So I have a rather embarrassing problem, which is not really a "problem", so much as a mental block I seem to be unable to overcome. I am trying to understand the "holonomy map" of a mapping torus. To keep things concrete I would like to understand the following basic example: Take a $\mathbb{T}^2$ bundle over $\mathbb{S}^1$. One way to get an example of such a fiber bundle is by a mapping torus construction: take the product an interval $I$ with the fiber $\mathbb{T}^2$ and "glue" the ends using an homeomorphism of $\mathbb{T}^2$, $f$, to get the mapping torus $M_f$.

$$M_f =\frac{(I \times \mathbb{T}^2)}{(1,x)\sim (0,f(x))}$$

$f$ is then called the *holonomy map* of the mapping torus. Now consider the following bundles: the trivial bundle $\mathbb{T}^2\times \mathbb{S}^1$ and the bundle given by "rotating" an $\mathbb{S}^1$ factor of the $\mathbb{T}^2$ $n$ times as we travel around the $\mathbb{S}^1$ base of the fibre bundle once i.e. the torus bundle with monodromy

$\left( \begin{array}{ccc} 1 & n \\ 0 & 1 \\ \end{array} \right)$

Question: what is the holonomy map for these basic examples? There should be a particular $f$ for each $n$ so that the mapping torus $M_f$ is associated to $f$ is the torus bundle by twisting $n$ times. But I don't know how a single homeomorphism of the torus can distinguish between the cases. How can a single homeomorphism capture that the torus has turned $n$ times? I mean, everytime you rotate $2\pi$ the map at the "end point" is back to being the identity. I hope I have made my problem clear, I am aware that it is a bit silly but I have been turning it around in my head for a while and still can't fully see what is happening.

(again, this was asked on stack exchange and I switched to here due to lack of response)