Let $Τ^2=S^1\times S^1$ be the standard torus, $f:T^2 \to T^2$ an orientation preserving homeomorphism that is the identity on a meridian $m$ and $A$ an $ε$-neighbourhood of $m$.
My question is, if we cut $f(T^2)$ along $f(m)$ do points in $A$ on the left of $m$ are still mapped on the left of $f(m)$?
I believe they do but I can't rigorously see why. If so, can someone please explain why?
EDIT: I have given it some thought and came up with something, I would like to check if it's correct.
Let's say, without loss of generality that a cylindrical neighbourhood of $m$ is mapped homeomorphically to another cylindrical neighbouhood of $m$. For connectivity reasons the left region will either be mapped homeomorphically on the left or right region of the image. Let's say it's mapped on the right region, then the right side will be mapped on the left side. Let's take a circle inside our neighbourhood crossing $m$ on points $a$ and $b$ and choose points $p$ and $q$ on our circle, $p$ on the left region and $q$ on the right region. If $\overrightarrow {pq}$ is the arc containing $a$ then it will be mapped to the arc $\overleftarrow {f(q)f(p)}$ that contains $a=f(a)$. But that would contradict $f$ being orientation preserving.