An embedding of $U(n)\to \text{Spin}^c(2n)$

Question

Let $(V,\langle,\rangle)$ be an $2n$-dimensional real inner product space and consider its Spin$^c$ group $\text{Spin}^c(V)\subset Cl(V)\otimes_{\Bbb R} \Bbb C$. Suppose there is a compatible (orthogonal) almost complex structure $J:V\to V$. We can then view $V$ as an $n$-dimensional complex vector space with a hermitian inner product $h$ given by $h(u,v)=\langle u,v\rangle -i\langle Ju,v\rangle$. In particular we can consider the unitary group $U(V)$. Define a map $\rho:U(V)\to \text{Spin}^c(V)$ as follows: Given $A\in U(V)$, we can choose a unitary basis $v_1,\dots,v_n$ for $V$ such that $Av_k=e^{i\theta_k}v_k$. Put $$\rho(A)=e^{(i\sum_k \theta_k)/2}\prod_{k=1}^n \left(\cos \frac{\theta_k}{2}+\sin\frac{\theta_k}{2}\cdot v_kJv_k \right)$$

In p.393 of Lawson & Michelson's book Spin Geometry, it is asserted that this map is a well-defined continuous group homomorphism, but I can't see why.

1. For well-definedness, we have to care about two choices: the choices of $\theta_1,\dots,\theta_n$, and the choice of basis. Clearly $\rho(A)$ is independent of the choice of the $\theta_k$'s. Also it was quite easy to show that $\rho(A)$ is independent of the order of $v_1,\dots,v_n$, because $v_kJv_k v_{k+1}Jv_{k+1}=v_{k+1}Jv_{k+1} v_kJv_k$. So now if $\{v_1,\dots,v_n\}$ and $\{w_1,\dots,w_n\}$ are two unitary bases with $Av_k=e^{i\theta_k}v_k$ and $Aw_k=e^{i\psi_k}w_k$, then we may assume $\theta_k=\psi_k$ for all $k$. If the $\theta_k$'s are all distinct (i.e. $A$ has $n$ distinct eigenvalues), then it is easy to see that $\rho(A)$ is well-defined. But I can't handle the case where $A$ has an eigenvalue with eigenspace $\dim>1$.

2. Also I can't see why $\rho$ is a continuous group homomorphism.

Any helps will be very appreciated.

Edit: There is a double cover $p:\text{Spin}^c(V)\to SO(V)\times S^1$. Consider the natural map $f:U(V)\to SO(V)\times S^1$ given by $f(A)=(A,\det A)$. Assuming $\rho$ is well-defined, I have shown that $\rho$ is a lift of $f$, i.e. $p\rho=f$. On the other hand, by covering space theory, there is a unique lift $\tilde{f}:U(V)\to SO(V)\times S^1$ of $f$ such that $\tilde{f}(\text{id})=1$. Thus, if $\rho$ is continuous, then we must have $\rho=\tilde{f}$.