The question is already in the title. For a Lie-group G and embedded Lie-subgroups $K < H < G$, prove that $\pi:G/K \rightarrow G/H$ is a surjective smooth submersion. Where it is meant that $K$ is an embedded Lie-subgroup of $H$ and $H$ is an embedded Lie-subgroup of $G$ respectively.
My thoughts so far:
It is clear that the projection $\pi$ is surjective. From the universal property of the quotient map $\pi_K$ we can also deduce that $\pi$ must be continuous. (Since $\pi \circ \pi_K = \pi_H$).
Once we show smoothness of $\pi$ we also know that the differential must be surjective, since
$d\pi(\pi_K)\circ d\pi_K = d\pi_H$ and $d\pi_H$ is surjective.
But I have no idea how to show smoothness of $\pi$.
Any help would be appreciated. Also if someone could tell me if my thoughts so far have been correct.
@Laz thank you for the hint. I think I got it now.
Denote the dimensions of $K,H,G$ by $l,m,n$ where $l \leq m \leq n$. The only thing missing is the smoothness of $\pi$. By your hint for any $g \in G$ there exists a neighbourhood U of g in G and linearly independent functions $g_{1},...,g_{n}:U \rightarrow \mathbb{R}$ that are smooth in the struct. of G such that:
$H \cap U = \{x \in U | \: g_1(x)=...=g_{n-m}(x)=0\}$ and
$K \cap U = \{x \in U | \: g_1(x)=...=g_{n-l}(x)=0\}$.
I now define charts on $G/U$ and $G/K$ around $gU$ and $gK$ respectively. Let $\phi: U/H \rightarrow \mathbb{R}^{n-m}, \: \phi(x) = (g_1(x),...,g_{n-m}(x)).$ This is obviously well-defined. Further, let $\psi:U/K \rightarrow \mathbb{R}^{n-l}, \: \psi(x) = (g_1(x),...g_{n-l}(x))$. If we let
$\kappa:U \rightarrow \mathbb{R}^n, \: \kappa(x) = (g_1(x),...,g_n(x))$, be a chart for G, then it is easy to see that both inclusions $i_1:U/K \rightarrow U/H$ and $i_2:U/H \rightarrow U$ are smooth (as inclusions $\mathbb{R}^{n-m} \hookrightarrow \mathbb{R}^{n-l}$ and $\mathbb{R}^{n-l}\hookrightarrow \mathbb{R}^n)$. So we get the usual smooth structures on $U/H$ and $U/K$. Further if we write $\pi$ in those charts:
$\pi(g_1(x),...,g_{n-l}(x)) = (g_1(x),...,g_{n-m}(x))$. Or if we write $y_{i} = g_i(x)$, $\pi(y_1,...,y_{n-l}) = (y_1,...y_{n-m})$ which is smooth.