I was reading the paper: "A fast algorithm for computing the Euler number of an image and its VLSI implementation (Dey et al., 2000)". On the first page they say:
Topological properties serve the purpose of representing the geometric
shape of an image. They remain invariant under any arbitrary
rubber-sheet transformation and hence are very useful in image characterization for matching shapes, recognizing objects, image
database retrieval and many other numerous image processing and
computer vision applications. An important topological feature of an
image is the Euler number (or genus) which is the difference between
the number of connected components and number of holes.
Now, let us suppose we represent our binary image as consisting of many tiny square pixels. The white clusters are our "connected components" and the black clusters are our "holes".
According to the definition given in the paper, the $\chi$ (Euler number) for both the figures is $\text{Number of Connected Components – Number of Holes} = 1 – 1 = 0$. However, to me, the first figure does not in any way seem like a rubber-sheet transformation or homeomorphism of the second figure! But then, how is $\chi$ a topological invariant? Am I missing something?
Best Answer
Invariants don't have much to do with homeomorphism per se - at least, not in that direction.
Homeomorphism (the rubber-sheet transformations - or, more properly, homeomorphism class) is a very strong kind of invariant. But there are stronger kinds (e.g. equality, diffeomorphism class, Lipschitz equivalence) and weaker kinds (e.g. homotopy class, $\pi_1$, cardinality).
Euler number is a relatively weak invariant: homeomorphic objects have the same Euler number, but not necessarily conversely.
There are, of course, also completely different, orthogonal kinds of invariant that have little to do with homeomorphism - e.g. two topological groups might be homeomorphic, but this tells you nothing about whether any of the homeomorphisms are group isomorphisms or not.