Degree of a map $S^0 \to S^0$ By definition the degree of a map $f: S^n \to S^n$ is $\alpha \in \mathbb Z$ such that $f_\ast(z) = \alpha z$ for $f_{\ast}:H_n(S^n) \to H_n(S^n)$.
What is the definition of the degree of $f: S^0 \to S^0$? 
(I need it to calculate the degree of the attaching map of a $1$-cell to two $0$-cells)
 A: Most definitions of degree require the domain to be connected. This is the case for the definition of degree for maps $S^n \to S^n$ you referenced, since it requires that $H_n(S^n) \cong \mathbb{Z}$. But $S^0 \cong \{-1,1\}$ is disconnected, with  $H_0(S^0) \cong \mathbb{Z} \oplus \mathbb{Z}$. 
Fortunately, this doesn't stop us from understanding the attaching map from the boundary of a 1-cell to the 0-skeleton. When it comes to, for example, computing cellular homology, degree is used to formalize the (signed) number of times the boundary of an $n$-cell "hits" each $(n-1)$-cell in the $(n-1)$-skeleton to which it's being attached. But in the case where $n=1$, we can extract that information directly from the map $\phi:\partial e^1 \to X^0$. Writing $e^1=[a,b]$ we let the natural orientation of $[a,b]$ correspond to assigning "-1" to $a$ and "+1" to $b$. Then, given $x_0 \in X^0$, we can define the "degree" of $\phi$ at $x_0$ to be $0$ if $f^{-1}(x_0)=\emptyset$, $-1$ if $f^{-1}(x_0)=\{a\}$, $1$ if $f^{-1}(x_0)=\{b\}$, or $-1+1=0$ if $f^{-1}(x_0)=\{a,b\}$. This is just the simplicial boundary map at the chain level.
A: It is defined using reduced homology group instead in Hatcher. There we don't have the porblem with $S^0$. 
