Connectivity and Euler characteristic for surfaces I learn the concept of connectivity from Hilbert's Geometry and the Imagination as follows:

A polyhedron is said to have connectivity $h$ (or to be $h$-tuply connected) if $h-1$, but not $h$, chains of edges can be found on it in a certain order that do not cut the surface in two, where it is stipulated that the first chain is closed and that every subsequent chain connects two points lying on the preceding chains.

Then he claimed the relationship between connectivity of a polyhedron and the Euler characteristic:
$$\chi=V-E+F=3-h$$
In the next section, he defined the connectivity on general surfaces (you can get it from google books), including surface-with-boundary and non-bounded surfaces. However, it seems that $\chi=3-h$ doesn't work for the general case. I want to see why the relation works for a polyhedron, but I didn't find the same concept via google. I need some reference for that concept. Thanks!

It may be hard to find because I'm not sure that this term is used very much; at the very least, I have never really heard it used. In fact, using it we find the unusual descriptions that


*

*$S^{2}$ is 1-connected

*$T^2 = S^1 \times S^{1}$ is 3-connected

*More generally a Riemann surface of genus $g$ is ${2g + 1}$ connected.


Anyhow, as to how one can see this:
Loosely speaking, choose a "triangulation" of your surface (which has no boundary) with only one vertex, and such that the ${h-1}$ edges which we will cut will be the 1-cells. If we do so, there will only be one face on the surface (since after you perform all of the cuts, the result is a polyhedron with ${2h-2}$ edges; otherwise you could cut it further). If we then put this together into the formula for the Euler characteristic, we have that
$$
V - E + F = 1 - (h-1) + 1 = 3 - h
$$
as claimed.
If this seems a bit weird, I recommend following through this argument with a torus. It's not too hard to pick the appropriate "triangulation", from which you can see the end result.
