Finite normal push-off count for an even-dimensional Euclidean immersion

The double point locus is finite. Choose for each $p\in M$ a neighbourhood $U_p$ on which $f$ is injective, as in [F2]; by compactness finitely many $U_{p_1},\dots ,U_{p_N}$ cover $M$, and the open set ${\bigcup }_i(U_{p_i}\times U_{p_i})$ is a neighbourhood of the diagonal on which $f(x)=f(y)$ forces $x=y$. Hence ${\mathrm{\Delta }}_2(f)$ is contained in the compact set $K=(M\times M)\setminus {\bigcup }_i(U_{p_i}\times U_{p_i})$ and is closed in $M\times M$. Self-transversality [F1] makes the derivative of $(x,y)\mapsto f(x)-f(y)$ an isomorphism at every point of ${\mathrm{\Delta }}_2(f)$: the two source tangent dimensions add to $4s$ and their image planes span ${\mathbb{R}}^{4s}$. The inverse function theorem [F5] therefore isolates each ordered coincidence. Thus ${\mathrm{\Delta }}_2(f)$ is compact and discrete, hence finite (its singleton cover has a finite subcover). Unordered branch pairs are the elements $\{x,y\}$ of $D(f)$ obtained by quotienting ${\mathrm{\Delta }}_2(f)$ by interchange. Since no image has three preimages, $D(f)\to \mathrm{\Sigma }(f)$ is bijective. Thus $\mathrm{\Sigma }(f)$ is finite and each image has the ordering-independent sign of its unique branch pair.

There is a smooth section $\sigma$ of ${\nu }_f$ that is transverse to the zero section and nonzero at all double-point preimages. Take finitely many trivializing charts of the rank-$2s$ bundle ${\nu }_f$ with relatively compact domains whose interiors cover $M$, and for each chart and each frame vector a smooth bump function supported in the chart, chosen so that the bump interiors still cover $M$; by [F5] such a finite family with nonnegative bumps exists. Extending bump times frame vector by zero gives finitely many smooth global sections $s_1,\dots ,s_L$ of ${\nu }_f$ that span ${\nu }_f(x)$ at every $x\in M$. The family $\mathcal{F}:M\times {\mathbb{R}}^L\to {\nu }_f$, $\mathcal{F}(x,b)={\sum }_j{b}_j{s}_j(x)$, valued in the total space ${\nu }_f$ of the bundle, is smooth and has surjective vertical derivative at every point (the $s_j(x)$ span), hence is transverse to the zero section $Z\subseteq {\nu }_f$; by parametric transversality [F5] the parameters $b$ for which ${\mathcal{F}}_b$ is not transverse to $Z$ form a null set of ${\mathbb{R}}^L$. For each of the finitely many points $x\in {\mathrm{pr}}_1{\mathrm{\Delta }}_2(f)$ the condition ${\mathcal{F}}_b(x)=0$ is the kernel of the surjective linear map $b\mapsto {\sum }_j{b}_j{s}_j(x)$, a proper linear subspace whence null; their finite union is null, and we choose $b$ outside it and outside the transversality-exceptional set. Then $\sigma :={\mathcal{F}}_b$ is a smooth section of ${\nu }_f$, transverse to the zero section; since transversality with a rank-$2s$ zero section in the $2s$-manifold $M$ makes the zero locus discrete and $M$ is compact, ${\sigma }^{-1}(0)$ is finite, and by construction $\sigma$ is nonzero at every double-point preimage.

For $t>0$ put $f_t(x)=f(x)+t\sigma (x)$, using the embedding ${\mathbb{R}}^{4s}$ as a vector space, and fix an auxiliary smooth bundle metric on $TM$ (which exists on the closed manifold). For small $t$, $f_t$ is an immersion: unit tangent vectors form a compact set, $df$ is fibrewise injective with $\min \mathrm{\parallel }df(v)\mathrm{\parallel }=c>0$ over unit vectors, while $d\sigma$ is bounded over the compact $M$ in any fixed finite family of charts, so $\mathrm{\parallel }td\sigma (v)\mathrm{\parallel }<c\le \mathrm{\parallel }df(v)\mathrm{\parallel }$ for small $t$ and every unit $v$, whence $d{f}_t=df+td\sigma$ is fibrewise injective. Near the diagonal, intersections of $f$ and $f_t$ are the zeros of $\sigma$: define the normal-addition map $E:{\nu }_f\to {\mathbb{R}}^{4s}$ by $E(x,v)=f(x)+v$, whose derivative at each $(x,0)$ is the isomorphism $u\oplus w\mapsto d{f}_x(u)+w$ from $T_{x}M\oplus {\nu }_f(x)$ onto ${\mathbb{R}}^{4s}$; by the inverse function theorem [F5], for each $p\in M$ there is a bundle neighbourhood over a source neighbourhood $U_p$ on which $E$ is injective. Choose finitely many smaller source neighbourhoods $V_i$ with $\overline{V_i}\subset U_i$ that cover $M$, and choose a uniform normal radius small enough that $E$ is injective on all vectors of that radius based in each $V_i$. The open set ${\bigcup }_i{V}_i\times V_i$ contains the source diagonal. If $(x,y)$ is in this set and $f(x)=f_t(y)=f(y)+t\sigma (y)$, then for uniformly small $t$ both $E(x,0)$ and $E(y,t\sigma (y))$ lie in the same injectivity neighbourhood. Their equality gives $(x,0)=(y,t\sigma (y))$, so $x=y$ and $\sigma (y)=0$. Conversely each zero $y$ of $\sigma$ gives the intersection $(y,y)$. At a zero, subtracting the $df$ columns from the $d{f}_t$ columns leaves the normal block $t{d}^{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}\sigma$, so the intersection determinant has the sign of $\det (d^{\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{t}}\sigma )$ because $t^m>0$; this sign, computed in the tangent-first orientation of $TM\oplus {\nu }_f$, is exactly the local contribution of the zero locus of the transverse oriented section $\sigma$; by [F5] the total contribution of these near-diagonal intersections is $⟨e({\nu }_f),[M]⟩$, the Koszul sign of the Euler-duality being $+1$ because the zero locus is $0$-dimensional (equivalently because the rank $2s$ is even).

Near each ordered pair $(x,y)$ with $f(x)=f(y)$, $x\mathrm{\ne }y$, and for small $t$ there is exactly one nearby intersection of $f$ with $f_t$, with the original local sign. Indeed $G(x^{\mathrm{\prime }},y^{\mathrm{\prime }},t)=f(x^{\mathrm{\prime }})-f(y^{\mathrm{\prime }})-t\sigma (y^{\mathrm{\prime }})$ satisfies $G(x,y,0)=0$, and its derivative in $(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$ at $(x,y,0)$ is $(u,w)\mapsto d{f}_{x}u-d{f}_{y}w$; by self-transversality [F1] the sum $d{f}_x(T_{x}M)+d{f}_y(T_{y}M)$ is all of ${\mathbb{R}}^{4s}$ and the dimensions add up, so this derivative is an isomorphism. The inverse function theorem with the parameter $t$ (apply the ordinary theorem to $(x^{\mathrm{\prime }},y^{\mathrm{\prime }},t)\mapsto (G(x^{\mathrm{\prime }},y^{\mathrm{\prime }},t),t)$) gives for small $t$ a unique solution $(x^{\mathrm{\prime }}(t),y^{\mathrm{\prime }}(t))$ near $(x,y)$, and the local sign at $t=0$ is $\epsilon (x,y)$; the sign is locally constant because a nonzero determinant of the ordered derivative pair persists for small $t$. Applying the same statement to the reversed ordered pair $(y,x)$ produces one further nearby ordered intersection with sign $\epsilon (y,x)$, and by [F1] for even $m=2s$ the two signs agree, $\epsilon (y,x)=\epsilon (x,y)=\epsilon (r)$, where $r$ is the double point. Hence each unordered double point contributes $2\epsilon (r)$.

Choose disjoint sufficiently small neighbourhoods of the diagonal and of the finitely many ordered double pairs, and small $t$ so that all the local statements apply. On the compact complement $K^{\mathrm{\prime }}$ of their union in $M\times M$ the distance $\mathrm{\parallel }f(x)-f(y)\mathrm{\parallel }$ has a positive minimum $c^{\mathrm{\prime }}>0$ because $f(x)\mathrm{\ne }f(y)$ for $(x,y)\in K^{\mathrm{\prime }}$; since $\sigma$ is bounded, $\mathrm{\parallel }{f}_t(y)-f(y)\mathrm{\parallel }=t\mathrm{\parallel }\sigma (y)\mathrm{\parallel }<c^{\mathrm{\prime }}$ for small $t$, so no further ordered pair satisfies $f(x)=f_t(y)$. Therefore the ordered intersections of $f$ and $f_t$ are exactly the near-diagonal zeros of $\sigma$ (counted with the signs of step 2.1) together with the two nearby ordered intersections contributed by each double point (step 2.2), and all of them are transverse because the derivatives computed in steps 2.1 and 2.2 are isomorphisms. Hence, summing, $I(f,f_t)=⟨e({\nu }_f),[M]⟩+2{\sum }_{r\in \mathrm{\Sigma }(f)}\epsilon (r)$ for every sufficiently small $t>0$. The empty cases are included: if $f$ is an embedding then $\mathrm{\Sigma }(f)=\mathrm{\varnothing }$ and the equation reads $I(f,f_t)=⟨e({\nu }_f),[M]⟩$, while a zero-free transverse section gives Euler number zero (characteristic-class vanishing, not an embedding conclusion). AC is used through the parametric transversality, the normal-bundle metric and the Euler dual-class supplier; the signs, the finite sums and the local inverse-function computations are choice-free.