Inverting the integration along a subgroup Given a locally compact group $G$ and a closed subgroup $H$, one often uses an operator of the form
$$P: C_c(G) \rightarrow C_c(H \backslash G), \qquad Pf(Hg) = \int_H f(hg) d_H h,$$
where $d_H h$ denotes a Haar measure on $H$. This map is surjective. Is there an explicit form for the right inverse $D: C_c(H \backslash G) \rightarrow C_c(G)$ of $P$?
Consider $\pi: G \mapsto H \backslash G$:
See comments of Daniel Litt for two nice solutions in the extremal cases:


*

*If $H$ is compact, $D \phi = \phi \circ \pi$.

*If $H$ is cocompact, then choose $q$ in the preimage of $1$ under $P$, and $D\phi = q \cdot (f \circ \phi)$.
So what about the groups inbetween?
 A: I have a bit of time, so I'll be explicit.  This construction assumes $\pi: G\to G/H$ is a fiber bundle, and that $G/H$ is paracompact Hausdorff (and so admits partitions of unity).  This should cover many examples that occur "in nature."  My comments below the original question also do the case where $H$ is compact, which I think need not be contained in this case.  
I'll need the following lemma.
Lemma.  Under the given assumptions, there is a continuous function $s: G\to \mathbb{R}$ such that $s$ has compact support when restricted to each fiber of $\pi$, and such that for each $g\in G$, $\int_H s(gh) d_Hh=1$.
Proof.  If $\pi: G\to G/H$ is trivial as a principle $H$-bundle (that is, it admits a section $t: G/H\to G$), this is easy; namely, pick any continuous function $s': H\to \mathbb{R}$ with compact support, satisfying $\int_H s'(h) d_Hh=1$.  Let $s(x)=s'(t(\pi(x))^{-1}x)$.
Now if $\pi: G\to G/H$ is a fiber bundle, we may cover $G/H$ by open $U_i$ such that the bundle is trivial over each $U_i$.  By the previous paragraph, we may choose $s_i: \pi^{-1}(U_i)\to \mathbb{R}$ with compact support on each fiber of $\pi$, and whose integral over each fiber equals $1$.  Now by assumption we may choose a partition of unity $\{\phi_j\}$ subordinate to the cover $\{U_i\}$.  Let $s=\sum_{i,j} s_i\cdot (\phi_j\circ \pi)$.  $\Box$
We now construct a right inverse $D$ to $P$.  Namely, for $f$ a compactly supported continuous function on $G/H$, let $D(f)=s\cdot (f\circ \pi)$.  It is clear that $P(D(f))=f$; one need only check that $s\cdot (f\circ \pi)$ has compact support, which I leave as an easy exercise (again using that $\pi$ is a fiber bundle).
