Follow up Do $SU(n) \to \operatorname{Spin}(2n)$ and $SU(n) \to O(2n)$ induces an embedding of $SU(n)$?, now I ask a more restricted thus more difficult version of question on $U(n)$ instead of $SU(n)$:
Does $U(n) \to Spin(2n)$ induces an embedding of $U(n) \subset Spin(2n)$?
Do $U(n) \to O(2n)$ induces an embedding of $U(n) \subset O(2n)$?
Here the spin group $Spin(2n)$ has a $\mathbf{Z}_2$ center such that $Spin(2n)/\mathbf{Z}_2=SO(2n).$ $$ \begin{array}{ccc} & & Spin(2n)\\ &\nearrow & \downarrow\\ U(n) & \longrightarrow & SO(2n) \end{array}. $$
Here the orthogonal group $O(2n)$ has a normal $SO(2n)$ subgroup such that $O(2n)/SO(2n)=\mathbf{Z}_2.$ $$ \begin{array}{ccc} & & O(2n)\\ &\nearrow & \uparrow\\ U(n) & \longrightarrow & SO(2n) \end{array}. $$
We can take the complex vector space $\mathbb C^n$ as $\mathbb R^{2n}$ and recall that the standard inner product on $\mathbb R^{2n}$ is the real part of the standard Hermitian inner product on $\mathbb C^n$. So this shows that $$U(n) \subset SO(2n)$$ (maybe this also proves that $U(n) \subset O(2n)$ as well? please clarify.)
I know the first homotopy group that $$\pi_1(U(n))=\mathbf{Z}, \; \forall n \geq 1$$ $$\pi_1(SO(2))=\pi_1(O(2))=\pi_1(Spin(2))=\mathbf{Z}, \; $$ $$\pi_1(SO(m))=\pi_1(O(m))=\mathbf{Z}_2, \; \forall m \geq 3$$ $$\pi_1(Spin(m))=0, \; \forall m \geq 3.$$ Perhaps the fundamental group is crucial to understand whether the lift and the embedding are possible (?).