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Definition: A marking on a sphere $S$ with $g$ handles ($g<\infty$) is a ordered collection $\Sigma_p=\{[\alpha_j],[\beta_j]\}_{j=1}^g$ of elements $[\alpha_j],[\beta_j]$ all of them in a same fundamental group $\pi_1(p,S),p\in S$ such that $\Sigma_p$ generates $\pi_1(p,S)$ and $\prod\limits_{j=1}^g[[\alpha_j],[\beta_j]]=1$, where $[[\alpha_j],[\beta_j]]$ denotes the comutator of $[\alpha_j]$ and $[\beta_j]$ (namely, $[\alpha_j][\beta_j][\alpha_j]^{-1}[\beta_j]^{-1}$).

Two markings $\Sigma_p=\{[\alpha_j],[\beta_j]\}_{j=1}^g$ and $\Sigma_{p'}=\{[\alpha'_j],[\beta'_j]\}_{j=1}^g$ on a sphere with $g$ handles are said to be equivalent, if exists a path $\gamma:[0,1]\to S$ joining $p$ to $p'$ such that the isomorphism $\tilde{\gamma}:\pi_1(p,S)\to \pi_1(p',S)$ induced by $\gamma$ (namely $\tilde{\gamma}([\alpha])=[\gamma^{(-1)}\cdot \alpha\cdot \gamma]$) is such that $\tilde{\gamma}([\alpha_j])=[\alpha'_j]$ and $\tilde{\gamma}([\beta_j])=[\beta'_j]$.

One can easily show that, if $\Sigma_p=\{[\alpha_j],[\beta_j]\}_{j=1}^g$ is a marking on $S$ and $f:S\to R$ is a homeomorphism, then $f_\star(\Sigma_p):=\{f_\star([\alpha_j]),f_\star([\beta_j])\}_{j=1}^g$ is a marking on $R$, with base point $f(p)$ (here, $f_\star:\pi_1(p,S)\to \pi_1(f(p),S)$ is the isomorphism induced by $f$).

I would like the following to be true:

Let $S$ be a sphere with $g$ handles and $\Sigma_p=\{[\alpha_j],[\beta_j]\}_{j=1}^g$ a marking on $S$. Then, if $f,h:S\to R$ are homotopic homeomorphisms, then the markings $f_\star(\Sigma_p)$ and $h_\star(\Sigma_p)$ on $R$ are equivalent.

Being $H:[0,1]\times R\to S$ the homotopy from $f$ to $h$, I think I should take (the natural choice) $\gamma:[0,1]\to S$, $\gamma(t)=H(t,p)$ as being the path joining $f(p)$ to $h(p)$ in order to get the equivalence between the markings. I've struggled with it without being successful so far...

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    You are on the right track. Try to construct the required homotopy of loops using $H$. – 2017-01-25
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    Try to think what will happend if $S=R$ and one of the map is identity...You are on right track. – 2017-01-25
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    Sorry @Anubhav. I couldn't see it, even doing $S=R$ and one of the maps the identity, namely, $h$. Then, I have to show that the marking $f_\star(\Sigma_p)=\{[f\circ \alpha_j],[f\circ \beta_j]\}_{j=1}^g$ is equivalent to $h_\star(\Sigma_p)=\Sigma_p$. Ok, in order to do this, I take $\gamma:[0,1]\to R$ as being $\gamma(t)=H(t,p)$. But then, for instance, $\tilde{\gamma}([f\circ \alpha_j])=[\gamma^{(-1)}\cdot (f\circ \alpha_j)\cdot \gamma]$: how do I show that this path is homotopic to $[\alpha_j]$? I don't see how taking such $\gamma$ can help... Can you give any hint? – 2017-01-25

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