Explain why the ring structure of $H^{*}(\mathbb{C}P^n; \mathbb{Z})$ is a truncated polynomial ring.
**I got a proof for it (without Poincare duality) but I did not understand it at all. And I knew that there is a proof using Poincare duality, so could anyone explain for me either of these 2 proofs in details? **
I know that the proof of Poincare duality is in AT, but still I do not understand it.
There is a proof I like in Milnor-Stasheff using the Gysin sequence (which is derivable from the Thom isomorphism theorem).
Suppose $E \to X$ is an oriented vector bundle of rank $n$ with Euler class $e \in H^n(X;\mathbb{Z})$, and let $E_0$ be the subspace of non-zero vectors, which is homotopy-equivalent to the sphere bundle $S(E)$. Then there is a long exact sequence
$$ \dots \to H^{i}(X) \stackrel{\cup e}{\to} H^{i+n}(X) \to H^{i+n}(S(E)) \to H^{i + 1}(X) \to \dots $$
(This is secretly the long exact sequence for the pair $(E, E_0)$.)
Consider $X= \mathbb{CP}^n$ and $E = \gamma_1$ the tautological complex line bundle (whose real rank is $2$). The space $(\gamma_1)_0$ is particularly nice in this case: it is given by pairs $(L, v)$ where $L$ is a complex line in $\mathbb{C}^{n+1}$ and $v$ is a non-zero vector in $L$ and hence determines $L$, and in fact $(\gamma_1)_0 \cong \mathbb{C}^{n+1} \setminus 0 \simeq S^{2n+1}$. Therefore for $0\leq i \leq 2n-2$ we have isomorphisms $$ \cup e(\gamma_1) \colon H^i(\mathbb{CP}^n) \cong H^{i+2}(\mathbb{CP}^n) $$
(For $n > 1$ this actually verifies that $e(\gamma_1) \neq 0$ and so $\gamma_1$ is non-trivial.) Since $H^0(\mathbb{CP}^n)\cong \mathbb{Z}$ and $H^1(\mathbb{CP}^n) \cong 0$, it follows that $H^{2k}(\mathbb{CP}^n) \cong \mathbb{Z}\langle e(\gamma_1)^k\rangle$ and $H^{2k+1}(\mathbb{CP}^n) \cong 0$ for $0\leq k \leq n$, and for dimension reasons $H^i(\mathbb{CP}^n)\cong 0$ for $i > 2n$.
For $\mathbb{CP}^\infty$ it's even easier, since the sphere bundle of the tautological bundle is $S^\infty \simeq *$ so $\cup e(\gamma_1)$ is an isomorphism in all degrees.
One nice feature of this argument is that it can be adapted to prove that $H^*(BU(n);\mathbb{Z}) \cong \mathbb{Z}[c_1, \dots, c_n]$ for cohomology classes with $|c_i| = 2i$. You proceed by induction on $n$, and use the fact that if $n > 0$ and $\gamma_n$ is the tautological complex $n$-plane bundle then the sphere bundle $S(\gamma_n)$ is homotopy-equivalent to $BU(n-1)$.
A similar argument also computes $H^*(\mathbb{RP}^n;\mathbb{Z}/2)$ using the fact that $\gamma_1^\mathbb{R}$ is $\mathbb{Z}/2$-orientable, and adapts to computing $H^*(BO(n);\mathbb{Z}/2)$.