If $f_n:X\to Y$ is a sequence of homeomorphism between two metric spaces $X$ and $Y$ which converges locally uniformly to $f:X\to Y$, then $f$ is also homeomorphism?
soumik: i have no idea, but part of it might be - you feel slightly more in control with your dominant hand free. your non-dominant hand might as well be steering the bike because if you needed to quickly react to something with one hand, you'd probably want it to be your dominant one.
@Thorgott I would change $\Sigma_2\times S^1$ to a mapping torus $M$ of $\Sigma_2$ and further require $\Sigma$ is transverse to the suspension flow of $M$.
Otherwise, $\Sigma_2\times\{1/2\}\cup\Sigma_2\times\{-1/2\}/\sim$ by connected sum along a round disk on $\Sigma_2$ whose boundary is nulhomotopic on $\Sigma_2$ is transverse to $S$ but embedded in $\Sigma_2\times (-1,1)$.
Requiring transversality with the suspension flow abandons this.
Also found a fact that could be helpful: $M$ has a natural taut foliation $\mathcal{F}$ with leaves $\Sigma_2$. In this case, if $\Sigma$ is incompressible, meaning $\pi_1(\Sigma)$ injects to $\pi_1(M)$, then $\Sigma$ can either be homotoped to a leaf or can be homotoped to intersect $\mathcal{F}$ in only saddle tangencies.
@onepotatotwopotato Limits of homeomorphisms are hard to be made into homeomorphisms, you can have piece-wise linear mappings for $X = Y = [0, 1]$ and the uniform limit not a homeomorphism
maybe better to describe this pictorially, basically you have $f_n$ and $f$ piece-wise linear on $[0, 1/3], [1/3, 2/3]$ and $[2/3, 1]$
$f(0) = f_n(0)= 0 $ and $f(1) = f_n(1) = 1$
$f(1/3) = 1/2$ and say $f_n(1/3) = (2-1/n)/6$ and $f(2/3) = 1/2$ and say $f_n(1/3) = (2+1/n)/6$
then $f_n$ converges uniformly to $f$, each $f_n$ is a homeomorphism but $f$ fails to be injective, namely on $[1/3, 2/3]$ where its constant
sometimes you can assure its a homeomorphism
the weird $G_{1/i}$ are maps which are surjective but not $1/i$-maps (i.e. fibers have diameters $\leq 1/i$
