= Solution
For $z\in U(1)$ and $U\in SU(n)$, the <matrix> $zU$ is unitary, since $(zU)^\dagger(zU)=|z|^2U^\dagger U=I_n$. 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 gives
$$
f((z,U)(w,V))=f(zw,UV)=zwUV=(zU)(wV)=f(z,U)f(w,V).
$$
Also $f(1,I_n)=I_n$. Hence
$$
\boxed{f:U(1)\times SU(n)\longrightarrow U(n),\quad(z,U)\longmapsto zU\text{ is a group homomorphism}.}
$$
The <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>.
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