What is the true relationship between impossible figures and cohomology?

Question

Penrose's paper On the cohomology of impossible figures suggests me that we can't draw such an impossible figure on a contractible part $Q$ of a sheet of paper so that it completely fills it, because the first cohomology group $H^1(Q,G)$ of such a domain is trivial (where G is the ambiguity group of the figure) (see also here) Still he does some, e.g. this (the picture is taken from here):

Why is it possible? And what is the general, true relationship between the impossibility of figures and cohomology? Cohomology of what, if not of the drawing domain?

Edit

I try to make my problem a bit clearer.

Here is a good cover of the solid disk on the paper containing the picture of the impossible staircase.

$Q_1$, $Q_2$ and $Q_3$ correspond to the open sets of Penrose (their overlapping areas are the thick radial blue lines), while $Q_4$, the solid disk bounded by the yellow circle, is an additional open set that overlaps with each other $Q$-s. I show the middle of the figure in big:

Here $d_{ij}$ stands for the same as in Penrose's paper. I think, that we have some problem with $d_{14}$, $d_{24}$ and $d_{34}$, but I don't know, what.

Answer 1

The drawing domain of most impossible figures is a circle or an annulus. The reason they're impossible is that there is no single "height function" that covers all of it, even though there is a "steepness function".

In other words, there exists a function that looks like it should be a derivative / gradient, but isn't. That's exactly what non-trivial cohomology is all about. This is impossible on a simply connected domain like the whole plane or a solid disc, but it is very much possible on a circle / annulus.

Answer 2

I think, I misunderstood the role of the annulus. My conclusion is that we can draw an impossible figure on any domain of a sheet. The annulus (or any non-contractible open set) is not a constraint. It is only a tool for testing the possibility of the picture. The test is that the cocycle $d_{ij}$ described in Penrose's paper is a coboundary, or not. That is, the true relationship between cohomology and the impossible figures is simply the following.

A locally realistic figure is globally impossible if and only if there is a non-contractible open subset of the drawing domain on which the cocycle $\{d_{ij}\}$ is not a coboundary.