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 $\mathbb{R}^{n}$.
Best Answer
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)