= Solution
Use the continuous determinant map
$$
\det:U(2)\longrightarrow S^1.
$$
It induces a homomorphism on fundamental groups. Under the standard identification
$$
\pi_1(S^1,1)\cong\mathbb Z,
$$
the class of $t\mapsto e^{2\pi it}$ corresponds to its <winding number> $1$. Since
$$
(\det\circ\gamma)(t)=e^{2\pi it},
$$
we obtain
$$
\det_*([\gamma]^k)=k\in\mathbb Z.
$$
For $k\ne0$ this is not the identity, so $[\gamma]^k$ could not have been the identity in $\pi_1(U(2),I)$.
Back to article page
