In "Clifford Modules" by Atiyah, Bott and Shapiro (p.10) or "Dirac Operators in Riemannian Geometry" by Friedrich (p.28) one finds some sort of a lift of the natural inclusion $\operatorname{U}(k)\to \operatorname{SO}(2k)$ to the group $\mathrm{Spin^c}(2k)$.
So if $f_{\mathit{can}}:\operatorname{U}(k)\to \operatorname{SO}(2k)\times S^1:A\mapsto (A,\det A)$ is given, they state the following $F_{\mathit{can}}:\operatorname{U}(k)\to \operatorname{Spin}^c(2k)=\operatorname{Spin}(2k)\times_{\mathbb{Z}_2}S^1$ is a lift.
$ A\mapsto F_{\mathit{can}}(A):= \prod_{j=1}^k\big(\cos \frac{\theta_j}{2} +\sin \frac{\theta_j}{2} \cdot \cdot u_j\cdot J(u_j)\big)\times e^{\frac{i}{2}\sum_{j=1}^k \theta_j }, $
where $J$ is the complex structure as a map $\mathbb{C}^k\to \mathbb{C}^k$, and $(u_1,J(u_k),\ldots, u_k,J(u_k))$ is a basis of $\mathbb{R}^{2k}$, some of the dots are Clifford multiplication and the "angles" $\theta_j$ come from writing $A$ as a matrix $ A=\begin{bmatrix} e^{i \theta_1} &&\\ & \ddots &\\ & & e^{i \theta_k} \end{bmatrix}$.
It is never explicitely shown why this is a group homomorphism. So I tried to do it. Let $B$ be another matrix representing an element $\operatorname{U}(k)$ with angles $\zeta_j$.
\begin{align*} &\big(\cos \frac{\theta_j}{2} +\sin \frac{\theta_j}{2}\cdot u_j\cdot J(u_j) \big)\cdot \big(\cos \frac{\zeta_j}{2} +\sin \frac{\zeta_j}{2} \cdot u_j\cdot J(u_j) \big)\\ \quad & = \cos \frac{\theta_j}{2}\cos \frac{\zeta_j}{2}+\sin \frac{\theta_j}{2} \cos \frac{\zeta_j}{2}\cdot u_j\cdot J(u_j)\\ \quad&\phantom{=}+\cos \frac{\theta_j}{2} \sin \frac{\zeta_j}{2} u_j\cdot J(u_j)+ \sin \frac{\theta_j}{2} \sin \frac{\zeta_j}{2}\cdot \underbrace{u_j\cdot J(u_j)\cdot u_j\cdot J(u_j)}_{=-1 \text{ by Clifford mult. law.}}\\ \quad & =\cos \frac{\theta_j+\zeta_j}{2}+\sin \frac{\theta_j+\zeta_j}{2}\cdot u_j\cdot J(u_j) \end{align*} If this is right I do not get why one needs the second factor with the determinant. Asked differently: What is wrong here and how does one do it right?