degree of a map in terms of fundamental classes confusion Hatcher makes the following definition in exercise 7 on page 258. 
For a map $f: M \rightarrow N $ between connected, closed, orientable $n$-manifolds with fundamental classes $[M]$ and $[N]$, the degree of $f$ is defined to be the integer $d$ such that $f_{*}[M]=d[N]$ (so the sign of the degree depends on the choice of fundamental classes).

I'm confused as to why $f_{*}[M]$ has be of the form $d[N]$ for arbitrary coefficient ring $R$. I know that $[M]$ will be a generator in the group that it lies in, namely $H_{n}(M;R) \approx R$ and similarly $[N]$ is a generator for $H_{n}(N,R) \approx R$. I showed that for any isomorphism $f:R\rightarrow S$ a generator of $R$ must map to a generator of $S$, where my definition of generator for $R$ is an element $u$ such that $Ru=R$. But with this I still can't see why $f_{*}[M]=d[N]$. 

 A: For the definition of degree, you should consider homology with integer coefficients. Now, $$f_*[M]\in H_n(N,\mathbb{Z})=\mathbb{Z}\langle[N]\rangle,$$and so, there is a unique $d\in\mathbb{Z}$ such that $$f_*[M]=d[N].$$ 
A: In the context of that definition in Hatcher's book, integer coefficients are understood.
Nonetheless, as your question suggests, one can also define degree with respect to any coefficient ring $R$ which I will assume to be a commutative ring with a multiplicative identity that I will denote $1_R$. So for any continuous function $f : M \to N$, in the fashion described in your question one defines the $R$-degree of $f$, namely the element $d_R(f) \in R$ which equals the image of $1_R \in R \approx H_n(M;R)$ under the $R$-module homomorphism
$$f_* : R \approx H_n(M;R) \to H_n(N;R) \approx R
$$ 
The relation between $d_R(f)$ and $d(f) = d_{\mathbb Z}(f)$ is that 
\begin{align*}
d_R(f) &= d(f) \cdot 1_R \\ &= \underbrace{1_R + \cdots + 1_R}_{\text{$d(f)$ times}} \quad \text{(if $d(f)$ is positive, otherwise switch signs)}
\end{align*}
This is a straightforward corollary of the fact that the short exact sequence in the Universal Coefficients Theorem is functorial; see Chapter 5, Section 2, Theorem 8 of Spanier's "Algebraic Topology".
So for example if you are using the coefficient ring $R=\mathbb{Z}/2\mathbb{Z}$ then $d_R(f)= 0$ if $d(f)$ is even and $d_R(f)=1$ if $d(f)$ is odd. (As a side light, one advantage of degree using $\mathbb{Z}/2\mathbb{Z}$ coefficients is that it is defined even if $M,N$ are not orientable and hence $d_{\mathbb{Z}}(f)$ is not defined).
