 Solution

First the group homomorphism is surjective. For any $W\in U(n)$, choose $z\in U(1)$ with $z^n=\det W$ and set $U=z^{-1}W$. Then $U$ is unitary and $\det U=z^{-n}\det W=1$, so $W=f(z,U)$.

Next $f(z,U)=I_n$ is equivalent to $U=z^{-1}{I}_n$, and the special unitary group condition becomes $z^n=1$. With $\zeta =e^{2\pi i/n}$, the kernel of a group homomorphism is thereforeIt is the cyclic central subgroup generated by $(\zeta ,{\zeta }^{-1}{I}_n)$; the scalar and special unitary group factors are paired with inverse phases. The first isomorphism theorem now givesExplicitly the isomorphism sends the coset $[(z,U)]$ to $zU$. The $n$ possible choices of the root $z$ differ precisely by multiplication by elements of $K$, which explains why the product before quotienting is not a unique factorization.