A question about Pontryagin product on $CP^{\infty}$ The homology of $\mathbb{C}P^{\infty}$ is $\mathbb{Z}$ for $i=0,2,4,...$ and zero otherwise. Pontryagin product is $H_i(\mathbb{C}P^{\infty}) \times H_j(\mathbb{C}P^{\infty})\longrightarrow H_{i+j}(\mathbb{C}P^{\infty})$. We denote the generator in $H_{2k}(\mathbb{C}P^{\infty})$ as $a_{2k}$.
Then why does $a_2^k = k! a_{2k}$?
The description here Pontryagin Product does not give proof of this. Please give elementary proof.
 A: The description I know of the $H$ space structure on $\mathbb{CP}^\infty$ is as the projectivisation of the countable vector space $\mathbb{C}[x]$, with multiplication given by multiplication of polynomials. (To check this is homotopic to your description, it suffices to check what the pullback on $H^2$ is.)
From this perspective, we have a natural basis of $2k$th homology as the fundamental class associated to the sub-projective space of polynomials of degree $\leq k$.
Now we can see what our multiplication does, taking the $k$th power of $a_2$ is the homology push forward of the top class of $(\mathbb{CP}^1)^k$ under the multiplication map. But this map factors through the projective subspace of degree $\leq k$ polynomials, and this top degree push forward from this subspace (by definition) gives our basis element $a_{2k}$.
Thus we are reduced to computing the degree (multiplication on top homology) of the map between these two smooth compact oriented manifolds: $$(\mathbb{CP}^1)^k\rightarrow \mathbb{CP}^k$$
In this nice setup (oriented smooth map of smooth manifolds) it is a classical result that the degree is equal to the cardinality of the generic preimage of the morphism. In our case, we have the open dense subset of degree $k$ polynomials with distinct roots, and for these we see that there are $k!$ preimages, given by choosing an ordering of the roots. Thus this coefficient is $k!$, and we are done.
