= Solution
First prove surjectivity. For any $A\in U(n)$, its <determinant> has modulus one. Choose $z\in U(1)$ with $z^n=\det A$ and define $U=z^{-1}A$. Then $U^\dagger U=I$ and $\det U=z^{-n}\det A=1$, so $A=f(z,U)$.
Next, $f(z,U)=I$ is equivalent to $U=z^{-1}I$. This lies in $SU(n)$ precisely when $z^n=1$. Therefore the <kernel of a group homomorphism> is the central <cyclic group>
$$
\boxed{\ker f=\{(\zeta^k,\zeta^{-k}I_n):k=0,\ldots,n-1\}\cong\mathbb Z_n,
\qquad\zeta=e^{2\pi i/n}.}
$$
Its generator is $(\zeta,\zeta^{-1}I_n)$, which has order $n$. The inverse phases in the two factors are essential: this is an explicitly embedded <subgroup>, not a quotient by scalar <matrices> in just one factor. The <first isomorphism theorem> now gives
$$
\boxed{U(n)\cong\bigl(U(1)\times SU(n)\bigr)/\mathbb Z_n.}
$$
This also holds as a <Lie group isomorphism>: the induced map is smooth and a continuous bijection from a compact <quotient space> to the Hausdorff <unitary group>, and its differential is the <Lie algebra isomorphism> $(it,X)\mapsto itI+X$. For $n=1$ the <kernel of a group homomorphism> is trivial, consistently with $SU(1)=\{1\}$.
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