As you know $$H^\ast (U(n);{\bf Z})=\bigwedge_{\bf Z}[x_1,x_3,...,x_{2n-1}]$$ where $|x_i|=i$
To prove this we use Leray-Hirsch Theorem for $$\tag{*}\ U(n-1)\rightarrow U(n)\rightarrow S^{2n-1}$$ To do this we must show that for inclusion $$f:S^{2n-1} \rightarrow U(n)$$ where $f^\ast (\alpha)$ is generator of $H^{2n-1}(S^{2n-1})$.
How can we prove this? Thanks in advance.
Answer
Here fiber is $U(n-1)$ so that the above statement is not true.
From Eq $(\ast)$ we have a lone exact seq. for homotopy so that $(U(n),U(n-1))$ is $(2n-3)$-connected.
For generator $x_{2k-1}\ (k=1, ... ,n-1)$ for $U(n-1)$, there exist $$i^\ast c_{2k-1}= x_{2k-1}$$ and $$c_{2k-1}\in H^\ast(U(n))$$ where $$i\colon U(n-1)\rightarrow U(n).$$