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Suppose we have a torus $S^1 \times S^1$ (or any other orientable surface) with non empty boundary (meaning, we have holes on the surface). If $a$ and $b$ are a meridian and a parallel respectively then by performing a sequence of Dehn twists along $a$ and $b$ and homeomorphisms isotopic to the identity we can move $a$ on to $b$. I know Dehn twists are supported on an annulus, that means that they will fix the boundary. Is that true for the homeomorphisms that are isotopic to the identity as well. For example, is there a way to slide things on the surface without moving the boundary? I can see it intuitevely but I haven't found a way to rigorously explain it (to myself). Can you please help me.

EDIT. I'll try and be more clear. In the picture it's clear that there is an isotopy that can slide the curve on to a meridian of the Torus. Is there a way to make that isotopic slide while keeping the boundary (black hole) fixed?

enter image description here

Amontillado
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  • You say "it's clear that there is an isotopy". If it's that clear to you then you should be able to write down an actual formula for the isotopy. Can you do that and add it to your question? I ask because if that formula does not already keep the black hole fixed, it should be easy for us to tell you how to alter it to get an isotopy that keeps the black hole fixed. – Lee Mosher Oct 17 '19 at 18:21
  • Isn't sliding the curve an isotopy? That's what I meant by clear. If it's not, then I'm much more in the dark than I thought I was. – Amontillado Oct 17 '19 at 18:28
  • Isotopies are functions, and functions are given by formulas. Yes, we learn to think about isotopies using intuitions like "sliding", and those intuitions can be useful. But for a question like this one, if you can formalize your isotopy as an actual function, the answer should be pretty easy. – Lee Mosher Oct 17 '19 at 18:32
  • Perhaps I can move the curve inside a disk that is away from the hole while keeping the boundary of the disk fixed? – Amontillado Oct 17 '19 at 18:33
  • I'm really sorry I don't think I can write such a formula explicitly, can you please help me? – Amontillado Oct 17 '19 at 18:53
  • @LeeMosher I have given it some thought and if it doesn't fix the boundary component, i.e. rotate it by $θ$, then I think I can rotate back the boundary with an isotopy in a annular neighbourhood of the boundary, while keeping the rest of the surface fixed. Will $(xe^{iθts},t)$ for $s,t \in [0,1]$ work when $(x,t) \in S^1 \times [0,1]$ is the annular neighbourhood? – Amontillado Oct 18 '19 at 08:27
  • That sounds like a good plan. – Lee Mosher Oct 18 '19 at 13:20
  • Thank you very much for you gave me the idea of moving the boundary back into place. – Amontillado Oct 18 '19 at 13:24

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