Calculation of the first Chern class of the canonical line bundle over $\mathbb{CP}^n$

Question

There are different ways of defining and thereafter calculating the Chern classes. Right now I'm studying from the lecture notes which introduce the first Chern class through the classifying spaces as follows:

The classifying bundle for $\mathbb{U}(1)$ is $\mathbb{S}^{\infty} \rightarrow \mathbb{CP}^{\infty}$. So the set of complex line bundles over $B$, which are $\mathbb{U}(1)$-bundles, is in the bijective correspondence with the maps from $B$ to $\mathbb{CP}^{\infty}$ taking up to homotopy. So for any bundle there exists a map induced on the cohomology rings from $\mathbb{Z}[t]$ to $H^{\bullet}(B)$. And the first Chern class is defined as the image of the generator $t$ in the latter map.

I have a canonical line bundle over $\mathbb{CP}^n$ and I want to calculate the first Chern class using classifying space theory. Do I have to follow the explicit construction of the map between $\mathbb{CP}^n$ and $\mathbb{CP}^{\infty}$ for this line bundle, because I couldn't do it, or there is another way using classifying spaces?

Answer 1

Moving from my own comment I have come to some idea which looks plausible to me. I would be grateful if you indicate mistakes or support.

The idea is to move the other way around from the given bijectivity between set of complex line bundles and $[\mathbb{CP}^n,\mathbb{CP}^{\infty}]$. The mentioned bijection is given by pull-backing the classifying bundle. So I wondered which bundle I will get if I pull-back it under the usual inclusion $\mathbb{CP}^n \hookrightarrow \mathbb{CP}^{\infty}$ and fortunately I get exactly the canonical line bundle! The total space of the pullbacked bundle is defined as $E=\{[l]\times x \in \mathbb{CP}^n\times S^{\infty}:x\text{ maps to } [x] \in \mathbb{CP}^k \text{ for } x\in S^{2k+1} \text{ under the usual quotient map and } [x]=[l] \}$

What can be said about $E$ ? It's second coordinate cannot exceed the sphere with dimension $2n+1$ because $[x]=[l]\in \mathbb{CP}^n = S^{2n+1}/ \sim$ so the set $E$ can be rephrased as $$\{[l]\times x \in \mathbb{CP}^n\times S^{2n+1}:x\text{ quotients to } [l] \in \mathbb{CP}^n \text{ for } x\in S^{2n+1}\}=$$ $$=\{[l]\times x \in \mathbb{CP}^n\times \mathbb{C}^{n+1}:x\text{ quotients to } [l] \in \mathbb{CP}^n \text{ for } x\in \mathbb{C}^{n+1}\}$$

But this is exactly the definition of the canonical line bundle!

The rest part of showing that the image of generator $t$ is again $t$ (actually $t+(t^{n+1})$) is rather a standart job, because the inducing map is inclusion.

Still can't say that I am pleased by this answer because it doesn't use constructive way but rather lucky guess.