I already started discussion there: One unclear detail in Baum 1968 paper "On the cohomology of homogeneous spaces". But I think that one is quite old, and this time the question is more specific, so let me ask again here.
There is a picture of two different inclusions of $U(n-1)$ into $U(n)$: one is aquired by just putting $1$ in the corner of a matrix, the second is by placing the inverse of determinant there instead. Then, if we consider the composition for arbitrary classifying map $X \longrightarrow BU(n-1) \longrightarrow BU(n)$, in the first case it would classify the new bundle which is the direct sum with a trivial bundle, in the second case - direct sum with the tensor inverse of the determinant bundle. This result is probably not surprising, but the question is - why so? What properties of $B$ as a functor should I consider to explain that?
In previous topic(link is above), Connor Malin suggested to start with maps $BU(n-1) \longrightarrow BU(n)$ which are given by the universal property that they pull bundles back the way I need. If I'm gonna go this way, how can I describe $BU$-maps in such a way, that they "fit" aforementioned inclusions of $U(n-1)$? In any case, I would be glad if you advise how to connect things that happen before and after applying $B$.
Instead of classifying any bundle by composition $X \longrightarrow BU(n-1) \longrightarrow BU(n)$, if talking about what's happening with cohomology I think it suffice to just say that unversal n-bundle pulls back to $EU(n-1)\oplus\varepsilon \longrightarrow BU(n-1)$ or $EU(n-1)\oplus \det(EU(n-1))^{-1} \longrightarrow BU(n-1)$.
Thank you! I kinda need this to give explanations in my course work, so if anyone can it to me I would appreciate that.