I wonder if someone could help me resolve the following contradiction.
I thought there was a fiber sequence $\mathbb{CP}^{n-1} \to BU(n-1) \times BU(1) \to BU(n)$. But since all the cohomology here is concentrated in even degrees, there's no room for any differentials in the associated Serre spectral sequence. This is a contradiction because the $E_2$-page $H^\ast(\mathbb{CP}^{n-1};H^\ast(BU(n))) = \mathbb{Z}[x_2]/x_2^n \otimes \mathbb{Z}[c_2,\dots, c_{2n}]$ doesn't even have the same rank in various degrees as $H^\ast(BU(n-1)\times BU(1)) = \mathbb{Z}[c_2,\dots, c_{2n-2}]\otimes\mathbb{Z}[x_2]$.
I concluded that there was a fiber sequence as follows. There is a well-known and very natural fiber sequence $U(n-1) \to U(n) \to S^{2n-1}$, given by taking the first column of a unitary matrix. We can enlarge the fiber to a sequence $U(1) \times U(n-1) \to U(n) \to \mathbb{CP}^{n-1}$, which we can loop around to a fiber sequence $\Omega \mathbb{CP}^{n-1} \to U(1) \times U(n-1) \to U(n)$. Applying a delooping functor $B$ which preserves finite products, we get $B\Omega \mathbb{CP}^{n-1} \to BU(1) \times BU(n-1) \to BU(n)$, and since $\mathbb{CP}^{n-1}$ is path-connected, this is just $\mathbb{CP}^{n-1} \to BU(1) \times BU(n-1) \to BU(n)$.
The only thing I don't think I'm sure of is that (nice enough) delooping functors preserve fiber sequences. Is this simply not true?