The Euler class of an oriented even-rank normal bundle controls self-intersection

First consider the embedded case. By [F1] one has $A\cdot A=⟨e({\nu }_A),[A]⟩$ for the closed oriented embedded $A^m\subseteq X$, computed with the tangent-first sign convention of the self-intersection number, and $e({\nu }_A)$ is Poincaré dual to the zero locus of any section transverse to the zero section: the Koszul sign of that duality is $(-1{)}^{m(n-m)}=(-1{)}^{m^2}=+1$ because $m$ is even. [F2] gives that an odd-rank oriented bundle has two-torsion Euler class, so on a closed oriented odd-dimensional source its evaluation vanishes, whereas no such vanishing is available in even rank, and $⟨e({\nu }_A),[A]⟩\mathrm{\ne }0$ excludes a nowhere-zero section of ${\nu }_A$ by [F2]. Over ${\mathbb{F}}_2$ no orientation is needed and the identity reads $A{\cdot }_{2}A=⟨w_m({\nu }_A),[A]{⟩}_2$ by [F1].

Now let $f:M^m\looparrowright {\mathbb{R}}^{2m}$ with $m$ even. The normal bundle ${\nu }_f$ of the formal immersion $(f,df)$ is oriented by the orientation of $M$ and the standard orientation of ${\mathbb{R}}^{2m}$, and by [F3] the orthogonal decomposition gives $TM\oplus {\nu }_f\cong {\epsilon }^{2m}$; hence $e({\nu }_f)$ is defined integrally and, by the section form of [F1] applied to the immersion normal bundle, $⟨e({\nu }_f),[M]⟩$ is the signed count of the zeros of any transverse section of ${\nu }_f$. If ${\nu }_f$ admits a nowhere-zero section then $⟨e({\nu }_f),[M]⟩=0$ by [F2]. We prove the asserted regular-homotopy invariance in the next step.

If $m=0$, every map $M\to {\mathbb{R}}^0$ is the same map, so its normal data and Euler number are unchanged by any regular homotopy. For $m>0$, let $H:M\times I\to {\mathbb{R}}^{2m}$ be a regular homotopy from $H_0$ to $H_1$, and write ${\mathcal{N}}_H:=H^{\ast }T{\mathbb{R}}^{2m}\mathrm{/}{d}_{M}H({\mathrm{pr}}_M^{\ast }TM)$ for the vertical quotient. The vertical derivative $d_{M}H$ is a smooth bundle map over $M\times I$ and is fibrewise injective because every $H_t$ is an immersion, so its image is a rank-$m$ subbundle and ${\mathcal{N}}_H$ is a smooth rank-$m$ vector bundle over $M\times I$ by [F4]. At each time $t$ the restriction of ${\mathcal{N}}_H$ to $M\times \{t\}$ is the normal bundle of $H_t$ under the canonical identification, and the orientations of $M$ and ${\mathbb{R}}^{2m}$ induce an orientation of ${\mathcal{N}}_H$ whose restrictions are the orientations of the endpoint normal bundles. With the endpoint inclusions $j_0,j_1:M\to M\times I$, naturality of the Euler class [F4] gives $e({\nu }_{H_t})=j_t^{\ast }e({\mathcal{N}}_H)$ for $t=0,1$, and $j_0,j_1$ are homotopic through $(x,t)\mapsto (x,t)$; by [F4] they induce the same map in cohomology, so $e({\nu }_{H_0})=e({\nu }_{H_1})$ and hence $⟨e({\nu }_{H_0}),[M]⟩=⟨e({\nu }_{H_1}),[M]⟩$. This is invariance under every regular homotopy, with no genericity assumption on the intermediate slices.

Suppose now $m=2s\ge 2$ and $f$ is self-transverse with no triple points, and put $D(f)={\sum }_{r\in \mathrm{\Sigma }(f)}\epsilon (r)$ with the ordering-independent signs of the self-transverse case. By [F6] there are a transverse section $\sigma$ and small $t>0$ with $I(f,f_t)=⟨e({\nu }_f),[M]⟩+2D(f)$. We show $I(f,f_t)=0$. The pair $(f,f_t)$ is transverse to the diagonal $\mathrm{\Delta }\subseteq {\mathbb{R}}^{2m}\times {\mathbb{R}}^{2m}$ at these intersections, and by [F5] the oriented intersection number of the pair equals $I(f\times f_t,\mathrm{\Delta })$. Here orient $\mathrm{\Delta }$ by $v\mapsto (v,v)$ and identify its normal quotient by $(u,v)\mapsto v-u$. At a coincidence the normal derivative of the product is $(-df,d{f}_t)$, so its determinant differs from that of the ordered pair $(df,d{f}_t)$ by $(-1{)}^m=1$, since $m$ is even. Translate the second factor: for $s\in [0,1]$ put $f_{t,s}(y)=f_t(y)+scv$ with a fixed nonzero vector $v$ and $c$ so large that $\mathrm{im}(f_t)+cv$ is disjoint from $\mathrm{im}f$; both images are compact, so such $c$ exists. This is a smooth homotopy of the product map $(f,f_{t,s})$ through the target of [F5], and the endpoint has no intersections with the diagonal, so its intersection number is $0$; by homotopy invariance [F5] the number $I(f\times f_t,\mathrm{\Delta })$ is $0$ as well. Hence $⟨e({\nu }_f),[M]⟩+2D(f)=0$.

It follows that $2D(f)=-⟨e({\nu }_f),[M]⟩$, so the Euler number $⟨e({\nu }_f),[M]⟩$ is even, $D(f)=-⟨e({\nu }_f),[M]⟩\mathrm{/}2$, and reducing the integer equality $D(f)=-⟨e({\nu }_f),[M]⟩\mathrm{/}2$ modulo $2$ gives $\mathrm{\#}\mathrm{\Sigma }(f)\equiv D(f)\equiv -⟨e({\nu }_f),[M]⟩\mathrm{/}2(\mathrm{m}\mathrm{o}\mathrm{d}2)$, since every sign is $\pm 1\equiv 1$ mod two. Finally, if $f$ were regularly homotopic to an embedding $g:M\hookrightarrow {\mathbb{R}}^{2m}$, then step 2.1 would give $⟨e({\nu }_f),[M]⟩=⟨e({\nu }_g),[M]⟩$, while for the embedding [F6] forces $e({\nu }_g)=0$; hence $D(f)=-⟨e({\nu }_f),[M]⟩\mathrm{/}2=0$. Contrapositively, a nonzero $D(f)$, and in particular an odd double-point parity, obstructs a regular homotopy to an embedding. The statement asserts no converse: vanishing of $D$ does not produce a regular homotopy to an embedding, and no odd-dimensional Whitney identification is claimed. AC is used through the normal-bundle and Euler-class suppliers and the oriented intersection theory; the regular-homotopy invariance of step 2.1 is valid for all $m$, while the push-off count of step 2.2 uses the even-dimensional hypotheses.