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Let $p\ \colon \widetilde X \longrightarrow X$ be a covering map such that any loop $\gamma$ in $X$ based at $x_0 \in X$ lifts uniquely to a loop based at $\widetilde {x_0} \in p^{-1} (\{x_0\}).$ Assume that $X$ is path connected. For any $x \in X,$ take a path $\gamma$ joining $x_0$ and $x$ (starting at $x_0$ and ending at $x$) and define a map $s : X \longrightarrow \widetilde X$ by $s (x) = \widetilde {\gamma}_{\widetilde {x_0}} (1),$ where $\widetilde {\gamma}_{\widetilde {x_0}}$ is the unique lift of $\gamma$ starting at $\widetilde {x_0}.$ Show that $s$ is well-defined.

My Attempt $:$ Let $x \in X$ be arbitrarily taken. Let $\gamma$ and $\delta$ be two paths in X starting at $x_0$ and ending at $x.$ In order to show that $s$ is well-defined we need only to show that $\widetilde {\gamma}_{\widetilde {x_0}} (1) = \widetilde {\delta}_{\widetilde {x_0}} (1),$ where $\widetilde {\gamma}_{\widetilde {x_0}}$ and $\widetilde {\delta}_{\widetilde {x_0}}$ are the unique lifts of $\gamma$ and $\delta$ to $\widetilde X$ starting at $\widetilde {x_0}.$ For that, consider the loop $\delta \ast \overline {\gamma}$ based at $x_0,$ where $\overline {\gamma}$ is the opposite path traversed by that of $\gamma$ i.e. $\widetilde {\gamma} (t) = \gamma (1 - t),$ $t \in [0,1].$ Let $\widetilde {{\delta} \ast \overline {\gamma}}_{\widetilde {x_0}}$ be the unique lift of $\delta \ast \overline {\gamma}$ starting at $\widetilde {x_0}.$ Then by the virtue of the unique path lifting lemma it follows that $\widetilde {{\delta} \ast \overline {\gamma}}_{\widetilde {x_0}} = \widetilde {\delta}_{\widetilde {x_0}} \ast \widetilde {\overline {\gamma}}_{\widetilde {\delta}_{\widetilde {x_0}} (1)}.$ By the given hypothesis this lift is loop based at $\widetilde {x_0}$ and hence $\widetilde {\overline {\gamma}}_{\widetilde {\delta}_{\widetilde {x_0}} (1)} (1) = \widetilde {x_0}.$ Does it help anyway? Any help would be highly appreciated.

Thanks for investing your valuable time on my question.

Paul Frost
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Akiro Kurosawa
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    I just want to acknowledge the work you did typesetting $\widetilde {\overline {\gamma}}{\widetilde {\delta}{\widetilde {x_0}} (1)} (1) = \widetilde {x_0}$. – JonathanZ Sep 22 '23 at 21:38

1 Answers1

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Your proof is almost there, you just need to observe that $\overline{(\widetilde{\bar \gamma}_{\widetilde{\delta}_{\widetilde{x_0}}(1)})}$ is a lift of $\gamma$ starting at $\widetilde{x_0}$ and therefore the unique lift of $\gamma$ starting at $\widetilde{x_0}$, ie $\widetilde{\gamma}_{\widetilde{x_0}}$.

ronno
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  • Oh! I see now. Thanks for your answer. Sorry for the late acceptance of the answer as I was asleep when you posted this answer. Another thing I just noticed is that $p \circ s = \text {id}{X}$ which implies that $s$ is injective and if we restrict $\widetilde {X}$ to the path component $\widetilde {X}{\widetilde {x_0}}$ containing $\widetilde {x_0}$ then $s : X \longrightarrow \widetilde {X}_{\widetilde {x_0}}$ is a bijective map. Can it be somehow shown that $s$ is continuous as well? – Akiro Kurosawa Sep 23 '23 at 04:53
  • I think for any open set $U \subseteq \widetilde {X}{\widetilde {x_0}}$ we have $s^{-1} (U) = p (U).$ The reason is the following $:$ The image of $s$ is $\widetilde X{\widetilde {x_0}}$ and $s$ is already injective. Therefore $s : X \longrightarrow \widetilde X_{\widetilde {x_0}}$ is bijective and since $p \circ s = \text {id}X$ it follows that $s^{-1} = p\ \big \rvert{\widetilde X_{\widetilde {x_0}}}.$ Now since $p$ is covering map it is open and hence for any open set $U \subseteq \widetilde X_{\widetilde {x_0}}$ we have...contnd – Akiro Kurosawa Sep 23 '23 at 06:55
  • $$\begin{align} p (U) & = p\ \big \rvert_{\widetilde X_{\widetilde {x_0}}} (U) \ & = s^{-1} (U) \end{align}$$ and therefore for any open subset $V \subseteq \widetilde X$ we should have $$\begin{align} s^{-1} (V) & = s^{-1} \left (V \cap \widetilde X_{\widetilde {x_0}} \right ) \ & = p \left (V \cap \widetilde X_{\widetilde {x_0}} \right ) \end{align}$$ which is open since $p$ is an open map. This shows that $s$ is continuous. – Akiro Kurosawa Sep 23 '23 at 07:13
  • N.B. $:$ Each path component of $\widetilde {X}$ is a clopen (closed as well as open) subset of $\widetilde X$ and hence given any open set $V \subseteq \widetilde {X}$ it follows that $V \cap \widetilde {X}_{\widetilde {x_0}}$ is also an open subset of $\widetilde X.$ – Akiro Kurosawa Sep 23 '23 at 07:21
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    This in turn implies that $\pi_1 (X, x_0) \cong \pi_1 \left (\widetilde {X}, s(x_0)\right ).$ – Akiro Kurosawa Sep 23 '23 at 07:32
  • How great it is. Believe me I haven't read it anywhere. I have discovered it myself. This is really a matter of great joy of exploration. Many many thanks for your help. – Akiro Kurosawa Sep 23 '23 at 07:34
  • It's not relevant though but let me tell you one thing which I can't refrain from doing. I have just browsed through your profile and found that you are a bengali and let me tell you that I am a bengali too. – Akiro Kurosawa Sep 23 '23 at 07:42
  • @AkiroKurosawa $s$ is in general not continuous. If you want, ask a new question concerning this issue. – Paul Frost Sep 23 '23 at 08:53
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    @ronno $p$ and $s$ are in general no homeomorphisms. This is only true under additonal assumptions (e.g. that $\tilde X$ is path-connected). – Paul Frost Sep 23 '23 at 15:14
  • @PaulFrost$:$ I get your point. I think the fact which could possibly go wrong here is that the path component of $\widetilde {x_0}$ in $\widetilde X$ i.e. $\widetilde X_{\widetilde {x_0}}$ is open which is not necessarily true unless $\widetilde X$ is locally path connected. – Akiro Kurosawa Sep 23 '23 at 15:29
  • @ronno$:$ I haven't thought about it yet. On the day after tomorrow we have an exam and I am preparing for the same. I would try to find one such after the exam gets over. Please don't mind. I think you would possibly have a counter-example in the meantime. – Akiro Kurosawa Sep 23 '23 at 21:17
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    @ronno Even if you only consider the path component of $\tilde x_0$ it is false. But that should not be discussed in comments. I suggest that somebody asks a new question. – Paul Frost Sep 23 '23 at 22:26
  • @ronno$:$ What will happen if we take topologists sine curve or quasi-circle and take the path component of that about $(0,0)\ $? – Akiro Kurosawa Sep 23 '23 at 22:31
  • @PaulFrost I see the problem with continuity of $s$ now, $p(U) \subseteq s^{-1}(U)$ is not clear. I still don't have an explicit counterexample but agree that that deserves to be a new question. – ronno Sep 24 '23 at 04:55
  • See https://math.stackexchange.com/q/4851864. – Paul Frost Jan 27 '24 at 23:35