[proofplan] The kernel is first identified with the fibre of the covering map $p$ over the identity $e_G$, and covering fibres are discrete because every point in a fibre has an evenly covered neighbourhood isolating it inside that fibre. For centrality, fix an element $k\in\ker p$ and measure its failure to commute with an arbitrary element $g\in\widetilde G$ by the commutator map $g\mapsto gkg^{-1}k^{-1}$. This map lands in the discrete space $\ker p$ and is continuous, so its image is connected and discrete, hence a singleton. Evaluating at $e_{\widetilde G}$ shows that singleton is $\{e_{\widetilde G}\}$, which is exactly the statement that $k$ commutes with every element of $\widetilde G$. [/proofplan] [step:Identify the kernel as a discrete fibre of the covering map] Since $p:\widetilde G\to G$ is a [group homomorphism](/page/Group%20Homomorphism), $\ker p=p^{-1}(\{e_G\})=\{x\in\widetilde G:p(x)=e_G\}$ is a subgroup of $\widetilde G$. We prove that this fibre is discrete in the [subspace topology](/page/Subspace%20Topology). Let $x\in\ker p$. Since the underlying map $p:\widetilde G\to G$ is a covering map, there exists an open neighbourhood $U\subset G$ of $e_G$ and an index set $A$ such that $p^{-1}(U)=\bigcup_{\alpha\in A}V_\alpha$, where the sets $V_\alpha\subset\widetilde G$ are pairwise disjoint open sets and, for each $\alpha\in A$, the restricted map $p|_{V_\alpha}:V_\alpha\to U$ is a homeomorphism. Let $\alpha_x\in A$ be the unique index such that $x\in V_{\alpha_x}$. If $y\in V_{\alpha_x}\cap\ker p$, then $p(y)=e_G=p(x)$. Since $p|_{V_{\alpha_x}}$ is injective, $y=x$. Hence $V_{\alpha_x}\cap\ker p=\{x\}$. Thus every point $x\in\ker p$ is isolated in the subspace topology on $\ker p$, so $\ker p$ is discrete. [/step] [step:Build a continuous commutator map with values in the kernel] Fix $k\in\ker p$. Define the map $c_k:\widetilde G\to\widetilde G$ by $c_k(g)=gkg^{-1}k^{-1}$. The multiplication and inversion maps of the Lie group $\widetilde G$ are continuous, so $c_k$ is continuous as a [composition of continuous maps](/theorems/4960). For every $g\in\widetilde G$, using that $p$ is a homomorphism and that $p(k)=e_G$, we compute $p(c_k(g))=p(g)p(k)p(g)^{-1}p(k)^{-1}$. Therefore $p(c_k(g))=p(g)e_Gp(g)^{-1}e_G=e_G$. Thus $c_k(g)\in\ker p$ for every $g\in\widetilde G$. Hence the same formula defines a continuous map $c_k:\widetilde G\to\ker p$ into the subspace $\ker p$, again given by $c_k(g)=gkg^{-1}k^{-1}$. [guided] Fix an element $k\in\ker p$. To prove that $k$ is central, we need to show that it commutes with every $g\in\widetilde G$. The expression that detects the failure of commutation is the commutator $gkg^{-1}k^{-1}$. It equals $e_{\widetilde G}$ exactly when $gk=kg$. We therefore define $c_k:\widetilde G\to\widetilde G$ by $c_k(g)=gkg^{-1}k^{-1}$. This map is continuous because it is built from the continuous multiplication and inversion operations in the Lie group $\widetilde G$, while $k$ and $k^{-1}$ are fixed elements. The important point is that $c_k$ actually lands in the kernel. For any $g\in\widetilde G$, the homomorphism property of $p$ gives $p(c_k(g))=p(g)p(k)p(g)^{-1}p(k)^{-1}$. Since $k\in\ker p$, we have $p(k)=e_G$, and hence also $p(k)^{-1}=e_G$. Substituting this into the preceding identity gives $p(c_k(g))=p(g)e_Gp(g)^{-1}e_G=e_G$. Thus $c_k(g)\in p^{-1}(\{e_G\})=\ker p$ for every $g\in\widetilde G$. Consequently we may regard $c_k$ as a continuous map $c_k:\widetilde G\to\ker p$, still given by $c_k(g)=gkg^{-1}k^{-1}$, where $\ker p$ carries its subspace topology from $\widetilde G$. [/guided] [/step] [step:Use connectedness and discreteness to force the commutator map to be constant] The image $c_k(\widetilde G)\subset\ker p$ is connected. Indeed, if it were separated by two nonempty disjoint open subsets in the subspace topology, their inverse images under the continuous map $c_k:\widetilde G\to c_k(\widetilde G)$ would separate the [connected space](/page/Connected%20Space) $\widetilde G$. Since $\ker p$ is discrete, every subset of $\ker p$ is discrete in the subspace topology. A connected discrete space has at most one point: if it contained distinct points $a$ and $b$, then $\{a\}$ and its complement would be nonempty disjoint open subsets separating it. Hence $c_k(\widetilde G)$ is a singleton. [/step] [step:Evaluate at the identity to prove centrality] Since $c_k(\widetilde G)$ is a singleton, it equals the value of $c_k$ at the identity element $e_{\widetilde G}$. We compute $c_k(e_{\widetilde G})=e_{\widetilde G}ke_{\widetilde G}^{-1}k^{-1}=kk^{-1}=e_{\widetilde G}$. Therefore $c_k(g)=e_{\widetilde G}$ for every $g\in\widetilde G$. Expanding the definition of $c_k$ gives $gkg^{-1}k^{-1}=e_{\widetilde G}$. Multiplying on the right by $kg$ gives $gk=kg$. Thus the fixed element $k\in\ker p$ commutes with every element $g\in\widetilde G$. Since $k\in\ker p$ was arbitrary, $\ker p\subset Z(\widetilde G)$. Together with discreteness from the first step, this proves that $\ker p$ is a discrete central subgroup of $\widetilde G$. [/step]