We may define the Euler characteristic of a topological space $X$ using singular integral homology by

$$\chi(X)=\sum_{i}(-1)^i\,\text{rank}\: H_{i}(X;\mathbb{Z}),$$

Then we have that $\chi(\mathbb{R}^n)=\chi(B^{n})=1$ since $\mathbb{R}^{n}$ and the $n$-ball $B^{n}$ are contractible and as defined $\chi$ is a homotopy invariant. Also, for the $n$-sphere $\chi(S^{n})=1+(-1)^{n}$.

On the other hand, the excision property tells us that $\chi(X)=\chi(C)+\chi(X-C)$ for any closed subset $C$ of $X$. Therefore, since $\mathbb{R}^{n}$ is homeomorphic to the interior of $B^{n}$ we have that $$\chi(\mathbb{R}^{n})=\chi(B^{n})-\chi(S^{n-1})=1-(1+(-1)^{n-1})=(-1)^{n}.$$

Which one is the right answer? Where am I going wrong?

Edit: See page 2 of Liviu I. Nicolaecu's notes on the Euler characteristic (Note that the Euler characterisric is definied there using compactly supported cohomology).

Edit: What about this? Add the point at infinity to $\mathbb{R}^{n}$, thus giving us a cell decomposition having one $0$-cell and one $n$-cell. But the one point compactification is an $n$-sphere thus having Euler characteristic $1+(-1)^{n}$. Then removing the added point leaves $(-1)^{n}$ as the Euler characteristic of $\mathbb{R}^{n}$.


Your definition of Euler characteristic does not agree with the notes you link to. Nicolaecu's notes talk about compactly supported cohomology groups, and it is true (as he calculates, that $H_c^k(R^n) \cong 0$ if $k\neq n$ and $\cong R$ if $k=n$, hence $\chi(R^n)=(-1)^n$ with his definition of Euler characteristic (there is also another notion (and more commonly used) of Euler characteristic, which is the one you define, but it is different from the compactly supported one)

  • $\begingroup$ By the way, note that this notion of Euler characteristic is only homotopy invariant for compact spaces, and $R^n$ is not compact (hence the fact that $R^n$ is compact does not tell you anything about its Euler char) $\endgroup$ – user17786 Sep 10 '12 at 14:44

The ball $B^n$ is not homeomorphic to the disjoint union of the boundary sphere and its interior. So the calculation $\chi(\mathbb{R}^n) = \chi(pt) = 1$ is the correct one.

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    $\begingroup$ @John: but the topology on $B$ is not the disjoint union topology, as $B$ is connected while the disjoint union $B^\circ \amalg \partial B$ isn't... $\endgroup$ – t.b. Sep 10 '12 at 11:15
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    $\begingroup$ But if we use the combinatorial definition of the Euler characteristic then $\chi(\mathbb{R}^{n})=(-1)^{n}$. $\endgroup$ – John Sep 11 '12 at 8:27

With the combinatorial definition, and the one using compactly supported cohomology we have $\chi(\mathbb{R}^{n})=(-1)^n$.

If we use singular homology then $\chi(\mathbb{R}^{n})=1$.

The confusion arose from the fact that since $\mathbb{R}^n$ is not compact, the topological definition of $\chi$ using homology and the usual combinatorial definition of $\chi$ (i.e. in terms of a decomposition into open cells homeomorphic to $\mathbb{R}^m$) do not coincide.

The excision property (aka inclusion-exclusion formula) does not hold in general when we define the Euler characteristic using homology. However,it does in the case of finite compact CW complexes. For the combinatorial definition and the one using compactly supported cohomology the excision propery holds.

The combinatorial definition and the definition using compactly supported cohomology do coincide. With either of these two definitions $\chi$ is homotopy invariant for finite compact cw complexes.

Hatcher's book, or Nicolaescu's notes are good references for the topological definitions of $\chi$. For the combinatorial one check this as well.

The argument I give in the second edit of the posted question is valid as long as we work with the combinatorial definition.

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    $\begingroup$ There are some mistakes in your exposition. The excision property does hold in general for the combinatorial and compactly supported Euler characteristics. In fact, that the inclusion-exclusion principle always hold is (in part) what makes the definition "combinatorial". You should also specify that Nicolaescu's notes uses the compactly supported version. $\endgroup$ – Willie Wong Sep 11 '12 at 9:15
  • $\begingroup$ Certainly, thank you and changes applied. $\endgroup$ – John Sep 19 '12 at 9:01

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