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:


1. If $H$ is compact, $D \phi = \phi \circ \pi$.
2. 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?



Inverting the integration along a subgroup
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).


