Orientations Poincare Lefschetz and Alexander Duality

In any chart containing $x$, restrict to an open ball about its image if that image is above the hyperplane, or to the intersection of an open ball with the half-space if it is on the hyperplane. Translate the center to $0$ (only parallel to the hyperplane in the latter case). Write $U$ for the resulting neighborhood of $x$. Since $M$ is Hausdorff, $M\setminus \{x\}$ is open: separate each other point from $x$ by open sets and take their union. The closed set $Z=M\setminus U$ omits $x$, so $\bar{Z}=Z\subseteq \mathrm{int}(M\setminus \{x\})$. Thus [F2] and [F6] identify the local pair homology with that of $(B,B\setminus \{0\})$ or $(B_{+},B_{+}\setminus \{0\})$, where the ball has radius $r>0$.

A point has exactly one singular generator in every degree. Its boundary is multiplication by ${\sum }_{i=0}^k(-1{)}^i$, namely identity for positive even $k$ and zero for odd $k$, with zero degree-zero boundary. Hence its homology is $R$ in degree zero and zero above. A nonempty contractible space has the same homology by [F5], with the degree-zero isomorphism induced by its map to a point. For a contractible $Y$ and nonempty $A\subseteq Y$, [F4] therefore identifies $H_k(Y,A;R)$ with ${\tilde{H}}_{k-1}(A;R)$ for $k\ge 1$ and gives $H_0(Y,A;R)=0$. At $k=1$ this is the kernel of the surjective augmentation $H_0(A;R)\to R$, and at higher degrees it follows from the two zero adjacent groups. Surjectivity uses any one point of the given nonempty $A$.

At an interior chart point with $n\ge 1$, $B$ contracts to $0$. For $\rho =r\mathrm{/}2$, the homotopy $h(y,t)=(1-t+\frac{t\rho }{\parallel y\parallel })y$ retracts $B\setminus \{0\}$ onto its radius-$\rho$ sphere: the norm is $(1-t)\parallel y\parallel +t\rho$, strictly between $0$ and $r$, and the sphere is fixed. Scaling identifies it with $S^{n-1}$. By [F3], [F5] and step 1.2, the pair group is $R$ exactly in degree $n$. This includes $n=1$, where the punctured interval has two components and the augmentation kernel is $\{(a,-a):a\in R\}$. When $n=0$, the chart pair is $(\{0\},\mathrm{\varnothing })$; its relative chain complex is the point complex calculated in step 1.2, giving the same conclusion.

At a boundary chart point necessarily $n\ge 1$. Choose $a=(0,\dots ,0,r\mathrm{/}2)$. Both $B_{+}$ and $B_{+}\setminus \{0\}$ contract to $a$ by $(y,t)\mapsto (1-t)y+ta$. The ball and half-space are convex; for $t>0$ the last coordinate is positive, so the punctured homotopy never meets $0$, and at $t=0$ the input already avoids $0$. Their inclusion induces an isomorphism on $H_0$ and on every higher group, as seen by their maps to a point. The exact sequence [F4] gives zero relative groups in every degree.

Now take $R=\mathbb{Z}$. The nonzero group in step 2.1 and the all-zero groups in step 2.2 are intrinsic to the pair $(M,M\setminus \{x\})$. Thus no point can be a boundary point in one chart and an interior point in another, even if different dimension labels are considered. This proves chart independence of [F1]'s boundary set and the formula in the statement. If a nonempty space has manifold dimension labels $n$ and $m$, some point is interior in an $n$-chart: every nonempty open subset of a positive-dimensional half-space meets its strict interior; for dimension zero every point is interior. The same point is interior for the other manifold structure by the zero/nonzero distinction. Its unique nonzero local degree is both $n$ and $m$, hence $n=m$. A homeomorphism identifies the local pairs by [F6], so preserves these data.

If the boundary is empty, [F1]'s restriction to small balls gives Euclidean neighborhoods. Conversely, if $M$ is locally Euclidean of dimension $n$, the interior calculation of step 2.1 at every point gives a nonzero integral local group and precludes a boundary chart by step 2.2. This proves both directions of the stated equivalence. For empty $M$ the pointwise assertion and boundary equivalence are vacuous, but every dimension label is allowed. Zero coefficients make all local groups zero, without affecting the integral argument for intrinsic properties. The contractions explicitly include their endpoints, and the point calculation uses unnormalized chains; no omitted degenerate-generator convention or choice principle is involved.

The ball $K$ is compact by [F6]; compactness of its chart preimage follows by pulling any open cover back to the Euclidean ball and taking a finite subcover there. It is closed in Hausdorff $M$. Thus $Z=M\setminus B$ is closed and contained in the open set $M\setminus K$. Excision gives $H_n(M,M\setminus K;R)\cong H_n(B,B\setminus K;R)$. It gives the analogous isomorphism for $\{x\}$, and these identifications commute with the restriction maps since all are induced by inclusions.

Put $\rho =(r+s)\mathrm{/}2$. The annulus $B\setminus K$ retracts onto $S_{\rho }=\{u:\parallel u\parallel =\rho \}$ by $u\mapsto (1-t+t\rho \mathrm{/}\parallel u\parallel )u$: throughout the homotopy its norm is strictly between $r$ and $s$. The inclusion $S_{\rho }\to {\mathbb{R}}^n\setminus \{x\}$ is a homotopy equivalence on homology. Indeed translate the target by $-x$; the resulting map $u\mapsto u-x$ is homotopic to $u\mapsto u$ by $u\mapsto u-tx$, which never vanishes because $\parallel tx\parallel <r<\rho$. The centered sphere is a radial deformation retract of ${\mathbb{R}}^n\setminus \{0\}$. Hence inclusion $B\setminus K\to {\mathbb{R}}^n\setminus \{x\}$ induces isomorphisms on reduced homology in all degrees, including the augmentation kernel in degree zero.

Choose $0<ϵ<s-\parallel x\parallel$. Both $B\setminus \{x\}$ and ${\mathbb{R}}^n\setminus \{x\}$ radially retract about $x$ to the sphere of radius $ϵ$: $h_x(y,t)=x+(1-t+\frac{tϵ}{\parallel y-x\parallel })(y-x)\mathrm{.}$ For the first space the homotopy lies in $B$, since it is the segment between $y$ and a point of that small sphere, both in the convex ball $B$. It avoids $x$ by positivity of the radial coefficient. These retractions commute with inclusion, so that inclusion induces isomorphisms. Factoring the isomorphism of step 1.2 through $B\setminus \{x\}$ proves that $B\setminus K\to B\setminus \{x\}$ induces reduced homology isomorphisms.

The ball $B$ is contractible. As calculated in [F1], the exact sequence [F3] identifies $H_n(B,A;R)$ with ${\tilde{H}}_{n-1}(A;R)$ for either nonempty subspace $A=B\setminus K$ or $B\setminus \{x\}$. At $n=1$ this is the kernel of $H_0(A;R)\to H_0(B;R)$; no unreduced degree-zero replacement is made. Naturality of the connecting map and step 2.1 therefore prove the desired restriction isomorphism, after step 1.1. Its target is $R$ by [F1].

If $J\subset \mathrm{int}K$, choose any $y\in \mathrm{int}J$. Restriction from $K$ to $y$ factors as restriction from $K$ to $J$ followed by restriction from $J$ to $y$, by [F5]. Both the composite and the latter map are isomorphisms by step 3.1 applied to the respective charts; hence the first map is an isomorphism. The same factorization holds for every $y$ in the interior of $J$, proving compatibility without any common-coordinate assumption on the two balls. For $n=0$ all these balls are singletons; excision reduces their groups and maps to the point group $R$ and its identity, as in [F1].

The statement requires an actual chart ball, so supplies no ball in an empty manifold. Zero coefficients give isomorphisms of zero modules. The radii have strict inequalities; neither radius zero nor a point on $\mathrm{\partial }K$ is claimed. All homotopies include $t=0,1$ and operate on unnormalized singular chains through [F4]. Only finitely many individual radii, points and charts are used; all comparison maps are canonical inclusions, and no AC is invoked.

Every point has an open neighborhood homeomorphic to a convex ball or half-ball, using a small ball in its chart. Line segments stay inside that model, so the neighborhood is path-connected, hence connected by [F3], and lies inside its point's component by [F4]. Taking the union of such neighborhoods over the points of a component proves it open. Components are disjoint and cover $M$: intersecting maximal connected sets have connected union (by the component definition), so coincide. Each component inherits the manifold structure, and its interior is its intersection with $\mathrm{int}M$, because chart-boundary membership is local.

Over ${\mathbb{F}}_2$, every interior stalk has precisely one nonzero element, its unique generator. Define the section to take this element. In any local module chart the section has constant coordinate $1\in {\mathbb{F}}_2$, so is continuous. Uniqueness of the fiber value makes this section canonical, without choosing any generators. It orients the interior and thus $M$ by [F1].

Any $R$-orientation restricts to one on each component. Conversely, given a family $({\mu }_C{)}_C$ of component orientations, define ${\mu }_x=({\mu }_C{)}_x$ for the unique component containing $x\in \mathrm{int}M$. This is a section of generators. It is continuous because its restriction to each of the open subsets $C\cap \mathrm{int}M$ is continuous: the inverse image of any open set in the total local system is the union of these open inverse images. Restriction and this union operation are inverse functions, proving the asserted bijection, including when the sets of orientations are empty.

If $M$ is compact, its open cover by components from step 1.1 has a finite subcover. Since components are nonempty and disjoint, any omitted component would contain a point not covered by that subcover. Thus the finite subcover lists all components. For empty $M$ this is the empty list.

An orientation on $M$ implies orientability of every component by restriction in step 2.1, without AC. Conversely, suppose each component is $R$-orientable and assume [A1]. For each component $C$, let $S_C$ be the set of its continuous generator sections, a nonempty subset of the set of functions on its interior into the local system. AC applied to this set-indexed family yields $({\mu }_C{)}_C$ with ${\mu }_C\in S_C$. Step 2.1 glues it to an orientation on $M$. This is the exact selection use of AC in the existence equivalence; the earlier bijection did not select a member of a product of nonempty sets.

For empty $M$ the empty section and empty family correspond. For $R=0$ the unique zero section is a generator section, as in [F1]. A zero-dimensional manifold has singleton components and the same gluing statement; a single component requires no family selection beyond its given existence. The construction in step 1.2 uses no sign choice since $1=-1$ over ${\mathbb{F}}_2$. Boundary points are not assigned local generators: orientation is on the interior exactly as stated in [F1]. There is no simplex nondegeneracy or numerical endpoint hypothesis in this proposition.

The singular simplex generators common to $D$ and $E$ are exactly the maps with image in $U\cap V$. Hence $D\cap E=C_{\ast }(U\cap V;R)$, including for the zero ring. The chain maps $0⟶C\mathrm{/}(D\cap E)\stackrel{j}{\to }(C\mathrm{/}D)\oplus (C\mathrm{/}E)\stackrel{q}{\to }C\mathrm{/}(D+E)⟶0,$ given by $j[c]=([c],[c])$ and $q([a],[b])=[a-b]$, are well-defined and commute with boundary because $D,E$ are subcomplexes.

The map $j$ is injective since a representative mapping to zero lies in both $D$ and $E$. The map $q$ is onto since $q([a],0)=[a]$, and $qj=0$. If $q([a],[b])=0$, write $a-b=d+e$ with $d\in D,e\in E$. Then $c=a-d=b+e$ has residues $[a]$ in $C\mathrm{/}D$ and $[b]$ in $C\mathrm{/}E$, so $j[c]=([a],[b])$. This proves exactness in every degree; the argument uses only the existence of a decomposition for one element of $D+E$.

By De Morgan's laws, $U\cap V=X\setminus (A\cup B)$ and $U\cup V=X\setminus (A\cap B)$. The first two nonzero complexes in step 1.1 therefore have exactly the relative homology groups displayed in the statement. Since $U,V$ are open, [F2] identifies the homology of the final quotient with $H_{\ast }(X\mathrm{\parallel }A\cap B;R)$. Applying [F3] to step 2.1 yields the asserted sequence. Composition of $q$ with this canonical quotient map is the difference of the two relative quotient maps, so the printed sign is precisely $a{\mathrm{\mid }}_{A\cap B}-b{\mathrm{\mid }}_{A\cap B}$.

If $A=\mathrm{\varnothing }$, then $U=X$, $D=C$, and the sequence reduces to identity maps on the groups supported in $B$, with zero groups for empty support; the other empty case is symmetric. If $A=B$, the diagonal and difference sequence has the stated exactness. For $X=\mathrm{\varnothing }$, $R=0$ or negative chain degrees all complexes concerned are zero. Degree zero follows from the same degreewise short exact sequence, with no reduced-group substitution. All singular generators, including degenerate ones, were retained in step 1.1. The quotient equivalence in [F2] uses prescribed small-chain operators; no AC or choice of a splitting is required.

Suppose $P(A)$, $P(B)$ and $P(A\cap B)$ hold. In [F4]'s exact sequence, $H_{n+1}(M\mathrm{\parallel }A\cap B)=0$ makes the diagonal map from $H_n(M\mathrm{\parallel }A\cup B)$ injective. If a class restricts to zero at all points of $A\cup B$, its two images vanish by pointwise injectivity for $A,B$, so it is zero. The classes $s_A,s_B$ have equal restrictions to $A\cap B$, because their point values there coincide and restriction on that intersection is injective. Thus $(s_A,s_B)$ lies in the kernel of the difference map and lifts to $s_{A\cup B}$; its point restrictions are the required ones. For $i>n$, the terms $H_{i+1}(M\mathrm{\parallel }A\cap B)$ and $H_i(M\mathrm{\parallel }A)\oplus H_i(M\mathrm{\parallel }B)$ are zero, so exactness also gives $H_i(M\mathrm{\parallel }A\cup B)=0$. This proves $P(A\cup B)$.

Let $A$ be a nonempty compact convex subset in a coordinate neighborhood identified with all of ${\mathbb{R}}^n$, first with $n\ge 1$, and fix $x\in A$. Choose $L>{\sup }_{a\in A}\parallel a-x\parallel$, which is finite by compactness. The radial homotopy about $x$ taking radius $\parallel y-x\parallel$ linearly to $L$ retracts both ${\mathbb{R}}^n\setminus A$ and ${\mathbb{R}}^n\setminus \{x\}$ to $S_L(x)$. To check the first assertion, if the initial radius is at most $L$, all movement is outward: if an outward multiple of $y-x$ met $A$, convexity with $x\in A$ would force $y\in A$, a contradiction. If the initial radius exceeds $L$, every intermediate radius is at least $L$ and is outside $A$. The same positive-radius formula works for the punctured space. The inclusion of complements therefore induces homology isomorphisms. The natural pair sequences [F5], together with the point-local calculation [F9], give isomorphisms $H_i(M\mathrm{\parallel }A)\to H_i(M\mathrm{\parallel }\{x\})$ for every $i$, by excision [F3]. They prove vanishing above $n$ and pointwise injectivity.

To realize a section over such an $A$, take a closed coordinate ball $D$ with $A\subset \mathrm{int}D$. Its ball trivialization from [F1] identifies $s$ over the connected open ball $\mathrm{int}D$ with a continuous function into a discrete module. Such a function is constant: any nonempty fiber and its nonempty complement would be a separation into open sets. The ball is connected by [F7]. The constant coordinate is a class in $H_n(M\mathrm{\parallel }D)$, whose restriction to $A$ realizes $s$. For $n=0$, a nonempty compact convex coordinate support is a singleton and the ball comparison is the identity on $R$. The empty support has zero relative complex and satisfies all three assertions. Hence $P(A)$ holds for every compact convex coordinate support. A coordinate ball may always be identified with all of ${\mathbb{R}}^n$: in centered radius-$r$ coordinates, $y\mapsto y\mathrm{/}(r-\parallel y\parallel )$ has continuous inverse $z\mapsto rz\mathrm{/}(1+\parallel z\parallel )$.

Property $P$ holds for every finite union of compact convex subsets in one coordinate ${\mathbb{R}}^n$. Induct on their number. To adjoin the last set, its intersection with the preceding union is a union of fewer compact convex sets, since each pairwise intersection is compact and convex, possibly empty. Apply the induction hypothesis to that intersection and preceding union, then step 1.1. This proves the induction without assuming intersections are balls.

Let $A$ now be any compact subset of one such coordinate ${\mathbb{R}}^n$, and let $\alpha \in H_i(M\mathrm{\parallel }A)$, $i\ge n$. By excision and [F2], represent it by a finite chain $z$ in the coordinate space with $\mathrm{\partial }z$ supported outside $A$. Let $C$ be the union of the images of the finitely many simplex maps occurring with nonzero coefficient in $\mathrm{\partial }z$. It is compact: a simplex is closed and bounded, hence compact by [F6], its continuous image is compact by pulling back covers, and finitely many compact images have compact union by taking finitely many finite subcovers. Thus $C$ is closed and disjoint from $A$. Take all closed balls centered at points of $A$ and small enough to miss $C$. Their interiors cover $A$; compactness gives finitely many with union $D$ containing $A$. Every chosen center belongs to $A$. Now $z$ represents ${\alpha }_D\in H_i(M\mathrm{\parallel }D)$ restricting to $\alpha$. If $i>n$, step 3.1 gives ${\alpha }_D=0$, hence $\alpha =0$.

If $i=n$ and $\alpha$ vanishes at every point of $A$, the image of ${\alpha }_D$ in the group supported on each ball of step 4.1 has zero restriction at its center in $A$. The convex-support isomorphism of step 1.2 makes that ball-supported class zero. Hence ${\alpha }_D$ vanishes at every point of each ball, and therefore at every point of $D$. Pointwise injectivity from step 3.1 gives ${\alpha }_D=0$, and so $\alpha =0$. This proves injectivity for arbitrary compact coordinate supports. Existence for these supports is the same restriction from a large ball as in step 2.1: compactness puts $A$ inside that ball and the section has constant coordinate on its interior. Thus $P(A)$ holds for every compact coordinate support. When $A$ is empty, no balls are needed and the relative group is already zero.

For a general compact $K\subset M$, consider all chart balls whose closures lie inside a larger chart ball. Their interiors cover $K$. These closures are compact by [F6] and continuous-image compactness, and are closed in $M$. Choose a finite subcover and put $K_j=K\cap {\bar{B}}_j$. These are compact, their union is $K$, and each lies in a chart homeomorphic to ${\mathbb{R}}^n$. Induct on the number of pieces. The intersection of the last piece with the preceding union is the union of fewer compact chart pieces $K_j\cap K_m$. Property $P$ for a single piece is step 5.1; the same induction and step 1.1 therefore prove $P(K)$. Only finite covers and finite selections have been used.

Apply realization to the orientation section from [F1] to define $[M{]}_K$. Its uniqueness is the pointwise injectivity just proved. If $K\subset L$ are compact, the restriction of $[M{]}_L$ to $K$ has the same point values, hence equals $[M{]}_K$. Components of $M$ are open: each point has a connected coordinate-ball neighborhood by [F7], and that neighborhood lies in its component by [F8]. Thus their cover of $K$ has a finite subcover; disjointness shows these are all the components meeting $K$. Each such component is also closed, since its complement is a union of the other open components. Thus its intersection with $K$ is compact. Repeatedly applying [F4] to these disjoint supports, whose intersections have zero relative groups, gives the direct-sum decomposition and sends $[M{]}_K$ to the individual orientation classes. For $R=0$ all groups and sections are zero, and for empty $K$ the result is the unique zero class. All component arguments used only this local connectedness, and no AC or nondegenerate-simplex convention was used.

The chain coefficient map ${\iota }_R:C_{\ast }(-;\mathbb{Z})\to C_{\ast }(-;R)$ sends $\sum m_j{\sigma }_j$ to $\sum (m_j{1}_R){\sigma }_j$. It commutes directly with boundary, passage to relative quotients, and maps of pairs. We now verify on the stated interfaces that its map on an integral local stalk takes a generator to an $R$-module generator, consistently with the abstract sphere and local-group computations in [F5] and [F6]. For $n\ge 1$, choose a chart carrying $x$ to $0$ in an open ball $B$, and a concentric closed disk $D$ with boundary sphere $S$. The radial formula $h(y,t)=(1-t+t\frac{\rho }{\parallel y\parallel })y$ deformation retracts $B\setminus \{0\}$ onto $S$, where $\rho$ is the radius of $D$: its norm is $(1-t)\parallel y\parallel +t\rho$, so it stays nonzero and inside $B$. The disk and $B$ are contractible, hence [F11] and the natural pair sequences show that inclusion $(D,S)\to (B,B\setminus \{0\})$ induces an isomorphism in degree $n$; for $n=1$ the connecting terms are the augmentation kernels in reduced $H_0$, exactly as in [F9]. By [F10], the chart map and excision then identify this ball-pair group with $H_n(M,M\setminus \{x\};G)$ for $G=\mathbb{Z}$ and for $G=R$. The sphere calculation [F5] gives a primitive positive boundary generator with coefficient $1$, and [F9] gives its unique preimage in $H_n(D,S;\mathbb{Z})$. The coefficient map sends the boundary generator to the same alternating cycle with coefficient $1_R$ (for $n=1$, it sends $(1,-1)$ to $(1_R,-1_R)$ in the augmentation kernel). Naturality of [F9] therefore sends its relative preimage to the preimage of the coefficient-$1_R$ boundary generator. Under the chart and excision isomorphisms this is an $R$-module generator of the local stalk, also when $R$ is the zero ring. Thus the induced integral local generator maps to an $R$-generator. For $n=0$, the same conclusion follows directly from the point class with coefficient $1$. Finally, the transports in [F2] and [F3] are composites of the same ball restriction isomorphisms, so their naturality with ${\iota }_R$ is the chain-level commutation just established. An integral stalk automorphism is multiplication by $+1$ or $-1$, since its value on a generator must again be a generator. Hence the corresponding loop transports on an $R$ stalk have those same signs.

Any two points of $M$ can be joined by a path. Indeed, the subset of points reachable from one fixed point is open, since paths can be extended within convex chart neighborhoods. Every other path-reachability class is open for the same reason, so its complement is open. Connectedness and nonemptiness force the reachable subset to be all of $M$. In a ball trivialization, a continuous local-system section is constant along a path segment: its discrete coordinate is continuous on a connected interval by [F7], and a nonconstant value would separate that interval by a fiber and its complement. Therefore a section is determined everywhere by its value at any one point, using finite path subdivisions as in [F2].

For any finite singular cycle $z$ in $M$, its image is compact: the finitely many simplex domains are compact by [F8], continuous images are compact by pulling back covers, and finite unions preserve compactness. Cover this image by interiors of closed coordinate balls lying inside larger charts, and take a finite subcover. The union $U$ of the smaller open balls contains the image of $z$ and has compact closure, since its closure lies in the finite union of the closed balls and is closed there. Put $V=M\setminus \bar{U}$, and $W=U\cup V$. These are open, with $U,V$ disjoint. Both $M\setminus V=\bar{U}$ and $M\setminus W=\bar{U}\setminus U$ are compact.

Fix a base point $x_0$ and one integral generator there, whose coefficient extension identifies the $R$ stalk with $R$ by step 1.1. A section's value $r$ is fixed by all loop transports. Conversely, for any such fixed $r$, define its value at $y$ by transport along any path from $x_0$ to $y$. Paths exist by step 1.2. Two choices give the same value because traversing one and reversing the other is a loop fixing $r$. Thus there is a unique prescribed value at each point, defining a function without choosing a path simultaneously for every point. On a ball its values are the local restrictions of one class, so the resulting section is continuous. This gives an $R$-linear bijection between sections and the common fixed submodule of the loop signs. It is $R$ if all signs are positive, and $R[2]$ if a negative sign occurs.

For the triple $V\subset W\subset M$, the quotient sequence $0⟶C_{\ast }(W;R)\mathrm{/}{C}_{\ast }(V;R)⟶C_{\ast }(M;R)\mathrm{/}{C}_{\ast }(V;R)⟶C_{\ast }(M;R)\mathrm{/}{C}_{\ast }(W;R)⟶0$ is degreewise exact: the first map is inclusion modulo $C_{\ast }(V)$, and the kernel of the last quotient consists exactly of the classes represented in $C_{\ast }(W)$. By [F4] it gives $H_{i+1}(M,W;R)⟶H_i(W,V;R)⟶H_i(M,V;R)\mathrm{.}$ Every singular simplex in $W$ lies in $U$ or $V$, since the inverse images of these disjoint open sets would otherwise separate its connected domain ([F7]). Thus $C_{\ast }(W)\mathrm{/}{C}_{\ast }(V)=C_{\ast }(U)$, compatibly with maps of the cycle $z$. By [F1] and the compactness in step 1.3, the left group is zero for $i\ge n$. Therefore $H_i(U;R)\to H_i(M,V;R)$ is injective for $i\ge n$. For $i>n$, its target also vanishes by [F1], so $z$ bounds already in $U$. This proves $H_i(M;R)=0$ for all $i>n$, whether or not $M$ is compact.

A section is an orientation exactly when its base-point value is a unit: all transports are module isomorphisms and [F3] characterizes generators as unit multiples of one generator. If the fixed submodule is $R$, it contains $1$ and $M$ is $R$-orientable. If it is $R[2]$ and contains a unit $u$, then $2u=0$ implies $2=0$ after multiplication by $u^{-1}$, so $R[2]=R$ and again $M$ is $R$-orientable. Consequently the fixed submodule is $R$ in the $R$-orientable case, and exactly $R[2]$ in the non-$R$-orientable case. No claim that the latter subgroup vanishes is justified without a restriction on $R$.

For compact $M$, any class $a\in H_n(M;R)$ yields a continuous section of point restrictions: over a coordinate ball, first restrict $a$ to its ball-supported group, and its subsequent point values form exactly a basic section of [F3]. By [F1] with support $M$, this map from classes to sections is bijective and all higher groups vanish. Evaluation at $x_0$ and steps 2.1–3.1 give the stated injective stalk map and its image. In the oriented case the section of generators corresponds to the fundamental class of [F3], and multiplying it by $r$ gives the class with local coefficient $r$.

Suppose $M$ is noncompact and $z$ is a degree-$n$ cycle. Its point restrictions define a continuous section just as in step 4.1. Outside its compact image the restriction is zero because the chain is in the omitted subspace. Such a point exists by noncompactness, and step 1.2 forces the section to be zero everywhere. By [F1], the image of $z$ in $H_n(M,V;R)$, supported on compact $\bar{U}$, is therefore zero. The injection of step 2.2 makes $[z]=0$ in $H_n(U;R)$ and hence in $H_n(M;R)$. This uses the triple injection, not just vanishing of all point restrictions.

A connected nonempty zero-manifold is one point, so is compact and has $H_0=R$, consistent with step 4.1. The zero ring has the unique orientation and every displayed module is zero; a nonorientable case does not arise for it. Characteristic two makes both signs act as identity. The two signs in step 1.1 include all monodromy possibilities, without a hidden assumption of integral orientability. Every global construction used unique values or a finite subcover; the base point and its generator are individual witnesses, not a family selected by AC. All chain calculations use unnormalized simplices.