi$ is surjective ?
I think I am missing something here. Thanks in advance for any help !
 A: Thank you @Vercassivelaunos for the comment.
I think I got the answer now.
Let $C_1 \subset G$ be any connected component. Let $C \subset G/H$ be the connected component that contains $\pi(C_1)$. By the above $\pi(C_1)$ is connected, so contained in only one connected component. Assume $\exists g \in G$ s.t. $gH \in C \setminus \pi(C_1)$. Then g is contained in another component $C_2$ of G. By the same argument $\pi(C_2) \subset C$. By iterating the argument we can write C = $\bigcup \limits_{i \in I} \pi(C_i)$.

*

*Since $\pi$ is induced by a smooth action it is an open map. (See Lee Lemma 21.1). Therefore $\pi(C_i)$ is open in $G/H \:  \forall i \in I$.


*I show now that this union is in fact disjoint.
Let's say that $\pi(C_i) \cap \pi(C_j) \neq \emptyset$. We can write $C_i = g_iG_0$ and $C_j = g_jG_0$ for some $g_i,g_j \in G$ where $G_0$ is the identity component in G (See Prop. 7.15 Lee). Let's say $a \in 
\pi(g_jG_0) \cap \pi(g_iG_0)$. Then $a = g_i \cdot g' \cdot h'$ and $a = g_j \cdot g'' \cdot h''$. For $g',g'' \in G_0$ and $h',h'' \in H$.
So $g_i = g_j \cdot $(something in $G_0$) $\cdot $(something in $H$).
But then $\pi(g_iG_0) = \pi(g_jG_0)$.
So we wrote C as disjoint union of open sets. This is a contradiction to our assumption that C is connected. And finally $\pi(C_1) = C$ is a connected component.
