# fibre bundle as a boundary of a fibre bundle

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?

• What is "the boundary of a CW complex"? – Misha Sep 14 '13 at 12:08
• Ok, what is the boundary of a topological space? – Misha Sep 14 '13 at 12:18
• There is not a good notion of the boundary of a general CW complex. Do you need spaces that are more general than manifolds? – Kevin Walker Sep 14 '13 at 13:58
• @Xiao-GangWen: Of course, there is a notion of boundary for manifolds, homology manifolds and pseudomanifolds. (Disk is a manifold.) But defining boundary for general CW complex let alone topological space, makes no sense whatsoever. – Misha Sep 14 '13 at 16:19
• The correct notion is the one of topological manifold, possibly with boundary. No smoothness is assumed here. – Misha Sep 15 '13 at 6:25

## 1 Answer

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.

• Regarding to your answer we can take $F_{n+1}$ to be the cone on $F_n$, can the monodromy $g:F_n \to F_n$ (needed to define the bundle over $S^1$) extended to $F_{n+1}$? Thank you very much. – Xiao-Gang Wen Sep 15 '13 at 8:31
• I wonder if $F_n$ is a simplicial complex that is equal to its closure, then we can define the boundary of $F_n$. I have always assumed that a complex is equal to its closure. (I wonder if there is name for such kind of complex.) – Xiao-Gang Wen Sep 15 '13 at 8:58
• You can extend g to the cone in the obvious way: the line between x and the cone point will be mapped to the line between g(x) and the cone point, for all $x\in F_n$. – ThiKu Sep 15 '13 at 9:09