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Let $A = [0,1] \times \mathbb{S}^1$ be the standard annulus and let $X$ be the space obtained from $A$ by identifying $\{0\} \times \mathbb{S}^1$ and $\{1\} \times \mathbb{S}^1$ by a map which represents twice the generator of $\pi_1(\mathbb{S}^1)$.

I am looking for a hint on how to find the $\Delta-$complex structure (cf Hatcher) of $X$.

24601
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1 Answers1

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The first hint is semantic: to write of the $\Delta$ complex structure on $X$ is an error. There are many different $\Delta$ complex structures on $X$, and your job is to construct just one of them.

And here's a hint to the construction: first choose vertices on $\partial A = \{0,1\} \times \mathbb S^1$ so that the identification map restricts to a two-to-one surjection from the chosen vertices on $\{0\} \times \mathbb S^1$ to the chosen vertices on $\{1\} \times \mathbb S^1$.

Lee Mosher
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  • Ah I think my confusion lies in the definition of the gluing map. I initially thought that this gluing map has the same effect as identifying the two boundary circles using the identity map, then applying (twice) Dehn twist around a closed curve ${p} \times \mathbb{S}^1$, for any $p \in [0,1]$. But your tip suggests to me that the actual effect is that of one of the boundary circles essentially wrapping twice around the other? – 24601 Jul 07 '19 at 14:25
  • I think that's the only sensible interpretation, although the wording is a little off. It would be better worded like this: "...by identifying ${0} \times \mathbb{S}^1$ and ${1} \times \mathbb{S}^1$ by a map from ${0} \times \mathbb{S}^1$ to ${1} \times \mathbb{S}^1$ which represents twice the generator of $\pi_1(\mathbb{S}^1)$. – Lee Mosher Jul 08 '19 at 01:26