I am preparing for my topology exam and going through some exercises in Munkres. This is section 79 problem 3:
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 $h(e_1) = e_2$.
For anyone not familiar with Munkres' notation:
- $p_*$ stands for induced homomorphism and for any pointed continuous map $h: (X, x_0) \to (Y, y_0)$ carrying $x_0$ to $y_0$, it is defined as map: $$h_∗ : \pi_1(X, x_0) \to \pi_1(Y, y_0)$$
such that:
$$h_∗([ f ]) = [h \circ f ]$$ - What he refers to as equivalence is just a deck or a covering transformation, i.e. a homeomorphism $h: E \to E$ such that $p = p \circ h$.
I am interested only in the backward direction of the proof i.e. showing that the subgroup is normal under the given assumptions. However I don't really know where to start from and how to use this condition of covering transformation that was given to me.
If I unpacked the definitions correctly, I think I have to obtain that for any $[f] \in \pi_1(E, e_0)$ and any $[g] \in H_0$ the following holds:
$$[g] * [p \circ f] = [p \circ f] * [g]$$
I assume that at some point I will have to lift these loops in $B$ to paths in $E$ and then somehow use the given condition but I am not sure how.
Any help is appreciated.
Best Answer
You have to show that for all $\gamma \in \pi_1(B,b_0)$ and all $\chi \in H_0$ one has $\gamma ^{-1} \chi \gamma \in H_0$ (not $\gamma \chi = \chi \gamma$ as you write).
With your notation this means $$[g]^{-1} * [p \circ f] * [g] \in H_0. $$
Let us lift the loop $g^{-1} * (p \circ f) * g$ in $B$ to $E$. Take the unique lift $\tilde g$ of $g$ such that $\tilde g(0) = e_0$. This is a path in $E$, but not necessarily a loop in $E$. Let $e_1 = \tilde g(1)$. The path $$F = \tilde g^{-1} * f * \tilde g$$ is a loop based at $e_1$.
Let $h : E \to E$ be a deck transformation such that $h(e_1) = e_0$. Then $h \circ F$ is a loop based at $e_0$, i.e. $[h \circ F] \in \pi_1(E,e_0)$. We get $$p_*([h \circ F]) = [p \circ h \circ F] = [p \circ F] = [(p \circ \tilde g^{-1}) * (p \circ f) * (p \circ \tilde g)] = [g^{-1} * (p \circ f) * g] \\= [g]^{-1} * [p \circ f] * [g]$$ which means that $$ [g]^{-1} * [p \circ f] * [g] \in H_0 .$$