Cohomology of a mapping torus How does the monodromy in a mapping torus $K_{\phi}$ affect the de Rham cohomology, if at all?  Maybe this is naive, but I don't see how twisting one of the ends of $K\times I$ via the diffeomorphism $\phi$ before gluing it to the other end changes $K_{\phi}$ from being homeomorphic to $K\times S^{1}$, in which case the Betti numbers are just $$b_{q}(K_{\phi})=b_{q}(K)+b_{q-1}(K)$$ no matter what the monodromy $\phi$ was.
 A: If $f:M\to M$ is a diffeo, then there is a linear map $f^p:H^pM\to H^pM$ and then we can consider the spaces $$H^p(M)_f=H^p(M)/(1-f^p)H^p(M)$$ and $$H^p(M)^f=\ker(1-f^p).$$ One can then show that there are exact sequences $$0\to H^{p-1}(M)_f\to H^p(S_fM)\to H^p(M)^f\to 0$$ with $SM$ the mapping torus of $f$.
One speedy way of getting this, by the way, is to use the fact that when a group $G$ acts on a space $X$, there is a spectral sequence with 2nd page $E_2^{p,q}=H^q(G,H^q(X))$ converging to $H^\bullet(X/G)$. In our situation, there is an action of $\mathbb Z$ on $M\times\mathbb R$ with quotient $(M\times\mathbb R)/G\cong S_fM$, so we get $$E_2=H^q(\mathbb Z,H^p(M\times\mathbb R))\Rightarrow H^\bullet(S_fM).$$ Of course $H^p(M\times\mathbb R)$ is the same as $H^p(M)$, one can check that the action of $\mathbb Z$ on this s just the one in which $1$ acts as $f$, and then we use the fact that $H^q(\mathbb Z,-)=0$ if $q>1$: the spectral seuquence degenerates in the third page, and writing out explicitly what this means, together with the concrete description of the cohomology of $\mathbb Z$  which is given in every textbook, we thet the result above.
One can also prove my claim above directly, and it is somewhat instructive — my students are going ot have to go through that by the end of this semester :-)
A: Consider the space $K$ given by the disjoint union of two circles. If $\phi$ is the map interchanging the two components (preserving the orientations), then we see that $K_{\phi}$ is a torus. However, $K\times I$ is the disjoint union of two tori, hence the Betti numbers are obviously different:
$$(b_0(K_\phi),b_1(K_\phi),\ldots)=(1,2,1,0,\ldots)$$
$$(b_0(K\times I),b_1(K\times I),\ldots)=(2,4,2,0,\ldots)$$

In general, to compute the Betti numbers of the mapping torus, one uses the long exact sequence in cohomology for $K\subset K_\phi$:
$$\cdots\to\mathrm{H}^d(K_\phi,K)\to\mathrm{H}^d(K_\phi)\to\mathrm{H}^d(K)\to\mathrm{H}^{d+1}(K_\phi,K)\to\cdots$$
The term $\mathrm{H}^{d}(K_\phi,K)$ is always the same as $\mathrm{H}^{d}(K \times \mathrm{S}^1,K)$, as can be seen by excision:
$$\mathrm{H}^{d+1}(K_\phi,K)=\mathrm{H}^{d+1}(K\times I,K\times \{0,1\})=\mathrm{H}^{d+1}(K\times \mathrm{S}^1,K)=\mathrm{H}^{d}(K).$$
The connecting map $\mathrm{H}^{d}(K)\to\mathrm{H}^{d+1}(K_\phi,K)=\mathrm{H}^{d}(K)$ is $\mathrm{id}-\phi^*$. So for example, if $\phi=\mathrm{id}$, then the connecting map is zero and
$$b_d(K_\phi)=b_d(K)+b_d(K_\phi,K)=b_d(K)+b_{d-1}(K).$$
If the connecting map is nontrivial, then $b_d(K_\phi)$ becomes less that $b_d(K\times \mathrm{S}^1)$.
Writing down the long exact sequence of the space in the above example should illustrate this phenomenon, so that one formally sees that the Betti number differ. In this case, the connecting map is represented by the matrix
$$\begin{pmatrix}1 & -1\\-1 & 1\end{pmatrix}.$$
