I was reading "A primer on Mapping Class Groups" by B. Farb and D. Margalit and I'm stuck at a point in the proof of Mod$(A)\cong \mathbb{Z}$ where $A$ is the annulus, where the restriction of $M$ on $\mathbb{R}\times\{1\}$ descends to a homeomorphism. I can't figure out how.
The question is as follows, Let $M$ be a $2\times 2$ real matrix defining a linear transformation from $\mathbb{R}^2$ to itself. If we restrict it to the universal cover of the annulus $(\mathbb{R}\times [0,1])$ such that it is equivariant under the group of deck transformations $(\mathbb{Z})$ then this restriction descends to a homeomorphism of the annulus.
In this case the matrix is$$ M= \left[ {\begin{array}{cc} 1 & n \\ 0 & 1 \\ \end{array} } \right] $$ I'm unable to prove this fact.
Thanks in advance.
Let $\pi:\mathbf R\times [0,1]\to A$ be the covering map. If $x\in A$, you can pick a $y\in \pi^{-1}(x)$. Thanks to the equivariance property, $f(x) = \pi(My)$ does not depend on the choice of $y$. Local triviality gives you the continuity of $f$.
In our case we can do the same things with $M^{-1}$, so this is an homeomorphism (I don't know what is needed in general).