Is the quotient functor of points of a Lie group with the subfunctor of a closed subgroup a sheaf? Let G be a Lie group and $H\subseteq G$ be a closed subgroup. It can be shown that $H$ has a unique differentiable structure such that the inclusion map $H\to G$ is an embedding of manifolds. 
The functor of points $h_G=\text{Hom}(-,G)$, $h_H=\text{Hom}(-,H)$ of the Lie groups G and H define group objects in the category of manifolds. The inclusion map $H\to G$ gives a natural transformation $h_H\to h_G$  which clearly makes $h_H$ into a subfunctor of $h_G$. Therefore we may take the quotient functor Q which takes each manifold M to the quotient $Q(M):=\text{Hom}(M,G)/\text{Hom}(M,H)$. 
Is this functor $Q$ a sheaf on the Grothendieck pretopology where the coverings of an arbitary manifold M are given by the sets of arrows {$\iota_i:U_i \to M $ } such that each $\iota_i$ is a diffeomorphism of $U_i$ onto an open subset of M and $\bigcup \iota_i(U_i)=M$?
(An analogous question is often posed in the context of algebraic groups, however, I am interested in this differentiable setting.) 
 A: No; in fact any sheaf $F$ in this topology is a sheaf on the open subsets of $M$ in the standard (continuous) topology as open subsets of $M$ map diffeomorphically into $M$, and conversely any manifold mapping diffeomorphically into $M$ is isomorphic to an open subset of $M$ (its image).  So if this were true the functor of global sections for sheaves on manifolds would be exact; but we know that this is not the case.

Added, (2/20/2011):  I figured I would, as per BCnrd's comment, add an example where the quotient fails to be a sheaf, using connected Lie Groups.  The construction is surprisingly elementary, and if I didn't make some silly mistake, I think quite nice!
Let $\rho: SO(3)\hookrightarrow U(n)$ be a faithful unitary representation of $SO(3)$; let $M=U(n)/SO(3)$.  I first claim that the quotient map $U(n)\to M$ is a non-trivial principal $SO(3)$-bundle.  Indeed $\pi_1(U(n))=\mathbb{Z}$, whereas $\pi_1(SO(3))=\mathbb{Z}/2\mathbb{Z}$, and thus $U(n)$ does not split topologically as a product of $M\times SO(3)$, as $\mathbb{Z}/2\mathbb{Z}$ is not a factor of $\mathbb{Z}$.  So $U(n)\to M$ represents a non-trivial class in $H^1(M, SO(3))$, (which is a pointed set, corresponding to isomorphism classes of principal $SO(3)$-bundles over $M$, via the Cech construction).
But note that this class is the image of the canonical element $i$ of $$H^0(M, U(n)/SO(3))=\operatorname{Hom}(M, U(n)/SO(3))=\operatorname{Hom}(U(n)/SO(3), U(n)/SO(3))$$ (namely, the identity map), through the boundary map $H^0(M, U(n)/SO(3))\to H^1(M, SO(3))$.   Thus $i$ maps to a non-trivial class, and so must not be in the image of the map $H^0(M, U(n))\to H^0(M, U(n)/SO(3))$.  
Thus this last map is not surjective, and so the sheaf of $U(n)/SO(3)$ valued functions on $M$ is not the presheaf $\operatorname{Hom}(-, U(n))/\operatorname{Hom}(-, SO(3))$.
