What is the induced map on fundamental group of the inclusion of unitary group in the orthogonal group?

Question

What is the induced map on fundamental group of the inclusion of unitary group $U(n)$ in the orthogonal group $SO(2n)$?(Note that the unitary group $U(n)$ can only embedded in the group $SO(2n)$, not $SO(n)$(its dimension is greater than $SO(n)$)!)

Answer 1

Given that $\pi_1(U(n))\cong \mathbb{Z}$, and $\pi_1(SO(n))\cong\mathbb{Z}/2\mathbb{Z}$, either $i_*\colon\pi_1(U(n))\to \pi_1(SO(n))$ is trivial or surjective.

Note that $U(1)\subset U(n)$ for all $n$ and in fact this inclusion induces an isomorphism on fundamental groups, and so we can instead consider the inclusion $U(1)\cong S^1$ into $SO(n)$ which factors through $U(n)$. As a subgroup, $U(1)$ can be seen as the set of linear maps representing rotations about some fixed axis, and in fact this subset represents the image of a loop which generates $\pi_1(SO(n))$ (as described in this answer). So we have a surjective group homomorphism (factoring through the fundamental group of $U(n)$) $$\pi_1(U(1))\stackrel{\cong}{\to}\pi_1(U(n))\stackrel{i_*}{\to}\pi_1(SO(n))$$ and then as the first map is an isomorphism $\pi_1(U(1))\to\pi_1(U(n))$, we see that $i_*$ must be a surjective homomorphism onto $\pi_1(SO(n))\cong\mathbb{Z}/2\mathbb{Z}$.