Yes, we know that if we let ascend the measure on $\Gamma \backslash SL_2(R) / SO(2)$ back to $SL_2(R) / SO(2)$ then we get the original measure back. The point is that the pushforward along the homeomorphism $\Gamma \backslash SL_2(R) / SO(2) \cong \Gamma^\pm \backslash G^1_R / SO(2)$ might destroy this knowledge. What happens if we let this measure ascend to $\Gamma^\pm \backslash G^1_R$? I do not see how we can conclude something about this pushed measure just from knowing how the original measure behaved under this ascending procedure.

The context is as stated above. Starting with a measure on the upper half plane $H$ , let it descend to $\Gamma \backslash H$. This measure has no invariance property as there is no action of, say GL$_2^+(R)$ on $\Gamma\backslash H$ (GL$_2^+(R)$ just acts on $H$ itself, not on the quotient)! Then we view this measure as a measure on $\Gamma^\pm \backslash G^1_R /SO(2)$ using that the spaces are homeomorphic. We let it ascend to a measure on $\Gamma^\pm \backslash G^1_R$. Question: is this the Haar measure of $G^1_R$ 'descended' to the quotient? This is not immediate!

That is why I said that regularity is NOT the problem here.

Notes on your edit: (1) What is the reason for $L^2(\Gamma \backslash G /K) \cong L^2(\Gamma \backslash G)^K$? This is precisely the assertion in the theorem stated in my post. [(a) or (b) depends on the direction from where you come] (2) So, what you are saying is that the second rightarrow in the diagram is a homeomorphism. How does that help? It is not at all clear that the measure on $\Gamma^\pm \backslash G^1_R$ is the natural quotient measure! All that one knows is that it arises as a measure from $\Gamma\backslash H$ as in the theorem in my post.

No, I meant regularity = outer/inner/... regularity of the measure. How to act an $\Gamma\backslash H$ by Moebius transformations? The action of $GL_2^+(R)$ is a left action and we have divided out something on the left that does not commute with $GL_2^+(R)$!

But this is precisely what the question is about: why is this pushforward measure [that originates in the long diagram described above] from the one on $H$ right invariant under the whole group $G_A$? Usually, the algebraic invariance is easy and the regularity is tricky, but here it is the other way around! What is the action of GL$_2(R)$ supposed to do on $\Gamma\backslash H$ ?

Side remark: The case 'space/group' is not harder in any sense than 'group/subgroup' as the proof is very much the same as for the case 'group/subgroup'...

Im sorry, I do not understand your answer. Let us do what you said: Then we end up with a function on SL$_2(R)$/SO(2). How to get from there to $G_QZ_R\backslash G_A$? In particular, we will have to get there while still keeping some kind of relation of the measures as we want the adelization to be an isometry. Secondly: If we just proceed to SL$_2(R)/SO(2)$, how to show that the map is an isometry here? The natural thing to view $f\overline{g}y^{2k}$ on is $\Gamma\backslash H$ and this is 'space/group' and not 'group/subgroup'!

Just to make sure I get the point: You are talking about the map $\tau \mapsto (Z_Rg, 1)$ for any $g$ such that $gi=\tau$, right? I do not really understand, this map is not well defined (i.e. it depends on choices!)...

Thanks. I have some questions still: 1) what is the right action on $\Gamma\backslash\mathbb{H}$? 2) When i remove what you said, how to get from $\Gamma\backslash \mathbb{H}$ to the adelic quotient? I have to put some homeomorphisms and removals of compact groups in between... the map $\tau \mapsto G_Q Z_R (1|g) \widehat{G_Z}$ for any $g$ such that $g.i=\tau$ is not a homeomorphism!