G mod compact subgroup is trivial bundle Let $ G $ be a lie group with finitely many connected components (for example the real points of any linear algebraic group). Then $ G $ deformation retracts onto it's maximal compact subgroup. Indeed, $ G $ is diffeomorphic to a cartesian product
$$
K \times F
$$
where $ K $ is a maximal compact and $ F $ is contractible. Suppose that $ K' $ is some closed subgroup of $ K $. Then is it true that the coset manifold $ G/K' $ is diffeomorphic to the cartesian product
$$
K/K' \times F ?
$$
The fact for lie groups with finitely many components follows from the fact for connected lie
groupshttps://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-55/issue-10/A-new-proof-of-E-Cartans-theorem-on-the-topology/bams/1183514165.full together with the fact that the maximal compact of of a lie group meets every connected component Are maximal compact subgroups determined by their connected component?
 A: I think this is true.
Given $K'\subseteq K\subseteq G$, one gets a fiber bundle $K/K'\rightarrow G/K'\rightarrow G/K$ where the the first map is induced by inclusion $K\rightarrow G$ and the projection is the obvious one sending $gK'$ to $gK$.
Assuming the existence of this bundle for the moment, since $G/K\cong F$ is contractible, the bundle must be trivial, so it is diffeomorphic to a product $K/K'\times G/K$.  But $G/K\cong F$.
So, where does this fiber bundle come from?
Well, for starters, we have a principal $K$ bundle $K\rightarrow G\rightarrow G/K$ coming from, say, right multiplication of $K$ on $G$.  (More generally, given any proper free action of a Lie group $K$ on a manifold $M$, one gets a principal bundle $K\rightarrow M\rightarrow M/K$.)
Now, note that $K$ acts on $K/K'$ by left multiplication.  So, we can form the space $(G\times K/K')/K$ where $K$ acts by right/left multiplication on both factors:  $k\ast( g, hK') = (gk^{-1}, khK')$.  This action is free and proper, so the quotient $(G\times K/K')/K$ is a smooth manifold.  Moreover, the projection $\pi:(G\times K/K')/K \rightarrow G/K$ given by $\pi( [ g, hK]) = gK$ is a fiber bundle map with fiber $K/K'$.  The inclusion $K/K'\rightarrow (G\times K/K')/K$ is obtained by fixed a point of $G$ (say, the identity $e\in G$), and mapping $hK'$ to $[(e,hK')]$.  This is called the associated bundle construction.
So, we have a bundle $K/K'\rightarrow (G\times K/K')/K\rightarrow G/K$.  The last step is to identify $(G\times K/K')/K$ with $G/K'$ and check that the maps work like we think.
So, consider $f:(G\times K/K')/K\rightarrow G/K'$ defined by $f([g,hK'])  = g h K'\in G/K'$.  Let's show that $f$ is a diffeomorphism.
First note that $f$ is well defined:  for any $k\in K$, $f([gk^{-1}, khK']) =  gk^{-1}khK'= ghK' = f([g,hK']$.  Moreover, $f$ is obviously smooth.
On the other hand, we can simply write down an inverse:  $f^{-1}(gK') = [(g, eK')]$.
I'll leave it to you to verify that this is the inverse.
To finish, I'll just verify that with respect to this diffeomorphism, the projection $G/K'\rightarrow G/K$ is $gK'\mapsto gK$; I'll leave the case of the inclusion of a fiber to you.  The projection $G/K'\rightarrow G/K$ is, via transport of structure, really the composition $G/K'\xrightarrow{f^{-1}} (G\times K/K')/K\rightarrow G/K$.  For an element $gK'\in G/K'$, this is $$gK'\mapsto [(g,eK')] \mapsto g/K.$$
