H^4(SO(3)xU(1),U(1)) with Borel cohomology I want to know what the above is. I would also like to know what H^5(B[SO(3)xU(1)],U(1)) is, where B[SO(3)xU(1)] is the classifying space of SO(3)xU(1). I don't know how to do the calculation myself
 A: If I'm understanding your notation right, then you're asking for the fourth cohomology of the Lie group $SO(3) \times U(1)$ with coefficients in the abelian group $U(1) = S^1$.  Using the exponential sequence $1 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 1$, we have
$$H^4(SO(3) \times U(1), U(1)) = H^5(SO(3) \times U(1), \mathbb{Z})$$
But the latter is 0, since $SO(3) \times U(1)$ is a 4-dimensional manifold.
Again, the exponential sequence shows that $H^5(B[SO(3)\times U(1)],U(1)) = H^6(B[SO(3)\times U(1)],\mathbb{Z})$.  Since $B[SO(3)xU(1)] = BSO(3) \times BU(1)$, and $BU(1) = \mathbb{C} P^\infty$ has cohomology equal to $\mathbb{Z}$ in every even dimension (and 0 in every odd), the Künneth formula gives
$$H^6(B[SO(3)\times U(1)],\mathbb{Z}) = H^6(BSO(3), \mathbb{Z}) \oplus H^4(BSO(3), \mathbb{Z}) \oplus H^2(BSO(3), \mathbb{Z}) \oplus H^0(BSO(3), \mathbb{Z}) \\
= (\mathbb{Z} \oplus \mathbb{Z} / 2) \oplus \mathbb{Z} \oplus 0 \oplus \mathbb{Z} = \mathbb{Z}^3 \oplus \mathbb{Z}/2.$$
Here, the cohomology of $BSO(3)$ is not a terribly difficult computation, using one of several Serre spectral sequences; see, for instance Ebert and Randal-Williams note "On the divisibility of characteristic classes of non-oriented surface bundles."
