I understand that if $f,g: X \to Y$ are maps and $f$ is homotopic to $g$, then the induced maps on the homology groups $f_*$ and $g_*$ are equal. Why does this imply that if $X$ is contractible then $H_0(X) = \mathbb{Z}$ and $H_n(X) = 0$ for $n > 0$?
[Math] Homology of contractible space
algebraic-topologyhomotopy-theory
Related Solutions
$\mathbb{R}^n$ is contractible therefore homotopy equivalent to a point and so $H_n(\mathbb{R}^m) = H_n(\{ \ast \})$.
$$ H_n(\{ \ast \}) = 0 , n > 0$$ $$ H_n(\{ \ast \}) = \mathbb{Z} , n = 0$$
You can find the construction of homology with general coefficients and the universal coefficient theorem in Hatcher's Algebraic Topology, which is available free from his website.
The answer to your third question is yes.
The answer to the second part of your first question is yes, especially in the case that we take $G$ to be a field, most often finite or $\mathbb{Q}$, or $\mathbb{R}$ in differential topology. Homology over a field is simple because $\operatorname{Tor}$ always vanishes, so you get e.g. an exact duality between homology and cohomology. Homology with $\mathbb{Z}_2$ coefficients is also the appropriate theory for many questions about non-orientable manifolds-their top $\Bbb{Z}$-homology is zero, but their top $\Bbb{Z}_2$ homology is $\Bbb{Z}_2$, which leads to the degree theory in Milnor you were mentioning.
Cohomology with more general coefficients than $\mathbb{Z}$ is even more useful than homology. For instance it leads to the result that if a manifold $M$ has any Betti number $b_i(M)<b_i(N)$, where $b_i$ is the rank of the free part of $H_i$, there's no map $M\to N$ of non-zero degree. This has lots of quick corollaries-for instance, there's no surjection of $S^n$ onto any $n$-manifold with nontrivial lower homology! Edit: This is obviously false, and I no longer have any idea whether I meant anything true.
But in the end $H_*(X;G)$ is more of a stepping stone than anything else; it gets you thinking about how much variety there could be in theories satisfying the axioms of homology. It turns out there's almost none-singular homology with coefficients in $G$ is the only example-but if we rid ourselves of the "dimension axiom" $$H_*(\star)=\left\{\begin{matrix}\mathbb{Z},*=0\\0,*>0\end{matrix}\right.$$ then we get a vast collection of "generalized (co)homology theories," beginning with K-theory, cobordism, and stable homotopy, which really do contain new information. In some cases, so much new information that we can't actually compute them yet!
Best Answer
If $X$ is a contractible space, you can take $f=Id_X:X\rightarrow X$ and $g(x)=y$ where $y$ is a fixed point of $X$ so $g$ is a constant map. The maps $f$ and $g$ are homotopic. If $n>0$, $f_*^n: H^n(X)\rightarrow H^n(X)$ is the identity, and $g_*^n=0$.