High normal Stiefel-Whitney classes obstruct low-codimension immersions

Suppose, for contradiction, that there is a smooth immersion $f:M\looparrowright {\mathbb{R}}^{m+k}$. By [F1] the hypothesis ${\mathrm{A}\mathrm{C}}_{\omega }$ of [F2] holds with $n=m+k>m$, so the normal bundle ${\nu }_f$ of the immersion is a rank-$k$ stable normal inverse of $M$.

By [F3] applied to the rank-$k$ inverse ${\nu }_f$, the class ${\bar{w}}_i(M)=w_i({\nu }_f)$ for the given index $i>k$. But [F4] gives $w_i({\nu }_f)=0$ because $\mathrm{rank}{\nu }_f=k<i$, a contradiction with ${\bar{w}}_i(M)\mathrm{\ne }0$. Hence no immersion into ${\mathbb{R}}^{m+k}$ exists.

The final sentence uses $k=i-1$. For $i\ge 2$ this satisfies $k\ge 1$, so step 2.1 excludes immersion in ${\mathbb{R}}^{m+i-1}$. For $i=1$, a nonzero class ${\bar{w}}_1(M)$ forces $M$ to be nonempty and $m\ge 1$. No nonempty compact positive-dimensional manifold immerses in ${\mathbb{R}}^m$: an equal-dimensional immersion is a local diffeomorphism by [F5], so its image is open; the image is also compact by [F6], hence closed in Hausdorff ${\mathbb{R}}^m$. Euclidean space is connected because any two points are joined by their straight segment, and it is noncompact for $m\ge 1$ because the cover by balls of integer radius has no finite subcover. Thus connectedness of ${\mathbb{R}}^m$ makes a nonempty open-and-closed image all of ${\mathbb{R}}^m$, contradicting its noncompactness. Thus the codimension-zero instance is excluded too. Negative codimension is impossible because the derivative could not be injective. Consequently every immersion has codimension at least $i$ under the nonzero-class hypothesis. No converse or classification is asserted.