Let $G$ be a (linear) algebraic group (over an algebraically closed field of characteristic zero - if you like you can assume $\mathbb{C}$). Let $K$ and $H$ be closed subgroups of $G$ such that $H\subseteq K\subseteq G$. Then we can consider the natural "projection" morphism: $\pi\colon G/H\to G/K$. I basically know that the morphism $\pi$ is separated and smooth, since the differential $d_x\pi$ is surjective at every (closed) point $x\in G/H$. Moreover since $\pi$ is smooth, we know that $\pi$ is open and closed.
My question is the following: is the morphism $\pi$ also proper and projective? This might be easy to see, but I am not sure how. Any help is appreciated. I already tried to look in books like the one of Timashev, but did not find the statement.
Note that the reasoning might not depend on the specific situation I have in mind. If you can abstract from the setup, it might be useful too. For this you may consider that $G/H$ and $G/K$ are both quasiprojective (but not necessarily projective (!)).
I am not sure, but maybe this question is relevant. Thanks!