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Let p:E→Bp: E \to Bp:E→B be a Covering, with p(e0)=b0p(e_0) = b_0p(e0)=b0. Then any path f:[0,1]→Bf: [0,1] \to Bf:[0,1]→B, with initial f(0)=b0f(0) = b_0f(0)=b0, has a unique lifting to a path f~:[0,1]→E\tilde{f}: [0,1] \to Ef~:[0,1]→E such that f~(0)=e0\tilde{f}(0) = e_0f~(0)=e0.
Proof.