I am having trouble understanding the following argument given in Spanier, Pages 515 ~ 516, regarding homotopy groups of even spheres.
I've attached the two pages; I'm having trouble understanding what alpha is, (highlighted in red) and why there is a C-isomorphism (highlighted in green).
Basically, my questions are as follows :
1) If $\pi_{2n-1}(W^{2n-1}) = \mathbb{Z} \bigoplus T$ where $T$ is a finite group, then does he mean $\alpha$ is $(0,x)$ as an element in $\mathbb{Z} \bigoplus T$, where $x$ is a generator for $T$? Or is it the other way around as $(1,0)$?
2) Why is $\alpha_{*}$ a C-isomorphism, as shown in the book?Page 515 is here
I first misread your questions, but I left my explanations here and edited the rest.
Let me first explain the "red" part. The author explained that we have $\pi_{2n-1}(W^{2n-1}) \cong C_\infty \oplus G$, where $G$ is a finite group. That in particular means that there is an element $a$ that generates the infinite cyclic group $C_\infty$. By definiton the higher homotopy groups are given as homotopy classes of continuous maps from $S^n$ to the space $X$ you are considering, which means in this situation there has to be a contiuous map $\alpha \colon S^{2n-1} \rightarrow W^{2n-1}$ that represents $a$.
Edit: As you used the notion $\mathbb{Z}$ for the infinity cyclic group, I can make it more precise. The author chose one representative $\alpha \colon S^{2n-1} \rightarrow W^{2n-1}$, such that $[\alpha] = (1,0)$ (or represents $(-1,0)$ in case that one for some reason wants to work with the other generator but makes no real difference).
The "green" part. Homology groups are functorial, which in particular means that they can be applied to a map. The map $\alpha_*$ is exactly the map induced by applying the functor $H_i$ to $\alpha$.
Edit: Since almost all homology groups on both sides vanish, every induced homomorphism yields an isomorphism there. Therefore we only have to look at the $i$ where the homologys do not vanish. As $\alpha$ represents the generator in the higher homotopy group, the map $\alpha_*$ should just be the map that sends the generator of $H_{2n-1}(S^{2n-1})$ to the generator $H_{2n-1}(W^{2n-1})$ and thus is an ismorphism as well. The case $i = 0$ is analogous.