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Suppose $M$ is a compact manifold so that it admits an embedding $f:M\to N\times\mathbb{R}^l$ with a splitting of the normal bundle as $\nu_f\simeq\nu\oplus\epsilon^l$ where $\nu$ is some $k$-dimensional vector bundle and $\epsilon^l$ is the trivial $l$-dimensional bundle. The mutli-compression theorem of Rourke and Sanderson, implies that $f$ is regularly homotopy to an embedding $g:M\to N\times\mathbb{R}^l$ so that the composition

$$i:M\to N\times\mathbb{R}^l\to N$$

is an immersion with $\nu_i\simeq\nu$. For the theorem, see

Rourke and Sanderson. The compression theorem I. Geom. Topol. 5, 399-429 (2001).

or Theorem 2.2 of

Eccles and Grant. Bordism groups of immersions and classes represented by self-intersections. Algebr. Geom. Topol. 7, 1081-1097 (2007).

My question is the following. It seems to me that having $\nu_f\simeq\nu\oplus\epsilon^l$ does not necessarily imply that the trivial part if all coming from the second component of $f$ and the theorem says that we may continuously, change $f$ so that stays in the same homotopy class and all trivial parts comes from the second component of the deformed map $g$, so that collapsing all of them will not damage $\nu$ ? Does this interpretation make sense?

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    $\begingroup$ Yes. If $f:M\hookrightarrow N\times \mathbb{R}^l$ is an embedding with an $l$-framed normal bundle, then you may perform an isotopy of $f$ until the $l$ vectors in the frame are parallel with the $l$ coordinate axes in $\mathbb{R}^l$. At least, that's how I interpret it. $\endgroup$ – Mark Grant Oct 22 '16 at 13:04
  • $\begingroup$ @MarkGrant I understand this is almost the statement in the paper of Yours and Eccles. Still, that you can deform $f$ so that all trivialised part somehow comes from an embedding of $M$ into some Euclidean space sounds very interesting/unusual to me. In the construction of bordism groups of immersion, the procedure of this theorem and its inverse seems to be used. That is, given an immersion $M\looparrowright N$, assuming $M$ is compact one get an embedding $i:M\to\mathbb{R}^{m+l}$ which does not necessarily have a trivial normal bundle (continued below). $\endgroup$ – user51223 Oct 28 '16 at 9:58
  • $\begingroup$ ...But, somehow $(f,i):M\to N\times\mathbb{R}^{m+l}$ seems to be isotopoc to an embedding where a part of its normal bundle is trivialised !?! $\endgroup$ – user51223 Oct 28 '16 at 9:58

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