I started learning about signature of a $4k$-manifold and one of the most common example is the signature of $\mathbb{C}P^{2n}$. The only reference I found is tom Dieck's Algebraic Topology.
Even though the "structure" of the proof (i.e. the idea behind it) seems clear there are some obscure passage in it, first of all, given that strange map $p \colon (\mathbb{C}P^1)^n\to \mathbb{C}P^{2n}$, realising what is the behaviour of $p_*$ or $p^*$ for me it's completely impossible.
So I thought about it and realised that maybe there is an easier (but surely sloppier) argument: Recall that $H^*(\mathbb{C}P^{2n};\mathbb{Z}) \cong \mathbb{Z}[\alpha]/(\alpha^{n+1}), |\alpha|=2$. Due to the fact that $H^{4n-1}(\mathbb{C}P^{2n};\mathbb{Z})=0$, then $H^{4n}(\mathbb{C}P^{2n};\mathbb{Z})\cong \hom_{\mathbb{Z}}(H_{4n}(\mathbb{C}P^{2n}); \mathbb{Z})$, so we can choose the generator of $H^{4n}(\mathbb{C}P^{2n};\mathbb{Z})$ as the dual of $[\mathbb{C}P^{2n}]$, and we can check wether $(\alpha^{n}\smile \alpha^{n})\frown [\mathbb{C}P^{2n}]$ is $1$ or $-1$, i.e. we check if $\alpha^{2n}:=\alpha^{n}\smile \alpha^{n}$ is the dual of the fundamental class or not. in the first case, $\sigma(\mathbb{C}P^{2n})=1$, in the other case $\sigma(\mathbb{C}P^{2n})=-1$. So it's just a matter of wether the element $\alpha$ "induces" the chosen orientation or not.
I'm aware that this reasoning is somewhat sloppy, BUT I really didn't find anything else, and I can't prove the passage tom Dieck does with the induced map $p_*,p^*$ in (co)-homology.
So I'm here asking if someone has a better proof of this fact OR can explain why (according to tom Dieck), $p^*(\alpha) = \alpha_1 + \cdots + \alpha_n$, where $\alpha_i$ is the dual of the fundamental class of the $i$-th copy of $\mathbb{C}P^1$ in $(\mathbb{C}P^1)^n$.