Oriented and Mod Two Intersection Numbers

By [F1] the set $f^{-1}(Z)$ is a $0$-dimensional embedded submanifold of $X$, hence discrete in its subspace topology: each of its points has a slice chart in which it is the only point of the set in that chart.

The set $f^{-1}(Z)$ is the preimage of the closed set $Z$ under the continuous map $f$, since smooth maps are continuous, hence closed in $X$ by [F2], hence compact by compactness of $X$ and [F3]. A compact discrete space is finite by [F4], so $\mathrm{\#}{f}^{-1}(Z)$ is a well-defined nonnegative integer; the empty case is included.

For the submanifold case, apply the map case to the inclusion of the compact factor: if $A$ is compact and $B$ is closed in $M$, then the inclusion $i_A:A\to M$ is smooth with $A$ compact, $i_A⋔B$ because $A⋔B$, and $i_A^{-1}(B)=A\cap B$, so 2.1 gives that $A\cap B$ is finite; the roles of $A$ and $B$ may be exchanged. No choice axiom is used.

On $S=f^{-1}(h(J))$ put $\phi =h^{-1}\circ f$. Differentiating $f=h\circ \phi$ and taking lengths gives $\mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }=1$. On each component $D$ of $S$, continuity makes ${\phi }^{\mathrm{\prime }}$ constant, so $\phi (s)={\epsilon }_{D}s+c_D$, with ${\epsilon }_D=\pm 1$. The graph is closed in $I\times J$: it is the inverse image of the diagonal of the Hausdorff manifold $M$ under $(f,h)$. Its segments are maximal intersections of their affine lines with $I\times J$, since closedness and the local diffeomorphism property prevent a segment from stopping where both coordinates remain in the relative interiors of their intervals. Included interval endpoints are retained in this assertion.

Each end of a segment therefore reaches an end of $I$ or $J$. At most one segment can reach any one of the four sides: two reaching an $I$-side would have overlapping $s$-projections, contradicting single-valuedness, and two reaching a $J$-side would have overlapping $t$-projections, contradicting injectivity. This includes unbounded ends, since two tails towards the same infinite end overlap. Every segment consumes two distinct sides, so there are at most two segments. With two segments their projections are disjoint on both axes; they must occupy opposite corners, joining left to top and bottom to right (slope $+1$), or left to bottom and top to right (slope $-1$). Thus their slopes agree.

If there is one component, extend its affine expression to $L:\mathbb{R}\to \mathbb{R}$. Maximality of the segment gives $S=I\cap L^{-1}(J)$. The union $I\cup L^{-1}(J)$ is an interval, and $f$ and $h\circ L$ agree on the overlap. Their glued map is smooth and unit-speed. If $f(s)=h(L(s^{\mathrm{\prime }}))$, then $s\in S$ and injectivity of $h$ gives $L(s)=L(s^{\mathrm{\prime }})$, hence $s=s^{\mathrm{\prime }}$. On each open domain piece it is the given local diffeomorphism $f$ or $h\circ L$, so the glued map is a diffeomorphism onto the open union, as required.

In the two-component case reflect a parameter if needed so both slopes are $+1$. The opposite-corner arrangement of 2.1 has $I=(a,d)$, transition expressions $s+p$ on $(a,b)$ and $s+q$ on $(c,d)$, and $J=(c+q,b+p)$, where $a<b\le c<d$ and $d+q\le a+p$. These endpoints are finite: $b,c$ lie inside $I$, while $a+p,d+q$ lie inside $J$. The endpoints of $I,J$ are excluded, since inclusion of one would equate a boundary point of one parametrization with an interior point of the other, by continuity of the transition and preservation of boundary under diffeomorphisms. Put $L=p-q>0$. On $\mathbb{R}\mathrm{/}L\mathbb{Z}$ define $H([t])=f(a+t)$ for $0<t<d-a$ and $H([t])=h(a+q+t)$ for $d-a\le t\le L$, identifying $L$ with $0$. These cover the circle because $d-a\le L$. At $t=d-a$ and $t=0$ the adjacent formulas agree in the $h$-chart with the same affine coordinate, so $H$ is smooth and unit-speed across both seams.

The first branch of $H$ parametrizes $f(I)$ injectively; the remaining arc parametrizes the part of $h(J)$ outside $f(I)$, including the two seams. The only identifications are the stated transition relations, so $H$ is injective and its image is $f(I)\cup h(J)$. It is a local diffeomorphism, hence has open image; its compact image is closed in the Hausdorff $M$. Connectedness forces its image to be all of $M$, and the bijective local diffeomorphism $H$ proves $M\cong S^1$. The given metric and parametrizations require no choice principle.

Fix a Riemannian metric $g$ on $W$ by [F1] under [A1]. Near any $x\in W$ choose a chart as in [F6]; in its coordinates the metric reads $h(t)d{t}^2$ with $h>0$ smooth, and $s(t):={\int }_{t_0}^t\sqrt{h}$ has $s^{\mathrm{\prime }}>0$, so by [F6] its inverse is smooth and $\sigma \mapsto (\text{chart}{)}^{-1}(s^{-1}(\sigma ))$ is a unit-speed reparametrization onto an open subset, i.e. an arc-length parametrization in the sense of [F7]. Hence every point of $W$ lies in an arc-length parametrization with some nondegenerate interval domain.

First suppose $W$ is connected. Fix one such parametrization $f:I\to W$ and let $\mathcal{E}$ be the set of arc-length parametrizations extending $f$. If two members $h_1,h_2\in \mathcal{E}$ did not agree at some point of their common domain, then the overlap of their images would have two components (with one component, [F2] makes $h_2^{-1}\circ h_1$ affine of slope $\pm 1$ on an interval containing $I$, where it is the identity, hence the identity everywhere on the overlap), and then [F2] gives $W\cong S^1$. If $W\cong ̸S^1$, all members of $\mathcal{E}$ therefore agree on overlaps, and the union formula defines a single arc-length parametrization $f^{\ast }$: the affine one-component transition between any two extensions is the identity, so their domain overlap is exactly their image overlap and the union is injective. Thus it defines $f^{\ast }$ on the interval $I^{\ast }:={\bigcup }_{h\in \mathcal{E}}\mathrm{dom}h$ extending every member; it is maximal by construction.

Let $f:I\to W$ be maximal and suppose $f(I)\mathrm{\ne }W$. Since $f(I)$ is open and $W$ is connected, its boundary in $W$ is nonempty; pick $x\in \mathrm{\partial }f(I)$, a limit point of $f(I)$ with $x\mathrm{\notin }f(I)$. By 1.1 choose an arc-length parametrization $h:J\to W$ near $x$; it satisfies $h(J)\cap f(I)\mathrm{\ne }\mathrm{\varnothing }$ and $h(J)\subseteq ̸f(I)$. Applying [F2] to the pair $f,h$: if the overlap has two components then $W\cong S^1$; if it has one component, [F2] exhibits an arc-length parametrization of $f(I)\cup h(J)$ over $I\cup L^{-1}(J)$ extending $f$, and this domain is strictly larger than $I$ because its image contains $x\mathrm{\notin }f(I)$; that contradicts maximality. Hence a maximal arc-length parametrization of a connected $W$ is onto, unless $W\cong S^1$.

If $W$ is connected and compact and $W\cong ̸S^1$, then by 3.1 a maximal arc-length parametrization $f:I\to W$ is a diffeomorphism; since $W$ is compact and $f$ is a homeomorphism, $I$ is a nondegenerate compact interval, hence diffeomorphic to $[0,1]$ by [F6], and $\mathrm{\partial }W$ consists of the two endpoints. Together with the circle alternative this gives: every connected compact smooth $1$-manifold is diffeomorphic to $S^1$ or to $[0,1]$.

For a general compact $W$ (the empty case included) the components are open by [F3] and cover the compact space $W$, so there are finitely many; each component is closed in $W$, hence compact, and with the restricted structure is a connected compact smooth $1$-manifold, so by 4.1 it is diffeomorphic to $S^1$ or to $[0,1]$. By [F4] the boundary $\mathrm{\partial }W$ is a closed $0$-dimensional embedded submanifold of $W$, hence a compact discrete space, hence finite by [F5]. Circle components contribute no boundary points and every interval component contributes exactly two, so $\mathrm{\#}\mathrm{\partial }W=2\cdot \mathrm{\#}\{\text{interval components}\}$ is even.

A circle contributes no boundary points. On $[a,b]$ with $a<b$ and positive tangent direction ${\mathrm{\partial }}_t$, the outward vector is $+{\mathrm{\partial }}_t$ at $b$ and $-{\mathrm{\partial }}_t$ at $a$. The outward-normal-first determinant rule gives the point signs $+1$ at $b$ and $-1$ at $a$, so $\mathrm{\partial }[a,b]=\{b\}-\{a\}$ and their sum is zero. Reversing the interval orientation reverses both point signs and preserves their cancellation.

By [F1] write $W$ as a finite disjoint union of such model components. The given orientation of $W$ restricts to an orientation of each component, and the outward-normal-first boundary orientation is computed componentwise, because a boundary point lies in exactly one component and the outward vectors of the component and of $W$ agree there. Adding the finitely many contributions of 1.1 gives ${\sum }_{p\in \mathrm{\partial }W}\epsilon (p)=0$; a circle component contributes no boundary point, an interval component contributes exactly $+1$ and $-1$, and the empty manifold contributes nothing.

Fix $p\in F^{-1}(Z)$ and put $z:=F(p)$; write $n:=\dim W$ and $n^{\mathrm{\prime }}:=\dim N$. At a boundary point choose a boundary chart $x:U\to x(U)\subseteq {\mathbb{H}}^n$ with $x(p)=0$; at an interior point use an ordinary Euclidean chart centred at $p$ and a slice chart $y:V\to y(V)$ of $Z$ at $z$ with $y(z)=0$, and let $g$ be the last $c$ coordinates of $y$ on $V$. Then $G:=g\circ F$ is defined near $p$ on a relatively open half-space set, and for $q$ near $p$ one has $G(q)=0$ exactly when $F(q)\in Z$. At such a $q$ the differential $d{G}_q=d{g}_{F(q)}\circ d{F}_q$ is surjective: $d{g}_{F(q)}$ is surjective with kernel $T_{F(q)}Z$, so its image of $d{F}_q(T_{q}W)$ is $(d{F}_q(T_{q}W)+T_{F(q)}Z)\mathrm{/}{T}_{F(q)}Z=T_{F(q)}N\mathrm{/}{T}_{F(q)}Z$, which is all of ${\mathbb{R}}^c$ by [F2]. At boundary points, on the face $\{x_n=0\}$ the same computation with $T_q\mathrm{\partial }W$ in place of $T_{q}W$ is surjective by the transversality of $F{\mathrm{\mid }}_{\mathrm{\partial }W}$.

(Interior points.) If $p\in F^{-1}(Z)\cap \mathrm{int}W$, then $F{\mathrm{\mid }}_{\mathrm{int}W}$ is smooth and transverse to $Z$ on the boundaryless manifold $\mathrm{int}W$, so [F4] shows that $F^{-1}(Z)\cap \mathrm{int}W$ is an embedded submanifold of $\mathrm{int}W$ of codimension $c$ with the stated tangent space; in the chart $x$ this exhibits $F^{-1}(Z)$ near $p$ as a coordinate subspace $\{x^{n-c+1}=\cdots =x^n=0\}$ intersected with the chart, so $p$ is an interior point of $F^{-1}(Z)$.

(Boundary points.) If $p\in F^{-1}(Z)\cap \mathrm{\partial }W$, then in the coordinates of 1.1 the differential $d{G}_p:{\mathbb{R}}^n\to {\mathbb{R}}^c$ is surjective and, by the face part of 1.1, its restriction to the face ${\mathbb{R}}^{n-1}\times \{0\}$ is surjective. Choose a set $A$ of $c$ coordinate directions inside the face on which the resulting square minor of $d{G}_p$ is invertible, let $B$ be the remaining $n-1-c$ face directions, and define $\mathrm{\Phi }:=(G,x_B,x_n)$ near $p$. Its differential is block triangular with invertible diagonal blocks, hence invertible; by [F6] the components extend smoothly to an open neighbourhood of $p$ in ${\mathbb{R}}^n$, so the inverse function theorem makes $\mathrm{\Phi }$ a local diffeomorphism at $p$. Since $\mathrm{\Phi }$ has last coordinate $x_n$, it carries $\{G=0,x_n\ge 0\}=F^{-1}(Z)$ near $p$ onto the relatively open subset $\{0\}\times {\mathbb{R}}^{n-1-c}\times [0,\mathrm{\infty })$ of the half-space ${\mathbb{H}}^{n-c}$; therefore $F^{-1}(Z)$ is near $p$ an embedded $(n-c)$-dimensional submanifold with boundary, with boundary exactly $F^{-1}(Z)\cap \mathrm{\partial }W$ and tangent space $\{v\in T_{p}W:d{F}_p(v)\in T_{z}Z\}$, and the chart $\mathrm{\Phi }$ simultaneously straightens the preimage and the face.

(Assembly.) The charts produced in 2.1 and 2.2 cover $F^{-1}(Z)$; any two of them are restrictions of charts of $W$, so their transitions are restrictions of smooth half-space transitions and the induced structure is a smooth structure on $F^{-1}(Z)$ making it an embedded submanifold with boundary of $W$, of dimension $n-c=\dim W-\mathrm{codim}Z$, with the tangent formula and with interior $F^{-1}(Z)\cap \mathrm{int}W$ and boundary $F^{-1}(Z)\cap \mathrm{\partial }W=(F{\mathrm{\mid }}_{\mathrm{\partial }W}{)}^{-1}(Z)$. If $n-c\ge 1$, the local model of 2.2 shows that the preimage is transverse to $\mathrm{\partial }W$ and meets it exactly in its boundary, so $F^{-1}(Z)$ is neat and $\mathrm{\partial }{F}^{-1}(Z)=F^{-1}(Z)\cap \mathrm{\partial }W$; the charts of 2.2 are precisely the boundary-adapted slice charts of [F5]. If $n-c=0$, boundary points are impossible because the face differential would have rank $c=n$ on an $(n-1)$-dimensional space. The preimage is discrete, lies in the interior, and has empty boundary. If $n<c$, even the full differential cannot be surjective, so the preimage is empty. The construction uses the charts named at each point and no choice principle; the atlas of the preimage is the set of all charts so obtained.

By [F1] and [F2] the trace $W:=F^{-1}(Z)$ is a compact embedded submanifold with boundary of $[0,1]\times X$ of dimension $1$, with $\mathrm{\partial }W=F_0^{-1}(Z)\bigsqcup F_1^{-1}(Z)$; compactness follows because $W$ is closed in the compact product $[0,1]\times X$. In particular the two slice preimages are finite by [F3].

By [F4] the boundary of the compact $1$-manifold $W$ has even cardinality, so $\mathrm{\#}\mathrm{\partial }W=\mathrm{\#}{F}_0^{-1}(Z)+\mathrm{\#}{F}_1^{-1}(Z)$ is even. Reducing modulo two and using [F5] gives $\mathrm{\#}{F}_0^{-1}(Z)\equiv \mathrm{\#}{F}_1^{-1}(Z)(\mathrm{m}\mathrm{o}\mathrm{d}2)$, that is, $I_2(F_0,Z)=I_2(F_1,Z)$ by [F3].

Given a continuous homotopy between transverse endpoints, reparametrize by a smooth function constant in end collars, extend constantly to $\mathbb{R}\times X$, and smooth it fixed on smaller closed end regions by [F6]. Call the smooth result $H$; it is constant at its transverse endpoints for $t\le \delta$ and $t\ge 1-\delta$, for some $0<\delta <1\mathrm{/}2$. Take the submersive parameter family $\mathcal{H}$ for $H$ from [F6], with centred ball $B$. Choose a smooth $\lambda :\mathbb{R}\to [0,1]$ equal to $0$ outside $(\delta \mathrm{/}2,1-\delta \mathrm{/}2)$ and positive inside, and set $G((t,x),a)=\mathcal{H}((t,x),\lambda (t{)}^{2}a)$. Where $\lambda >0$ the parameter derivative is surjective; where $\lambda =0$, $d({\lambda }^2)=0$ and the source derivative is that of the already transverse endpoint map. Thus $G$ is transverse to $Z$. By parametric transversality choose a good $a\in B$ (also when $B$ is zero-dimensional); its slice, restricted to $[0,1]\times X$, is transverse and has the original endpoints. Applying 2.1 to this fixed-endpoint transverse homotopy proves equality of the parities of any homotopic transverse representatives. Representatives exist by [F6], so $I_2$ is well defined and homotopy invariant. For two compact-source maps, product homotopies and the diagonal definition give invariance under deformation of either or both factors. Countable Choice is inherited from the classification and approximation suppliers; the finite parity computation adds no choice.

For positive-dimensional factors take positive bases $u,w$. The swap sends the domain's ordered basis $(u,w)$ to the list consisting first of the $U$-vectors in the second summand, then of the $W$-vectors in the first. Reordering that image to the codomain's positive list $(w,u)$ takes $kl$ transpositions, so its determinant sign is $(-1{)}^{kl}$. This is the exterior-algebra identity $w\wedge u=(-1{)}^{kl}u\wedge w$.

The same exterior identity applies to arbitrary positive determinant elements, including signed scalars for a zero-dimensional factor; if $k=0$ or $l=0$ its sign is $+1$. For an internal sum, the two addition maps satisfy $a_{W,U}\circ s=a_{U,W}$, so transporting their product rays to $V$ gives the same comparison sign. Thus both statements hold in every dimension, and for $k=l=1$ the swap reverses orientation.