Assuming that $E$ and $B$ are path-connected spaces, let $p:(E,e)\rightarrow(B,b)$ be a cover map and $p_*:\pi_1(E,e)\rightarrow \pi_1(B,b)$ with $[\gamma]\mapsto[p\ \circ \gamma]$, the usual induced homomorphism. I'm trying to prove that if $p_*$ is a group isomorphism, then $E_b:=\{x\in E:p(x)=b\}$ is a singleton set. I couldn't make a reasonable attempt, I only proved the converse. If someone could help, I'd be grateful. Thank you!
If induced homomorphism is isomorphism, then $p^{-1}(b)$ is a singleton set
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algebraic-topology
homotopy-theory
covering-spaces
fundamental-groups
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0What do you mean by a "loop by $x$"? – 2017-01-19
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0Sorry, in fact this element isn't in the domain of $p_*$. I'am going to edit it. – 2017-01-19
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1Also, I assume $E$ and $B$ are path-connected? – 2017-01-19
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0Yes! Sorry again, I forgot it. – 2017-01-19
2 Answers
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Let $p(e)=b$. Recall that for $H=p_*(\pi_1(E,e))\subseteq \pi_1(B,b)$ we have a bijection $$\varphi:\pi_1(B,b)/H\to p^{-1}(b),\quad \varphi([\omega]H) = \tilde{\omega}(1)$$ where $\tilde{\omega}(1)$ is the endpoint of the unique lift of $\omega$ starting at $e$. Hence, $$[\pi_1(B,b):p_*(\pi_1(E,e))] = |p^{-1}(b)|.$$ What does this imply if $p_*$ is an isomorphism?
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Another proof proceeds of two facts that are not so complicated:
$(i)$ The function $f:E_b\times\pi_1(B,b)\rightarrow E_b$ given by $(x,[\gamma])\mapsto\widetilde{\gamma}(1)$, where $\widetilde{\gamma}$ is the lifting of $\gamma$, is a transitive action.
$(ii)$ In the action above, if $x\in E_b$, then $St(x)=p_*(\pi_1(E,e))$, where $St(x)$ is the stabilizer of $x$.
Then, let $x,y\in E_b$. As $p_*$ is surjective, $St(x)=\pi_1(B,b)$. By $(i)$, $(x,[\delta])\mapsto y$ for same $[\delta]\in\pi_1(B,b)$. But since $[\delta]\in St(x)$, $x=y$.