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In the proof of unoriented cobordism ring being isomorphic to homotopy group of Thom spectra, one considers a large enough dimensional Euclidean space where a given manifold has all the embeddings isotopic. This is needed to show the Pontrjagin Thom collapse map does not depend on the embedding.

Here is my question: Why does one need isotopy of embeddings? Given a Euclidean space any embedding will have normal bundle isomorphic to a trivial bundle modulo the tangent bundle of the given manifold. It does not depend on embedding in that particular Euclidean space. Any help will be appreciated.

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Well, why would "a trivial bundle modulo the tangent bundle" be uniquely defined? That depends on a choice of an embedding of the tangent bundle in a trivial bundle, and there may be multiple different such embeddings (even into trivial bundles of the same rank) which give quotients which are not isomorphic.

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  • $\begingroup$ I was exactly thinking this. But I am unable to find an explicit example. It will be helpful if you provide one. Thank you. $\endgroup$ – Paladin Nov 29 '19 at 17:04
  • $\begingroup$ I guess an easy example can be given by using some stably trivial bundle. $\endgroup$ – Paladin Nov 29 '19 at 17:13
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    $\begingroup$ Right, stably trivial bundles that are not trivial show that you can't uniquely "subtract" vector bundles in general. I don't know off the top of my head of an example specifically with normal bundles though. $\endgroup$ – Eric Wofsey Nov 29 '19 at 17:19

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