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I have an inclusion of topological spaces (actually manifolds with corners) $X \to Y$. I can show that for every $x \in X$ there is a neighborhood of $x$ in $Y$ of the form $U \times V$. Also, the intersection of $U \times V$ with $X$ is $U \times V'$.

In fact, $V$ is a neighborhood of the origin in $[0,1)^k$. Also, $V' \subset V$ is given by a set of subsets $\mathcal I$ of $\{1, \ldots, k\}$. It is the union

$$ \bigcup_{I \in \mathcal I} \{ x_i = 0 \text{ for all } i \in I \}. $$

That is, $V'$ is made up of some half-lines, quarter-planes, octants, etc. in $[0,1)^k$.

Question: Is this enough to show that $X \to Y$ is a cofibration of topological spaces?

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Can you explain where the set U is coming from? –  Dan Ramras Feb 17 '10 at 20:03
    
I confess I did not entirely understand where $U$ was coming from when I wrote the question, but now I do. I can show there is a neighborhood of $x$ in the manifold with corners $Y$ which is of the form [open set in interior of $\mathbb R^k$ ] $\times$ [neighborhood of 0 in $\mathbb R^l$] U is the open set in the interior of $\mathbb R^k$. –  mpdude Feb 19 '10 at 23:59
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2 Answers

up vote 3 down vote accepted

The inclusion of a CW subcomplex $K$ into a CW complex $L$ is a cofibration. Briefly, we can extend a homotopy from the $n-1$-skeleton to the $n$-skeleton by projecting $e\times I$ to $e\times\{0\}\cup\partial E\times I$ for any $n$-cell $e$ not in $L$.

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I see, so it just boils down to the fact that I can pick some CW structure on a manifold with corners. Do I need to show that the map $X \to Y$ takes dimension $d$ faces to dimension $d$ faces? –  mpdude Feb 20 '10 at 0:09
    
mpdude -- if you have a submanifold (with corners) of a manifold (with corners), one can equip both with CW structures so that the submanifold becomes a subcomplex, in which case the inclusion is a cofibration. This happens whenever you have a "reasonable" space "of geometric origin" sitting inside another "reasonable" space "of genoetric origin" as a closed subspace. –  algori Feb 20 '10 at 1:04
    
This seems a bit optimistic, in general. Just for ordinary manifolds, it's still an open question whether every closed topological 4-manifold has a CW structure! This is mentioned in Hatcher's book, after Corollary A.12 in the appendix. If your manifolds are smooth, then you're in better shape: all smooth manifolds (even C^1 manifolds) are triangulable. One proof is due to Cairns. Does anyone know the status of these questions for manifolds with corners? –  Dan Ramras Feb 27 '10 at 0:35
    
Dan -- indeed, I was a bit imprecise on that account. But all imaginable manifolds admit a triangulation (hence, a CW-structure, but not necessarily a PL-structure). And yes, I think the standard proof of the existence of triangulations for smooth manifolds can be adapted to the case with corners. –  algori Feb 27 '10 at 1:09
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I'm not sure whether this will be helpful to mpdude, but I thought I'd just point out that given a smooth manifold M and a smooth submanifold N, you can see very easily that the inclusion $N\to M$ is a cofibration, without any cell structure.

Simply observe that a tubular neighborhood gives you a mapping cylinder neighborhood of N inside M (if you choose a metric, then the disk bundle is the mapping cylinder of the projection from the sphere bundle down to N). The homotopy extension property is then easy to verify, because in general for a map $f: X\to Y$, the pair $(M_f, X\cup Y)$ has the homotopy extension property ($M_f$ is the mapping cylinder) so given a map on M and a homotopy on the submanifold M, you can extend your homotopy trivially outside the disk bundle of the tubular neighborhood, and use the HEP to extend it on the disk bundle itself (leaving the homotopy constant on the sphere bundle).

I suppose this procedure would also work for infinite-dimensional manifolds (with appropriate hypotheses).

Closely related is the characterization of (closed) cofibrations in terms of neighborhood deformation retracts, which appears for example in Arne Strom's Note on Cofibrations II.

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