Meaning of monodromy of mapping torus Let $f: X \to X$ a homeomorphism. The space $T(f):=(X\times I)/(0,f(x))\sim (1,x)$ is called the mapping torus of $f$.
Sometimes it is also called mapping torus with monodromy map $f$. 
Is this simply a definition or is there a concept of monodromy which actually makes $f$ a monodromy map?
If $X$ is discrete, then $T(f)$ is a cover of $S^1$ with fiber $X$ whose monodromy is $f$.
Is there still a notion of monodromy which makes sense for a more general case? For example, is there a canonical connection for fiber bundles over $S^1$ which makes $f$ a monodromy map?
 A: The structure of the mapping torus enforces a connection, namely: in $(x,t)$ coordinates on $X \times T$, you have the tangent vector $\frac{\partial}{\partial t}$ at every point. Projecting this tangent vector under the quotient map $X \times I \mapsto T(f)$ gives a vector field on $T(f)$ which is the connection in question. It follows that $f$ is the monodromy map, or "first return map", of the flow generated by that vector field, when $X$ is identified with the image in $T(f)$ of $X \times 0$ (equivalently, the image of $X \times 1$).
Edited to address the comment
You can use vector fields to define flows in settings where everything in sight is smooth, in particular $X$ is a smooth manifold and $f$ is a smooth map. 
But instead of pursuing the vector field point of view, I'll just use the mapping torus structure more directly to define the flow/connection. 
The idea is to use an alternate description of the mapping torus $T(f)$, as follows (and I've cleaned up the notation a bit):
$$T(f) = (X \times \mathbb R) \, / \, (x,t) \sim (f(x),t-1)
$$
On the space $X \times \mathbb R$ there is a natural flow $\Phi : (X \times \mathbb R) \times \mathbb R \mapsto X \times \mathbb R$ defined by the formula 
$$\Phi((x,t),s) = (x,t+s)
$$
This flow $\Phi$ descends to a well-defined flow $\phi : T(f) \times \mathbb R \to T(f)$ via the quotient map $X \times \mathbb R \mapsto T(f)$. Furthermore, the projection map $X \times \mathbb R \mapsto \mathbb R$ induces the fiber bundle structure $p : T(f) \mapsto \mathbb R / 2 \pi \mathbb Z \xrightarrow{t \mapsto \exp(it)} S^1$. When you chase through the diagram of quotient maps and projections, you will see that for each $s \in \mathbb R$, the time $s$ map on $T(f)$ defined by the formula $m \mapsto \Phi(m,s)$ takes each fiber $\pi^{-1}(\exp(it))$ to the fiber $\pi^{-1}(\exp(i(t+s)))$. This gives the desired connection. 
And since $\Phi((x,0),1) = (x,1) \sim (f(x),0)$, one sees that the time $1$ map on the fiber $\pi^{-1}(exp(i \cdot 0)) \approx X$ is simply the map $f$.
