Bundle over circle; Monodromy action on cohomology of fiber I want to understand the following sentence:

Let $\pi: M \to S^1$ be a fiber bundle with path connected fiber $F$.
  Then its monodromy action on $H^k(F; \mathbb C)$ satisfies...

How is the monodromy action defined? Its clear that the idea is to look at the standard loop in $S^1$ and investigate how a cohomology class changes along this loop. But I cannot see how to do this.
So how is the monodromy action defined? How can I calculate it and where can I read more about it? Thank you for your help.
 A: One of the lemmas about fiber bundles is that the restriction of every bundle to a contractible subspace is trivial. 
Let's choose four points $p,q,r,s$ equally spaced around $S^1$, and $p$ will be the base point. Since $S^1-\{q\}$ is contractible, using the trivialization over it we obtain a homeomorphism $\mu : \pi^{-1}(p) \to \pi^{-1}(r)$. And since $S^1 - \{r\}$ is contractible, using the trivialization over it we obtain a homeomorphism $\nu : \pi^{-1}(r) \to \pi^{-1}(p)$. By composition we obtain the homeomorphism
$$f : F = \pi^{-1}(p) \xrightarrow{\mu} \pi^{-1}(r) \xrightarrow{\nu} \pi^{-1}(p) = F
$$
This homeomorphism $f : F \to F$ is by definition, the monodromy map (there's something to prove, namely that it is well-defined up to homotopy). 
And now once you have it, you can start looking at what it induces. For instance $f$ induces by the usual pullback process the isomorphism $f_* : H^k(F;\mathbb C) \to H^k(F;\mathbb C)$.
Added after some comments: Here's a more general point of view which works when the base space $S^1$ is replaced by a more arbitrary space $B$. One should assume that $B$ is path connected, and perhaps also that $B$ is a reasonable nice kind of space such as a manifold or a CW complex. As indicated in the comments, in the case $B=S^1$ this definition is equivalent to the one given above, but it is longer and more complicated; nonetheless, as long as one understands both definitions, the proof of equivalence should be easy.
Pick a base point $p \in B$. Choose $F \subset M$ to be the "base fiber" $\pi^{-1}(p)$.
One associates to the bundle a "monodromy homomorphism", which one can think of as a group homomorphism whose domain is the fundamental group $\pi_1(B,p)$. The target group of this homomorphism can be thought of as a "mapping class group" of the fiber $F$. For instance, if the only structure on $F$ is a topology then this is the group of self-homeomorphisms of $F$ modulo the normal subgroup of homomorphisms that are isotopic to the identity. (With more structure around, one can restrict to mapping class groups that respect that structure).
Here's how the monodromy homomorphism is defined. Consider a closed path $\gamma : [0,1] \to B$ based at $p$. Let $\gamma^* M$ denote the pullback bundle: it is the subset of $[0,1] \times M$ consising of all $(t,x)$ such that $\gamma(t) \in \pi(x)$, and the projection $\gamma^*M \mapsto [0,1]$ defines a bundle with fiber $F$. Since the base space is contractible, this bundle is trivial. Pick a trivialization $\tau : [0,1] \times F \to \gamma^*M$ which has the property that that the map 
$$F \xrightarrow{x \to (0,x)} \{0\} \times F \xrightarrow{\tau} \gamma^* M \xrightarrow{\text{projection to $M$}} F
$$
is the identity. The value of the monodromy map on the element of $\pi_1(B,p)$ represented by $\gamma$ is the element of the mapping class group of $F$ that corresponds to the homeomorphism
$$F \xrightarrow{x \to (1,x)} \{1\} \times F \xrightarrow{\tau} \gamma^* M \xrightarrow{\text{projection to $M$}} F
$$
