Euler characteristic of an $n$-sphere is $1 + (-1)^n$. I am using the textbook Guillemin and Pollack's Differential Topology but I am asked to solve a question need this fact that Euler characteristic of sphere is $1 + (-1)^n$.
So may I ask if this is introduced in the text, or this topic is discussed somewhere else?
Thanks~
 A: I don't have a reference for you, but the computation isn't bad at all.  As a CW-complex, $S^n$ is an $n$-cell whose boundary is glued to a $0$-cell.  The Euler characteristic of an $n$-dimensional CW-complex is the alternating sum $d_0-d_1+d_2-d_3+\ldots+(-1)^nd_n$ where $d_i$ is the number of cells of dimension $i$.  It follows that the Euler characteristic of $S^n$ is $1+(-1)^n$.
A: A good reference is the Hirsch's book "Differential Topology", page 134. If you don't have the book, I will write here the proof:
" Let $P$ the north pole and $Q=-P$ the south pole. Let $\sigma: S^n - P \rightarrow \mathbb{R}^n$ and $\tau: S^n - Q \rightarrow \mathbb{R}^n$  be the stereographic projections. The coordinate change $\tau \sigma^{-1}= \sigma \tau^{-1}: \mathbb{R}^n-0 \rightarrow \mathbb{R}^n-0$ is given by $x\mapsto \frac{x}{|x|^2}$.
Let $f$ be the vector field on $S^n-P$ whose representation via $\sigma$ is the identity vector field on $\mathbb{R}^n$. Then $f(x) \rightarrow 0$ as $x \rightarrow P$ and we define $f(P)=0$. Thus, $f:S^n \rightarrow TS^n$ has zeroes only at $P$ and $Q$.
In $\tau$ coordinates $f$ corresponds to the vector field $x \mapsto -x$ on $S^n-Q$. Thus $f$ is $C^{\infty}$.
The idendity map of $\mathbb{R}^n$ has degree $1$ or $0$, the  antipodal map has degree $(-1)^n$. Therefore $Ind_P f=1$, $Ind_Q f=(-1)^n$.
Thus, we have proved $\chi(S^n)=1+(-1)^n$ "
A: In Hatcher's algebraic topology, he has a formula $$\chi(M)=\sum_n(-1)^n dim_{\mathbb{Z}}(H_n(M))$$ with $H_n$ being the $n$-th singular homology group. For an $n$-sphere $S^n$, we have $H_0(S^n)=H_n(S^n)=\mathbb{Z}$ and is $0$ otherwise. So, $$\chi(S^n)=(-1)^0\cdot 1+(-1)^n\cdot 1=1+(-1)^n.$$
