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In geometric topology, what does it mean for a curve to be homotopic to a puncture?

Let $F$ be a hyperbolic surface of finite type, i.e. $F$ is a surface of genus $g$ with $b$ boundary components and $n$ punctures such that, $2-2g-b-n< 0.$ Let $\pi_1(F)$ be the fundamental group of $F$. We identify $\pi_1(F)$ with a discrete subgroup of $PSL_2(R)$, the group of orientation preserving isometries of the upper half plane $H$. The action of $\pi_1(F)$ on $H$ is properly discontinuous and does not fix any point. Therefore the quotient space is isometric to $F.$ Henceforth by an isometry of $H$, we mean an orientation preserving isometry and by a closed curve we mean an oriented close curve.

A homotopically non-trivial closed curve in $F$ is called essential if it is not homotopic to a puncture. By a lift of a closed curve $g$ to $H$, we mean the image of a lift $R\rightarrow H$ of the map $g\circ\pi$ where $\pi:R\rightarrow S^1$ is the usual covering map.

[1] http://arxiv.org/abs/1609.05891

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OregonTrail
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  • A curve is (say) a $1$-dimensional subspace that probably has some reasonably nice embedding; a puncture is a point removed from a space. What would it mean for one to be homotopic to the other? Are you just talking about a null-homotopic path? – anomaly Sep 23 '16 at 06:02
  • In the context of the question, a curve most likely means a map $S^1 \mapsto F$ with no properties other than continuity. @anomaly – Lee Mosher Sep 23 '16 at 16:45
  • @LeeMosher: Probably, yeah. When I originally posted the comment, the shaded box hadn't been added. – anomaly Sep 23 '16 at 17:55

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For a curve $g : S^1 \to F$ to be homotopic to a puncture means that there exists a closed subset $A \subset F$ homeomorphic to $\mathbb{D}^2 - \{(0,0)\}$ such that $g$ is homotopic in $F$ to a curve in $A$. Here $\mathbb{D}^2 \subset \mathbb{R}^2$ denotes the closed unit disc.

If $F$ is explicitly given as $F = \overline F-P$ where $\overline F$ is a compact surface-with-boundary and $P$ is a finite subset of the interior of $F$, then $g$ is homotopic to a puncture if and only if there exists $p \in P$ such that for every neighborhood $U \subset \overline F$ of $p$ the curve $g$ is homotopic in $F$ to a curve in $U-p$.

Lee Mosher
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  • I appreciate it – OregonTrail Sep 24 '16 at 13:13
  • @LeeMosher I think the second paragraph of your answer is not equivalent to the following: There is a homotopy $H:\Bbb S^1\times [0,1]\to \overline F$ such that $H(z,0)=g(z)$ and $H(z,1)=p$ for all $z\in \Bbb S^1$. That is, $p\in P$ is thought as a marked point on the surface $F=\overline F-P$. Am I right? – Sumanta Feb 01 '21 at 16:35
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    No, it is not equivalent. That set of curves in your comment represents the kernel of the inclusion induced homomorphism $\pi_1(F) \to \pi_1(\overline F)$, which is much larger than just the curves which are homotopic to the pucture: that kernel is the normal closure in $\pi_1(F)$ of elements represented by those curves. – Lee Mosher Feb 01 '21 at 17:26