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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.


(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 $H^*(K(\mathbb Z,2r);\mathbb Q)=\mathbb Q[a_{2r}]$.

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up vote 1 down vote accepted

2) The defining property of $K(\mathbb Z,2r)$ is that for any $X$, we have $H^{2r}(X, \mathbb Z)= [X, K(\mathbb Z,2r)]$. In particular for $X=K(\mathbb Z,2r)$ we know that $H^{2r}(X, \mathbb Z)= \mathbb Z$ by Hurewitz and is generated by the element corrsponding to the identity map. This is the fundamental class. Similarly with $\mathbb Q$ coefficients.

1, 3) $H^i(K(\mathbb Z,2r);\mathbb Q)\stackrel{\cup e}{\rightarrow}H^{i+2r}(K(\mathbb Z,2r);\mathbb Q)$ means that $\cup e$ is an isomorphism from $H^0(K(\mathbb Z,2r);\mathbb Q)$ to $H^{2r}(K(\mathbb Z,2r);\mathbb Q)$. The later is generated by the fundamental class. Hence $e$ is plus or minus the fundamental class.

Now, by induction on $i$, all the $H^i$ with $i$ not multiple $2r$ are zero, while the others are all isomorpic to $\mathbb Q$ and are generated by $e$, $e^2$ etc. or equivalently by $a_{2r}$, $a_{2r}^2$ etc. That is, $H^*(K(\mathbb Z,2r);\mathbb Q)=\mathbb Q[a_{2r}]$.

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thanks that explains it all.. – palio May 25 '11 at 16:52
@palio You could accept the answer, then. – Max May 27 '11 at 7:26

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