In the book Introduction to Teichmüller Spaces, by Imayoshi & Taniguchi, we finde the following definition of the Teichmüller space of a Riemann surface $R$, denoted $T(R)$:

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I want to draw attention for the term homotopic in this definition. I have tried to show that such relation as defined above is an equivalence relation (as it should be). I could show it is reflexive and transitive, but I could not to show it is symetric. What I have done:

Suppose $(S_1,f_1)\sim (S_2,f_2)$. Then, exists a conformal mapping $f:S_1\to S_2$ and a continuous fuction (this is the homotopy) $H:[0,1]\times S_1\to S_2$ such that $H_0=f_2\circ f_1^{-1}$ and $H_1=f$, where $H_t(p)=H(t,p)$, $p\in S_1$. Then, my first (and, I think, the most natural) thought was trying to show that $f_1\circ f_2^{-1}:S_2\to S_1$ is homotopic to $f^{-1}:S_2\to S_1$ (which is also a conformal mapping). If I do so, then I would have $(S_2,f_2)\sim (S_1,f_1)$.

I wish I could write $K:[0,1]\times S_2\to S_1$ as $K_t=H_t^{-1}$ because I would have $K_0=H_0^{-1}=f_1\circ f_2^{-1}$ and $K_1=H_1^{-1}=f^{-1}$. But I can't do that since I don't know if the inverse $H_t^{-1}$ exists, for all $t\in [0,1]$. I mean, I don't know what happens "along the process" of the homotopy $H$... Even I could write it, I still would have to show that $K$ is a homotopy.

Then I have started to think:

1) Is the

"If a map $g$ is homotopic to a map $h$ then $g^{-1}$ is homotopic to $h^{-1}$"

result TRUE and maybe a basic fact in homotopy theory?; and

2) Instead of HOMOTOPY at that definition, could not be ISOTOPY, maybe? Or asking for $H_t$ be quasiconformal, for all $t\in [0,1]$, anything like this?

Anyway, I need to show that such relation is an equivalence relation: HOW DO I SHOW THAT THE RELATION IS SYMMETRIC?

Thank you!

  • 1
    $\begingroup$ I know (but could not give you a reference right now) that you can take "homotopic" or "isotopic" in the two definitions and you would get the same thing. I believe the proof of the equivalence is not trivial and probably hard to find in the literature $\endgroup$
    – Albert
    Oct 4, 2016 at 14:30
  • $\begingroup$ Thank you, @Glougloubarbaki! $\endgroup$
    – Derso
    Oct 4, 2016 at 15:53

1 Answer 1


I got an answer here. It's pretty simple and indeed 1) holds:

Consider topological spaces $X$ and $Y$, two homotopic homeomorphisms $f,g:X\to Y$ and $I=[0,1]\subset \mathbb{R}$. Let $H:I\times X\to Y$ be the homotopy with $H(0,x)=f(x)$ and $H(1,x)=g(x)$. Then define $$\begin{array}{llll}K:&I\times Y&\to&X\\ &(t,y)&\mapsto &f^{-1}\big(H(1-t,g^{-1}(y))\big)\end{array}.$$

This is clearly continuous, $K(0,y)=f^{-1}(y)$ and $K(1,y)=g^{-1}(y)$.

All the credit to John Rognes, author of the answer.

  • $\begingroup$ all right, I guess I misread your question 1). but is it obvious that if $f$ and à$g$ are homotopic then they are isotopic? also in the definition of Teichmüller space I usually use one is requesting that the isotopy consists of K qc homeos, with K > 1 uniform in t. It seems not obvious to me that this is equivalent to homotopic, but sadly this is glossed over by most authors $\endgroup$
    – Albert
    Oct 10, 2016 at 9:12
  • $\begingroup$ @Glougloubarbaki In fact isotopic is equivalent to homotopic, in this context. Obviously this is not trivial. I did not see the details of the proof yet, but, if I get it right, there is a proof of this fact in the Hubbard's Teichmüller Theory, page 258, in the Proposition 6.4.9. $\endgroup$
    – Derso
    Oct 15, 2016 at 15:25

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