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By the classical Pontryagin-Thom construction, we know that the cobordism group $\Omega^n_U(X)$ is represented by cocyles

$$ M\hookrightarrow X\times \mathbb{R}^{2k}\rightarrow X,$$ where $M$ is a manifold with dimension $\dim X-n$, and the normal bundle of the first map is equipped with a complex structure.

I am wondering that if there is any way to turn each representative of $\Omega^n_U(X)$ into a

  • inclusion $ M\hookrightarrow X$, when $n\geq0$,
  • fibering $M\rightarrow X$, when $n\leq0$.

I know that we may turn a map into a inclusion or a fibering with the mapping cylinder/pathspace construction, however, this seems not working directly because of dimension reasons.

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    $\begingroup$ TeX note: $\operatorname{dim} X$ \operatorname{dim} X spaces better than $\mathrm{dim} X$ \mathrm{dim} X. Actually, in this case, it's pre-defined, so you can just use $\dim X$ \dim X. I edited accordingly. $\endgroup$
    – LSpice
    Commented Sep 9, 2023 at 3:55
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    $\begingroup$ $M$ need not be a manifold. See Section 1 in Elementary proofs of some results of cobordism theory using Steenrod operations (Quillen, AIM 7 (1971) 29-56). Also, if you speak of cobordisms, usually you put $n$ into superscript and if of bordisms, then into subscript. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 9, 2023 at 4:47
  • $\begingroup$ @მამუკაჯიბლაძე Typo fixed, and I was referring to this note, where (on page 447) the representatives are considered as manifolds. $\endgroup$
    – timaeus
    Commented Sep 9, 2023 at 5:11
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    $\begingroup$ Well you can represent homotopy type of any finite CW-complex by a manifold with boundary (as Quillen does), what I want to say is that in some cases, in particular for cobordism cocycles, it is sometimes convenient not to do so. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 9, 2023 at 5:14
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    $\begingroup$ Are you familiar with Thom's work on cobordism? He shows that there are integral cohomology classes that aren't realized by submanifolds. Together with the map from complex cobordism to integral cohomology, I think this gives a no answer to your first question. $\endgroup$
    – Mark Grant
    Commented Sep 9, 2023 at 8:02

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