Let $M\_{n+1}$ be a fibre bundle with $S\_1$ as the base and $n$-dimensional CW complex $F\_n$ as the fibre. 
Assume $M\_{n+1}$ is oriented. 


(1) Can one show that **$M\_{n+1}$ is always a boundary of a CW complex $M\_{n+2}$,
where $M\_{n+2}$ is a fibre bundle with $S\_1$ as the base and $(n+1)$-dimensional CW complex $F\_{n+1}$ as the fibre**?


(2) Can one show that **$M\_{n+1}$ is always a boundary of a CW complex $N\_{n+2}$,
where $N\_{n+2}$ is a fibre bundle with a disk $D\_2$ as the base and $n$-dimensional CW complex $F\_{n}$ as the fibre**?


Edit: Based on the discussions below, one needs to replace the term "CW complex" above by term "manifold" or "topological manifold". This maybe the key issue of my question: when (1) will be valid?



fibre bundle as a boundary of a fibre bundle
I'll answer your questions with "CW complex" replaced by "manifold".


The answer to question (1) is no. If I understand you correctly, you want the boundary of $F\_{n+1}$ to be $F\_n$. But some manifolds cannot be realized as boundaries of other manifolds. (The simplest examples are 4-manifolds with non-zero signature.)


(On the other hand, if we allow $F\_{n+1}$ to be a CW complex (and ignore the difficulties of defining what "boundary" means in this context), then the answer to (1) is clearly yes, since we can take $F\_{n+1}$ to be the cone on $F\_n$.)


The answer to question (2) is also no. The bundle over $S^1$ is determined by the monodromy $g:F\_n\to F\_n$. If $g$ is not isotopic to the identity, then the bundle does not extend over the disk. And if $g$ is isotopic to the identity, then the bundle is a product.


