One unclear detail in Baum 1968 paper "On the cohomology of homogeneous spaces" I was asking the same kind of question about a year ago, haven't got any distinct answer then, afterwards I got distracted, and now when I'm thinking about that question again - I still can't come up with any clue. Basically everything what I wanted to understand is on the screenshot:

We have two embeddings $U(n-1) \longrightarrow U(n)$ which give us different actions on cohomology of classifying spaces. But how did Baum come up with these expressions? I'm not even asking why they are different -  just how he calculated action of $\rho$ in each case? I have a slight understanding of the fact that $BU(n)$ is a base space of a universal bundle, and that $H^*(BU(n))$ is generated by  characteristic classes - can it help anyhow in deducing underlined expressions? I tried to think what cocycles in $BU(n)$ generate classes $x_{2i}, y_{2i}$ in $H^*$, but now I don't feel like it would help.
Thanks a lot!
 A: The idea is to use properties of characteristic classes. In the first case, one can verify that the induced map $BU(n) \rightarrow BU(n+1)$ has the property that if $X \rightarrow BU(n+1)$ classifies a vector bundle, then the composition $X \rightarrow BU(n)$ classifies the direct sum with a trivial line bundle. The characteristic classes of the new bundle are the same as the old bundle according to the theory of Chern classes, so applying this to the identity $BU(n) \rightarrow BU(n)$ (which represents the universal vector bundle whose Chern classes are the $c_i$), we get the first formula.
In the second case, one can verify that the induced map $BU(n) \rightarrow BU(n+1)$ has the property that if $X \rightarrow BU(n)$ classifies a vector bundle, then the composition $X \rightarrow BU(n+1)$ classifies the direct sum with the tensor inverse of the determinant bundle. The determinant bundle has its single possible nontrivial class given by $c_1$ of the original bundle, so its inverse has its first class $-c_1$. Applying the Whitney sum formula, we get the second formula.
