= 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 therefore
$$
\boxed{K=\{(\zeta^k,\zeta^{-k}I_n):k=0,1,\ldots,n-1\}\cong\mathbb Z_n.}
$$
It 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 gives
$$
\boxed{U(n)\cong\big(U(1)\times SU(n)\big)/K.}
$$
Explicitly 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.
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