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Let $X$ be a topological space and $*$ be the base point of $X$. How does $\chi(X-*)$ relate to $\chi(X)$ do we have $\chi(X-*)=\chi(X)-\chi(*)=\chi(X)-1$?

palio
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3 Answers3

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This additivity property is true for the compactly supported Euler characteristic $\chi_c$, that is, the alternating sums of the dimensions of the cohomology groups with compact support. More generally one has $$\chi_c(X) = \chi_c(X \setminus Z) + \chi_c(Z)$$ for a closed subset $Z$ of $X$. In this sense the compactly supported Euler characteristic is nicer, but there are other drawbacks, like that the compactly supported cohomology groups is not a homotopy invariant.

On a manifold $M$ of even dimension $2n$, the compactly supported and the ordinary Euler characteristic coincide by Poincaré duality, interpreted as the assertion that there is a perfect pairing between $H^i(M)$ and $H^{2n-i}_c(M)$. This generalizes Poincaré duality on a closed manifold, since on a closed manifold the ordinary cohomology and the cohomology with compact support coincide. This also tells you that on a manifold of odd dimension, the two Euler characteristics only differ by a sign.

To learn about this and much more, see e.g. the book by Bott and Tu.

Dan Petersen
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  • If the manifold is open nonoriented then the poincare duality is not hold. For example in case of open Mobius band the compactly supported cohomology is trivial. – King Khan Jun 03 '18 at 14:59
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From the Mayer-Vietoris sequence for (singular) homology it follows that if $U,V\subset X$ are open subsets and $X=U\cup V$, $$\chi(X) = \chi(U) + \chi(V) - \chi(U\cap V).$$ In your example you can take $U=X\backslash \{x\}$ and $V$ a small neighbourhood of $x$. Then you'll have $$\chi(U) = \chi(X) - \chi(V) + \chi(U\cap V).$$ In general this numbers can vary a lot (think on a point in a space not having a contractible neighbourhood, or the vertex of the cone $CY$ of some space $Y$). In the particular case when $X$ is a $n$-manifold, and therefore, locally homeomorphic to $\mathbb{R}^n$, taking $U=X\smallsetminus \{x\}$ and $V$ a small neighbourhood of $x$ homeomorphic to a ball we get that $U\cap V$ is homeomorphic to a ball minus a point, hence homotopy equivalent to $S^{n-1}$, so we get $$\chi(U) = \chi(X) - 1 + 0$$ if $n$ is even and $$\chi(U) = \chi(X) - 1 + 2$$ if $n$ is odd.

user17786
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  • Did not you lose the $\chi(V)$ term? – Anixx Apr 25 '23 at 01:59
  • $\chi(V) = 1$ and $\chi(S^{n-1})$ is either $2$ or $0$. – user17786 Apr 26 '23 at 10:07
  • You are wrong, a ball minus a point has Euler characteristic $0$, it is not equivalent to a sphere. – Anixx Apr 26 '23 at 10:49
  • χ(X) = χ(C) + χ(X \ C), for every closed subset C ⊂ X https://www3.nd.edu/~lnicolae/EulerChar.pdf – Anixx Apr 26 '23 at 16:27
  • The equation χ(X) = χ(C) + χ(X \ C) holds for the compactly supported Euler characteristic, as was already mentioned in the answer by Dan Petersen. The paper you are linking is about the compactly supported Euler characteristic, which is not the standard notion of Euler characteristic. – user17786 Apr 27 '23 at 12:56
  • A ball minus a point is definitely homotopy equivalent to a sphere, though not properly homortopy equivalent. Depending on whether you use proper homotopies or ordinary homotopies you get different homology theories and different notions of Euler characteristic. – user17786 Apr 27 '23 at 12:57
  • This rule is universal: χ(X) = χ(C) + χ(X \ C). If you take a point from a set, its Euler characteristic always reduces by 1. – Anixx Apr 27 '23 at 13:04
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Let X be a para-compact Hausdorff space. Consider the compactly supported Euler characteristic $e_c(X) = \sum_{i} (-1)^i \dim \ \mathrm{H}_c^i(X,\mathbf{Q})$. Let $U\subset X$ be open with complement $Z$. Since you have a Mayer-Vietoris sequence for compactly supported cohomology groups, we have $$ e_c(X) = e_c(U) + e_c(Z).$$ I don't know how the non-compactly supported Euler characteristic behaves.

Haki
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