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!

  • $\begingroup$ A basic theorem of Epstein (refining earlier work of many people, eg. Baer) says that with only a few exceptions homotopic homeomorphisms surfaces are isotopic. See Epstein, D. B. A. Curves on 2-manifolds and isotopies. Acta Math. 115 1966 83–107. $\endgroup$ Sep 29, 2016 at 2:05
  • $\begingroup$ A simple fix is to replace "homotopic" by the equivalent condition "induce the same isomorphism of fundamental groups" (up to conjugation). With this in mind, the statement reduces to the obvious fact that if an isomorphism $\phi$ of fundamental groups is induced by a conformal map, the inverse of the conformal map induces the inverse isomorphism of fundamental groups. $\endgroup$
    – Misha
    Sep 29, 2016 at 13:34

1 Answer 1


If $H$ is a homotopy from $f$ to $g$, with $f$ and $g$ invertible, then $f^{-1} H g^{-1}$ is a homotopy from $g^{-1}$ to $f^{-1}$.

  • $\begingroup$ thank you very much. Please, if you also has an account at Math Stack Exchange and if you want, post your answer here math.stackexchange.com/questions/1940853/…, so you can win the bount! $\endgroup$
    – Derso
    Oct 5, 2016 at 0:51
  • $\begingroup$ I've posted your answer there, but you can just copy and paste my answer as yours and I'll delete my own, and give you the bounty (+100 of reputation!) $\endgroup$
    – Derso
    Oct 5, 2016 at 1:24

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