Consider the path fibration: $K(\mathbb Z,2r-1)\rightarrow PK(\mathbb Z,2r)\rightarrow K(\mathbb Z,2r).$
Suppose that $H^*(K(\mathbb Z,2r-1);\mathbb Q)=H^*(S^{2r-1};\mathbb Q).$
We want to show that $H^*(K(\mathbb Z,2r);\mathbb Q)=\mathbb Q[a_{2r}]$.
The Gysin sequence gives that we have an isomorphism $H^i(K(\mathbb Z,2r);\mathbb Q)\stackrel{\cup e}{\rightarrow}H^{i+2r}(K(\mathbb Z,2r);\mathbb Q)$ where $\cup e$ is the cup product with the rational euler class.
Questions:
(1) Why is it that $e$ and the fundamental class $a_{2r}\in H^{i+2r}(K(\mathbb Z,2r);\mathbb Q)\cong \mathbb Q$ are non-zero multiples of each other?
(2) What does the "fundamental class" mean in this context? and finally,
(3) I'm not clear as to how to deduce that $H^*(K(\mathbb Z,2r);\mathbb Q)=\mathbb Q[a_{2r}]$.