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Following this question Euler's formula for tetrahedral mesh, but also wikipedia, we can see that the euler characteristic can be generalized and is actually an alternating sum depending on the dimension. The question is the following:

For example in the surface of a $3D$ ball we get the: $\chi = V-E+F = 2 = 2-2g$, where $g = 0$. In higher dimensions is there still such a simple connection between $\chi$ and some generalization of genus, for example Betti numbers, or some other invariant?

In the aforementioned question there is a formula of $V-E+F-C=1$ but without explanation, so I do not see how can you get such formulas for other n-manifolds.

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    I'm not sure what you're asking. You clearly already know about the alternating formula in terms of Betti numbers. What are you hoping for? Would it help to mention that it is extremely difficult to classify closed manifolds in higher dimensions? If I remember correctly, the classification of closed 3-manifolds has only recently been completed (and even then it may only be a subclass, I forget), whereas in dimension 2, you only need to know orientability and then one of a long list of possible invariants in order to get a classification (fundamental group, genus, Euler characteristic, etc....) – 2017-02-06
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    Ok. Let me be more precise. Let's stick to closed 3-manifolds. $V-E+F-C=1$ for the 3D ball. How is the 1 derived? I was looking for a generalization of the genus actually to get $\chi$ in 3 dimensions. Is it for example $2-2betti_{dim-1}$? Apparently not because in that case $V-E+F-C=0$. Is there an equivalent of $\chi=V-E+F-C=2-2*something$ in 3D? – 2017-02-06
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    The problem is that the genus is defined in terms of the classification theorem for 2 manifolds. Every orientable closed 2-manifold is homeomorphic to the $g$-fold connect sum of tori, and we define the genus to be this value $g$. I don't think you'll be able to do much better than the formulae you already have - $\chi = V-E+F-C = \sum_{i=0}^n(-1)^i b_i$ where $b_i$ is the $i$th Betti number. If it helps, closed $3$-manifolds always have $b_0 = b_3 = 1$ so you also have $\chi = b_2-b_1$. If you're in the orientable setting, you also have $b_2 = b_1$ and so the Euler characteristic is $0$. – 2017-02-06
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    Obviously, if you are not orientable then the above is no longer true, and also if you are not closed (as in the case of the example in the question you linked) then you can also get changing Euler characteristic. The $1$ is derived in the above by just noting that the shape is homeomorphic to a solid $3$-ball, which we could model as a cube for instance, with $V=8,E=12, F=6, C=1$ giving $\chi = 8-12+6-1 = 1$. The Euler characteristic is a topological invariant, so after noticing the homeomorphism, it doesn't matter how we model that space. – 2017-02-06
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    Euler characteristic is a topological invariant, how can you get changing Euler characteristic as you say? Second question: Is the mesh exhibited in the example I linked not closed? Isn't it homeomorphic to the 3-D solid ball? Feel free to write what we discuss here as an answer so I can accept it please – 2017-02-06
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    When I say changing, I meant non-zero. As I said, the Euler characteristic is always 0 for an odd-dimensional orientable closed manifold, so it can only differ from 0 if one of those properties does not hold (orientability, odd-dimension or closed). – 2017-02-06

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