Let $\Sigma_{\phi}$ be the mapping torus of $\phi$, i.e., $\Sigma \times [0,1] / \sim$ where $(\phi(x),0) \sim (x,1)$ for all $x \in \Sigma$. Also let
$$M_{\phi} = \Sigma_{\phi} \cup_{\psi} (\coprod_{|\partial\Sigma|} S^1 \times D^2)$$
where $|\partial\Sigma|$ is the number of boundary components of $\Sigma$. For each boundary component $l$ of $\Sigma$, the map $\psi: \partial(S^1 \times D^2) \rightarrow l \times S^1$ is defined to be unique (up to isotopy) diffeomorphism that takes $S^1 \times \{p\}$ to $l$ where $p \in \partial D^2$, and $\{q\} \times \partial D^2$ to $(\{ q'\} \times [0,1] / \sim) = S^1$ where $q \in S^1$ and $q' \in \partial \Sigma$.
Now suppose $\Sigma$ is a surface of genus $g$ with $n$ boundary components and $\phi$ is the identity map on $\Sigma$. What I want to prove is that $M_{\phi} = \#_{2g+n-1} S^1 \times S^2$.
I could not prove the assertion for the general case. But for $g=1, n=0$ case I found the following.
Since $M_{\phi} = \Sigma_{\phi} \cup_{\psi} (\coprod_{|\partial\Sigma|} S^1 \times D^2)$ and $n=0$, for this case we have $M_{\phi} = \Sigma_{\phi}$. Then $\Sigma_{\phi} = \Sigma \times [0,1] / \sim = S^1 \times S^1 \times S^1$. But the question suggests the answer $\#_{2g+n-1} S^1 \times S^2$, that is, in our case, $S^1 \times S^2$.
Questions: Is the specific case solution correct? Is the assertion of the question correct (general case)?
The question and the above definitions are from Lectures on open book decompositions and contact structures by John Etnyre.