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I have some questions from the following text:

enter image description here

$\bullet$ It is clear to me that $f_\alpha$ is a loop and represents a generator of $\pi_1(M_\alpha)$ as covering induces injective map on fundamental groups, but why $f_\alpha$ is shortest?

Recall that a loop on a surface equipped with a Riemannian metric is called shortest if it has the shortest length in its free homotopy class.

$\bullet$ I think here $l:\Bbb R\to \widetilde M$ is obtained as the lifting of the following composition map $\Bbb R\xrightarrow{\exp}\Bbb S^1\xrightarrow{f_\alpha} M$ w.r.t. the universal covering $p:\widetilde M\to M$ using the fact $\Bbb R$ is simply-connected. Now, $l$ is simple as $f_\alpha$ is simple. Am I right?

$\bullet$ Why the map $l:\Bbb R\to \widetilde M$ is properly embedded in $\widetilde M$? I don't know the meaning of "proper" here. I think the term embedding comes to form the fact that $l$ is simple.

$\bullet$ What will be the full pre-image of $f_\alpha(\Bbb S^1)$ in universal cover i.e. what is $p^{-1}\big(f_\alpha(\Bbb S^1)\big)$? I guess it will be $\pi_1(M)$-tanslates of $l$, but have no proof.

Any help will be appreciated. It will be better if someone tells me some text containing all these basic facts.

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    $\begingroup$ Seems we might need to know lemma 5.2.1. Can you give a reference for this (book and location)? $\endgroup$
    – Steve D
    Commented Jan 17, 2021 at 19:00
  • $\begingroup$ Lemma 5.2.1 Let $A$ be an open annulus with a Riemannian metric. Let $\gamma$ be a shortest loop on $A$ representing a generator of $π_1(A).$ Then, $\gamma$ is simple. Also, if $\gamma'$ a second such loop on $A$, then $\gamma'$ equals $\gamma$ or is disjoint from $\gamma$. This is on page 101. $\endgroup$
    – Someone
    Commented Jan 17, 2021 at 19:14

1 Answer 1

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In the first bullet, suppose that $M_\alpha$ contained a loop $f'_\alpha : \mathcal S^1 \to M_\alpha$ in the same free homotopy class as the lift $f_\alpha$ but with length shorter than $f_\alpha$. Then the loop $f' : \mathcal S^1 \to M$ obtained by composing $f'_\alpha$ with the covering map $M_\alpha \mapsto M$ would be a shorter loop in the same homotopy class as $f$, which is a contradiction (the homotopy from $f_\alpha$ to $f'_\alpha$, when composed with the covering map $M_\alpha \mapsto M$, would be a homotopy from $f$ to $f'$).

In the second bullet, some care is needed. First one needs to verify that $\alpha$ has infinite order, but this is true in the fundamental group of any oriented surface: every nontrivial element has infintie order. Also, however, your interpretation does not quite make sense because the statement says that $l$ is the preimage of $f_\alpha(S^1)$, so $l$ itself is a set, not a function. However, if instead you state your interpretation to say that $l$ is the image in $\widetilde M$ of the map $\mathbb R \mapsto \widetilde M$ obtained by lifting the map $\mathbb R \mapsto \mathbb S^1 \mapsto M$, then your interpretation does make sense and is correct.

In your third bullet, the meaning of "properly embedded" is that the parameterization $\mathbb R \mapsto \widetilde M$ of $l$ is a proper function, meaning one for which the inverse image of every compact subset of $\widetilde M$ is a compact subset of $\mathbb R$. Intuitively this means that as points in $\mathbb R$ go to $-\infty$ or to $+\infty$, their images in $\widetilde M$ also "go to infinity".

In your fourth bullet, your guess is correct, if you restate it slightly: the full preimage will be the union of the $\pi_1(M)$ translates of $l$.

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  • $\begingroup$ Thank you for your answer. Just for clarification: So, in the first bullet you are using covering is a local isometry. In the second and third bullets, a parameterization $\Bbb R\to \widetilde M$ of $l$ is simple as $f_\alpha$ is simple, and that is why the text says $l$ is embedded in $\widetilde M$. Am I right? $\endgroup$
    – Someone
    Commented Jan 18, 2021 at 4:07
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    $\begingroup$ That sounds right $\endgroup$
    – Lee Mosher
    Commented Jan 18, 2021 at 4:29
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    $\begingroup$ except that showing $l$ is embedded in $\widetilde M$ does make use of the fact that $\alpha$ is of infinite order. $\endgroup$
    – Lee Mosher
    Commented Jan 18, 2021 at 4:30
  • $\begingroup$ Can you elaborate on your last comment whenever you have time? or any reference will also be helpful. $\endgroup$
    – Someone
    Commented Jan 18, 2021 at 5:01
  • $\begingroup$ I think it follows from the lemma 5.2.1(as stated above) as the image of $l$ in $M_\alpha$ that is $\text{im}(f_\alpha)$, represents a generator of $\pi_1(M_\alpha)$. Am I right? $\endgroup$
    – Someone
    Commented Jan 18, 2021 at 5:07

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