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A surface is a two-dimensional topological manifold. The term can also be used to describe a smooth surface, depending on the context.

4 votes
2 answers
132 views

Characterization of a non-trivial non-peripheral element of the free homotopy classes of a c...

Let $\Sigma$ be a compact orientable connected $2$-manifold with a non-empty boundary. Let $\widehat \pi(\Sigma)$ denote the set of free homotopy classes of curves in $\Sigma$. We say $x\in \widehat \ …
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  • 1,097
3 votes
1 answer
234 views

Classification of degree one map between two closed orientable surfaces without using induct...

A theorem of Edmonds (see Theorem 3.1. of "Deformation of Maps to Branched Coverings in Dimension Two") says that Theorem 1: A degree-one map between closed orientable surfaces is homotopic to a pinch … Theorem 2: A degree-one map between closed orientable surfaces, when it induces an injective map between the fundamental groups, is homotopic to a homeomorphism. …
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  • 1,097
3 votes
1 answer
114 views

A closed curve can be homotopic to remove all intersections with a filling $\Gamma$ if it ha...

Let $\Sigma$ be a compact oriented connected bordered surface other than the pair of pants. Let $\Gamma:=\{\gamma_i\}$ be a finite collection of simple closed curves on $\Sigma$ such that each compone …
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  • 1,097
4 votes
1 answer
184 views

Inequivalent free $\Bbb Z/n\Bbb Z$-actions on orientable compact bordered surface

Let $S_{g,b}$ denote the orientable connected compact surface of genus $g$ with $b$ boundary components. A group homomorphism $\varphi\colon G\to \text{Homeo}^+(S_{g,b})$ is said to be free $G$-acti …
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  • 1,097
2 votes
1 answer
149 views

Isotopic homeomorphisms of surface induces same map on the space of ends

Let $\Sigma$ be a non-compact orientable connected two-manifold without boundary. Let $f,g\colon \Sigma\to \Sigma$ be two homeomorphisms. Suppose there is a homotopy $H\colon \Sigma\times [0,1]\to \Si …
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  • 1,097
5 votes
0 answers
130 views

Earliest known proof of "Any degree one self-map of an orientable connected finite-type non-...

I attended a talk where the speaker said the following is due to Nielsen. I searched here and there but couldn't find the corresponding paper, if any. So, what is the earliest known proof of the follo …
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  • 1,097
2 votes
0 answers
104 views

Vanishing of Goldman bracket requires simple-closed representative?

Let $\Sigma$ be a connected oriented surface, and $[-,-]\colon \Bbb Z\big[\widehat\pi(\Sigma)\big]\times Z\big[\widehat\pi(\Sigma)\big]\to Z\big[\widehat\pi(\Sigma)\big]$ be the Goldman Bracket. Note …
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