D. J. Simms in his book "Lie groups and quantum mechanics" (page 9) says that: any representation of $\sigma$ of $\tilde{G}$ (the simplgy connected covering group of the Lie group) in $U(H)$ such that $\sigma (K)\subset U(1)$ (in which $K$ is the kernel of covering map $p:\tilde{G}\to G$) will defines a unique projective representation $T$ of $G$ in $U(\hat{H})$ such that $T\circ p=\pi \circ \sigma$ (where $\pi :U(H)\to U(\hat{H})$ sends every unitary operator $U:H\to H$ to the corresponding symmetry $S_U :PH\to PH$).
I would really appreciate if somone could explain the reason of the claim above. Thank you in advance.
I assume "$U(1)$" here refers specifically to the centre of $U(H)$.
Note all we need to know is that $T$ is well defined as it will have to be unique automatically from the condition $T \circ p = \pi \circ \sigma$.
Then observe that for $\tilde{h}, \tilde{g} \in \tilde{G}$, $p(\tilde{h})=p(\tilde{g})$ if and only if $\tilde{h} = k\tilde{g}$ for $k \in K$. So take $g\in G$ with $g = p(\tilde{g})=p(\tilde{h})$ so that $T(g) = \pi(\sigma(\tilde{g})) = \pi(\sigma(\tilde{h}))$.
Then $\sigma(\tilde{h}) = \sigma(k\tilde{g}) = \sigma(k)\sigma(\tilde{g})$ but since $\sigma(k)$ is in the centre of $U(H)$ it acts by scalars so $\pi(\sigma(\tilde{h})) = \pi(\sigma(k)\sigma(\tilde{g})) = \pi(\sigma(\tilde{g}))$. Thus $T(g)$ is well defined.