The image is in the unitary group, since

(z U)^{†} (z U) = \overset{z}{ˉ} z U^{†} U = I

for

z \in U (1)

and

U \in S U (n)

(z, U) (w, V) = (z w, U V)

. Scalar matrices commute with every matrix, so

f ((z, U) (w, V)) = z w U V = (z U) (w V) = f (z, U) f (w, V) .

Thus

f

f (1, I) = I

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