 Solution

For $z\in U(1)$ and $U\in SU(n)$, the matrix $zU$ is unitary, since $(zU{)}^{†}(zU)=|z{|}^2{U}^{†}U=I_n$. Thus the map has its image in the unitary group $U(n)$. Scalar matrices commute with all matrices, so the direct product of groups multiplication givesAlso $f(1,I_n)=I_n$. HenceThe special unitary group condition $\det U=1$ imposes no requirement that $zU$ itself have unit determinant; its determinant is $z^n$, as expected for a general element of the unitary group.