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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.