Let \(p: E \to B\) be a covering map; let \(p(e_0) = b_0\). Show that \(H_0 = p_*(\pi_1(E, e_0))\) is a normal subgroup of \(\pi_1(B, b_0)\) if and only if for every pair of points \(e_1, e_2\) of \(p^{-1}(b_0)\), there is an equivalence \(h: E \to E\) with \(h(e_1) = e_2\).
Solution
Assume \(E\) is connected and locally path connected, as in the covering-space classification theorem. Fix \(e_0\in p^{-1}(b_0)\) and put \[ H_0=p_*\pi_1(E,e_0)\le G:=\pi_1(B,b_0). \] For any \(e\in p^{-1}(b_0)\), choose a path \(\lambda\) in \(E\) from \(e_0\) to \(e\), and let \[ g=[p\lambda]\in G. \] Changing basepoint along \(\lambda\) gives \[ p_*\pi_1(E,e)=g^{-1}H_0g \] (up to the consistent convention for concatenation; only conjugacy matters below).
Suppose first that \(H_0\triangleleft G\). Then for every \(e\in p^{-1}(b_0)\), \[ p_*\pi_1(E,e)=H_0. \] Given \(e_1,e_2\in p^{-1}(b_0)\), the two pointed coverings \((E,e_1)\to(B,b_0)\) and \((E,e_2)\to(B,b_0)\) therefore have the same associated subgroup. By the equivalence theorem for pointed connected coverings, there is a covering equivalence \[ h:E\to E,\qquad p\circ h=p,\qquad h(e_1)=e_2. \]
Conversely, suppose that for every pair \(e_1,e_2\) in the fiber there is such an equivalence. For any \(g\in G\), lift a representative loop at \(b_0\) beginning at \(e_0\), and let its endpoint be \(e_g\). The basepoint-change formula gives \[ p_*\pi_1(E,e_g)=g^{-1}H_0g. \] By hypothesis there is an equivalence \(h:E\to E\) with \(h(e_0)=e_g\). Since \(p h=p\), \[ p_*\pi_1(E,e_g) =p_*h_*\pi_1(E,e_0) =p_*\pi_1(E,e_0)=H_0. \] Thus \(g^{-1}H_0g=H_0\) for every \(g\in G\). Hence \(H_0\) is normal.