Orientations Poincare Lefschetz and Alexander Duality — Examples

The spaces here satisfy the manifold hypotheses. The sphere is a closed bounded subset of ${\mathbb{R}}^{n+1}$, and the torus is the closed bounded subset of ${\mathbb{R}}^{2m}$ defined by one unit-circle equation in each coordinate pair; both are compact by [F9] and Hausdorff as metric subspaces. On the sphere the sets where one coordinate is strictly positive or strictly negative are graph charts over an open unit ball, with the omitted coordinate $\pm \sqrt{1-\mathrm{\mid }v{\mathrm{\mid }}^2}$. On the torus take products of open arc charts. Both have finite chart covers; pulling back rational Euclidean ball bases in these finitely many charts gives a countable base. The sphere for $n\ge 1$ is path connected: normalize the straight segment between nonantipodal points, and for antipodal points concatenate via any fixed perpendicular unit vector. Each circle is path connected by an arc, and finitely many such paths give paths in the product. The sphere carries its outward boundary orientation. Increasing angular coordinates on each circle, in factor order, give the product orientation; transitions between angular lifts are translations by integers and preserve it. At $m=0$ use the positive point, and $S^0$ is two open points with the two boundary signs.

By [F1], the only nonzero circle homology groups are $H_0=\mathbb{Z}$ and $H_1=\mathbb{Z}[z]$. The Ext terms in [F2] vanish: $0$ has the zero resolution and a finite free group has its identity augmentation as a length-zero free resolution, whose Hom has zero degree-one cohomology. Thus evaluation gives $H^0(S^1)=\mathbb{Z}1$, $H^1(S^1)=\mathbb{Z}u$ with $u(z)=1$, and no higher groups. In particular $u^2=0$. A point has $H^0=H_0=\mathbb{Z}$ and no higher groups: its unnormalized chain differential is identity in positive even degrees and zero in odd degrees, and dualizing has the corresponding zero positive cohomology.

For $n\ge 1$, take the alternating facet cycle of an oriented $(n+1)$-simplex containing the origin and transport it by radial projection to $S^n$. The radial projection is a homeomorphism from the simplex boundary onto the sphere: every ray meets that boundary once, and its piecewise radial inverse is continuous, with matching values at facet boundaries. The facet signs give precisely the boundary orientation. In the simplicial calculation underlying [F1], the alternating facet cycle is primitive: every simplicial top cycle has equal signed facet coefficients, there are no higher chains in the boundary complex, and comparison carries this generator to singular homology. It has positive local coefficient at the interior of any outward oriented facet, so its class and the fundamental class in [F6] have the same positive restriction at one such point and are equal by the injectivity in [F6], using step 1.1. For $S^0=\{-1,+1\}$ the outward endpoint signs of the interval give $[+1]-[-1]$ by [F6]'s componentwise definition. Both point classes are independent by [F1], so this element does not generate the entire $H_0$.

Inductively apply [F3] to $T^{m-1}\times S^1$, always using the last circle as the finite-free homology factor from step 1.2. Its cohomology is the graded tensor of the preceding ring with $\mathbb{Z}1\oplus \mathbb{Z}u$, $u^2=0$. Thus $x_I$ for increasing subsets $I$ is a basis; repeated factors square to zero, and interchanging distinct degree-one factors changes the sign. There are no groups above degree $m$. This is the exterior presentation: the tensor algebra modulo all squares maps to this ring, the relations sort and remove repeated generators, and the resulting increasing monomials have independent images by the tensor bases. Similarly [F4] gives a homology basis $b_J$ by induction. Its Tor terms vanish since both factors' homology groups are finite free, and their length-zero free resolutions tensor to complexes with no degree-one homology. The cycle representatives insert point cycles at the missing coordinates and $z$ at the others, using the prescribed shuffle. The initial induction case is the point in step 1.2.

For any two factor cycles $c,d$ and matching-degree cocycles $\phi ,\psi$, tensor evaluation $J$ is closed and [F5] gives $JAS(c\otimes d)=J(c\otimes d)+J(dL+Ld)(c\otimes d)=\phi (c)\psi (d)\mathrm{.}$ Indeed $Jd=0$ follows by the cocycle equations on the two tensor summands, and $d(c\otimes d)=0$ follows by the signed tensor boundary. Thus iteration, starting with $u(z)=1$, gives $⟨x_1\cdots x_m,b_{\{1,\dots ,m\}}⟩=1$. More generally, let $\mathrm{\mid }I\mathrm{\mid }=\mathrm{\mid }J\mathrm{\mid }$ and pull $x_I$ back to the coordinate subtorus indexed by $J$. By [F10], a factor $x_i$ with $i\mathrm{\notin }J$ pulls back to zero: its projection is constant and factors through a point, whose positive cohomology is zero by step 1.2. If $I=J$, the same iteration gives value one. As equal-size distinct subsets have an element of $I\setminus J$, this proves $⟨x_I,b_J⟩={\delta }_{IJ}(\mathrm{\mid }I\mathrm{\mid }=\mathrm{\mid }J\mathrm{\mid })\mathrm{.}$ Evaluation therefore detects every homology coefficient in this basis.

The full shuffle cycle representing $b_{\{1,\dots ,m\}}$ has the product orientation. To check its sign, decompose each circle into the signed arcs of $z$. The product of one arc from each circle is a cube with parameter order $1,\dots ,m$. Iterated shuffle divides it into the simplices whose vertex paths increment each coordinate once, in permutation order. The edge matrix for such a simplex has determinant equal to the sign of that permutation: subtract successive columns to get the ordered coordinate-unit columns. This is exactly its shuffle coefficient; induction on the last inserted coordinate gives the same sign for the left-associated shuffle. Multiplying by the arc-orientation signs consequently makes every signed simplex positive in the product orientation. Internal faces cancel by [F5], as do the outer faces from the circle cycles. At an interior point of any one simplex, the cycle has positive local coefficient one. By [F6] and the connected compact manifold verification in step 1.1, its class equals $[T^m]$. For $m=0$ the empty product is the positive point class and already satisfies [F6].

For a top-degree cochain $\theta$ and a top-dimensional cycle $c$, the augmentation of $\theta \cap c$ is $\theta (c)$: [F8] retains the last vertex with coefficient given by evaluation. Apply this to the cap/cup identity in [F8], with $\alpha =x_I$ and any $\beta \in H^{m-\mathrm{\mid }I\mathrm{\mid }}(T^m)$, to get $⟨\beta ,x_I\cap [T^m]⟩=⟨x_I\beta ,[T^m]⟩\mathrm{.}$ By step 2.2 the product $x_I{x}_J$ is zero if $I\cap J\mathrm{\ne }\mathrm{\varnothing }$. If $\mathrm{\mid }J\mathrm{\mid }=m-\mathrm{\mid }I\mathrm{\mid }$ and the intersection is empty, then $J=I^c$. Sorting the concatenation of the increasing lists $I,I^c$ to $1,\dots ,m$ requires one interchange for each element of $I^c$ less than an element $i_r$ of $I$. There are $i_r-r$ such elements for $i_r$, so $x_I{x}_{I^c}=(-1{)}^{{\sum }_r(i_r-r)}{x}_1\cdots x_m\mathrm{.}$ Steps 3.1 and 3.2 then make the right side evaluate to that sign. Since step 3.1 detects homology coefficients, the cap image is exactly the displayed signed complementary basis element.

By [F7] and step 1.1, these cap maps are the Poincaré-duality isomorphisms on each compact oriented torus. The explicit matrix in step 4.1 also shows bijectivity directly, since complementation permutes the finite bases and every coefficient is $\pm 1$. On $S^n$ for $n\ge 1$, step 1.2's length-zero resolution argument with [F1] and [F2] gives a normalized top class $v$ evaluating to one on $[S^n]$ and no intermediate cohomology. Thus $1\cap [S^n]=[S^n]$ and $v\cap [S^n]$ is the positive point class by [F8]'s augmentation calculation and $H_0=\mathbb{Z}$. On $S^0$, a zero-cocycle with values $(a_{+},a_{-})$ caps to $a_{+}[+1]-a_{-}[-1]$, an isomorphism of the two degree-zero groups.

For $I=\mathrm{\varnothing }$ or $I=\{1,\dots ,m\}$ the exponent is zero, giving respectively the whole fundamental class or the positive basepoint class. At $m=1$ these are the only two cases. At $m=2$, the formula says $x_1\cap [T^2]=b_{\{2\}}$ and $x_2\cap [T^2]=-b_{\{1\}}$, fixing the cohomology-first sign convention. The $m=0$ and $n=0$ cases were computed separately, and zero classes have zero images by bilinearity. Empty spaces and zero coefficient rings are outside this fixed integral example. Every chain calculation retains unnormalized degeneracies. AC in [F11] supplies the arbitrary-rank PID cycles, projections and sections for [F2]–[F4], the simultaneous finite-free homology sections/bases for [F3], and the coordinate-neighborhood selections and local UCT lifts used by [F7]. The displayed basis evaluation, permutation signs and cap computation themselves add no choice use.

The cycle $Z$ of [F1] represents $[{\mathrm{\Sigma }}_g]$. For $g\ge 1$, its signed fan is the positive characteristic-disk cycle, and at a point in the interior of one fan triangle its local coefficient is $+1$ in the given orientation: the affine triangle has the polygon orientation when its coefficient is $+1$, and the opposite vertex order and coefficient $-1$ compensate on the other triangles. Restriction to that point discards the other triangles, whose images omit it. Thus $[Z]$ and the class of [F2] have the same local restriction. By [F3] they are equal. For $g=0$, the positive characteristic two-cell used in [F1] has the same normalization at an interior point, giving the same conclusion. The hypotheses of [F3] hold since ${\mathrm{\Sigma }}_g$ is the given nonempty connected closed oriented surface.

Substituting step 1.1 into [F1]'s calculation gives $I(\alpha ,\beta )={\sum }_{i=1}^g(r_i{s}_i^{\mathrm{\prime }}-s_i{r}_i^{\mathrm{\prime }})\mathrm{.}$ Consequently $I(a_i,a_j)=I(b_i,b_j)=0$, $I(a_i,b_j)={\delta }_{ij}$ and $I(b_i,a_j)=-{\delta }_{ij}$. These are exactly the entries of the stated block matrix $M=\mathrm{diag}(J,\dots ,J)$. Setting $\beta =\alpha$ gives $I(\alpha ,\alpha )={\sum }_i(r_i{s}_i-s_i{r}_i)=0$ for every integral vector, proving alternation directly, including vectors with negative or zero coordinates.

Direct multiplication gives $J^2=-I_2$, so $M^{-1}=-M$ is an integral matrix. In particular the map from the basis module to its dual defined by either slot of the pairing has an integral inverse (the other slot uses the transpose matrix). Also $\det J=0\cdot 0-1\cdot (-1)=1$, hence $\det M=1$. This proves unimodularity over $\mathbb{Z}$, not only nonsingularity over a field.

At $g=0$ the module $H^1$ is zero and $M$ is the empty matrix. Its determinant is the empty product $1$, and the unique map $0\to 0$ is an isomorphism, so both conclusions still hold. At $g=1$ there is one block $J$, already checked in step 3.1. Orientation reversal sends the fundamental class to its negative by [F2], so for a fixed basis the pairing and matrix change sign. No assertion concerns an empty surface or a zero coefficient ring, since the example fixes a connected surface and integral coefficients. Representatives and degenerate singular simplices are handled by the actual cocycle and cycle calculation in [F1]; the matrix computation does not replace that argument. AC is used only in [F1]'s UCT construction via [F4]; the fundamental-class identification and finite matrix inversion need no additional choice.

For $n\ge 1$ the disk is compact by [F9], Hausdorff and second countable with its Euclidean subspace topology. Interior points have Euclidean charts. At a boundary point, rotate that point to the last positive coordinate axis and write nearby points as $(u,\sqrt{1-\mathrm{\mid }u{\mathrm{\mid }}^2}-t)$ with $\mathrm{\mid }u\mathrm{\mid }$ small and $0\le t$ small. The inverse is $(y^{\mathrm{\prime }},y_n)\mapsto (y^{\mathrm{\prime }},\sqrt{1-\mathrm{\mid }{y}^{\mathrm{\prime }}{\mathrm{\mid }}^2}-y_n)$; after restricting to a small open neighborhood the disk condition is exactly $t\ge 0$, since the opposite graph is bounded away. Thus these are half-space charts and its boundary is $S^{n-1}$. The supplied standard orientation on the interior gives the orientation required by [F1] and [F2]. The map $(x,s)\mapsto (1-s)x$ contracts $D^n$ to $0$. By [F6], $H_k(D^n;\mathbb{Z})$ is $\mathbb{Z}$ in degree zero and zero in positive degrees: at a point the unique simplex in degree $k>0$ has boundary coefficient ${\sum }_{j=0}^k(-1{)}^j$, alternately $1$ and $0$, giving this calculation. In particular augmentation identifies the class of any disk point with $1$. The case $n=0$ has these same groups directly.

Straight segments join any two disk points. By [F7], a zero-cocycle has equal values at their endpoints and hence is constant; conversely constants are cocycles. Since there are no degree-zero coboundaries, $H^0(D^n;\mathbb{Z})=\mathbb{Z}$ with generator the constant $1$. Apply the first isomorphism of [F2] at $p=0$. The formula [F8] evaluates this constant on the initial vertex of each simplex and retains the entire simplex, so $1\cap z=z$. Consequently $H_n(D^n,A;\mathbb{Z})$ is generated by $z$. This also proves its positive normalization by the supplied interior orientation of [F1].

For $n\ge 2$, the pair sequence [F3] and the disk calculation in step 1.1 identify the connector $H_n(D^n,A)\to {\tilde{H}}_{n-1}(A)$ as an isomorphism; both groups are $\mathbb{Z}$ by [F4] and step 2.1. For $n=1$, the exact sequence instead identifies $H_1(D^1,S^0)$ with the kernel of $H_0(S^0)\to H_0(D^1)$. This map sends $(a,b)$ to $a+b$, so its kernel is generated by $[+1]-[-1]$. The oriented interval simplex $s\mapsto 2s-1$ has exactly that boundary and represents the positive interior local generator. By [F1] it represents $z$. In every dimension [F1] identifies the connector with the outward-normal-first boundary orientation, so $\mathrm{\partial }z=[A]$ with the stated signs. At $n=0$ its target is a negative homology group and the empty boundary class is zero.

Apply the second isomorphism of [F2] with $p=n$, using $H_0(D^n)=\mathbb{Z}$ from step 1.1. Thus $H^n(D^n,A;\mathbb{Z})$ is infinite cyclic. Its generator with cap image the origin is characterized by evaluation $1$ on $z$, not by an unspecified sign choice. Indeed, for a relative cocycle $\phi$ and a relative cycle $c={\sum }_j{a}_j{\sigma }_j$ representing $z$, [F8] gives the zero-chain $\phi \cap c={\sum }_j{a}_j\phi ({\sigma }_j)[{\sigma }_j(v_n)]$. Its augmentation is ${\sum }_j{a}_j\phi ({\sigma }_j)=\phi (c)$. Therefore the cap isomorphism followed by augmentation is precisely evaluation on $z$, proving existence and uniqueness of the normalized class and its asserted image.

When $n=0$ both cap maps are the identity of $\mathbb{Z}$ for the positive point orientation, so the formulas agree. The boundary is empty only in that case; no undefined sphere homology in degree $-1$ is invoked. The disconnected two-point boundary at $n=1$ was treated by the augmentation kernel in step 3.1. Zero cohomology classes map to zero by the displayed linear formulas, and degenerate singular simplices are included in the augmentation computation in step 3.2. The straight contraction has the required time endpoints, and the relative boundary signs are those of [F1], with the explicit interval calculation fixing the low-dimensional convention. AC in steps 2.1 and 3.2 is only [F2]'s local free-module/UCT selections and countable coordinate exhaustion; all disk charts, chains, signs and the normalized generator calculation require no further selection.

The sphere is compact by [F8], so $P$ is compact: pull an open cover back under the continuous surjection $q$ and project a finite subcover. The map $q$ is open because $q^{-1}q(V)=V\cup (-V)$ for open $V\subset S^n$. Distinct orbits $\{x,-x\}$ and $\{y,-y\}$ have positive minimum distance $d$, the minimum of finitely many positive distances. The unions of radius-$d\mathrm{/}3$ balls about the points in the two orbits are disjoint, open and antipodally invariant. Their quotient images are disjoint open neighborhoods. Hence $P$ is Hausdorff. For each $0\le i\le n$, the open subset $U_i=\{[x]:x_i\mathrm{\ne }0\}$ has the affine chart $[x]\mapsto (x_j\mathrm{/}{x}_i{)}_{j\mathrm{\ne }i}$ onto ${\mathbb{R}}^n$. Ratios are continuous and invariant on the sphere preimage, so descend continuously through the open quotient restriction. The inverse sends $u$ to the orbit of $(u_0,\dots ,1,\dots ,u_n)\mathrm{/}\sqrt{1+\mathrm{\mid }u{\mathrm{\mid }}^2}$, with the $1$ in coordinate $i$. These are continuous inverse maps. For $n=0$ this is the one-point chart and its singleton basis. For $n>0$ the pullbacks of rational balls in these finitely many charts form a countable basis: any open neighborhood intersects one of the covering charts, whose rational-ball basis refines it. Thus [F1] and [F9] make $P$ a boundaryless $n$-manifold.

This manifold is nonempty and connected. For $n=0$ both sphere points form one orbit, so $P$ is a point. For $n\ge 1$, any two sphere points $x,y$ not antipodal are joined by the path $((1-t)x+ty)\mathrm{/}\mathrm{\mid }(1-t)x+ty\mathrm{\mid }$, whose denominator cannot vanish unless $y=-x$. If they are antipodal, choose a unit vector $v$ perpendicular to $x$ and concatenate such paths through $v$. Such a vector exists explicitly: take a standard basis vector not parallel to $x$, subtract its projection on $x$ and normalize; at least one of the $n+1\ge 2$ standard vectors is not parallel. These paths project to paths between all projective points, proving path connectedness and hence connectedness by [F12]. No family of path choices is needed. Apply [F2] to the manifold in step 1.1: at every stalk take its unique nonzero mod-two value, which is continuous in every local trivialization. By [F11], compactness then gives its canonical fundamental class $[P{]}_2$. These steps require no AC.

By [F4], the integral cellular groups have one oriented generator in each degree $0,\dots ,n$, with positive-degree differentials alternating $0,2$. Reduction of chain coefficients modulo two commutes with boundary, relative quotient, and the connecting-map formula taking a relative cycle to its boundary. By [F10] it sends each integral characteristic-disk generator to the coefficient-one generator of the corresponding ${\mathbb{F}}_2$ cellular group. Thus the cellular differentials modulo two are the reductions of $0,2$, both zero. Therefore each cellular homology group in degrees $0,\dots ,n$ is ${\mathbb{F}}_2$, and the groups in all other degrees are zero. By [F5] these are the singular homology groups. In top degree $[P{]}_2$ is their nonzero generator because by step 2.1 its point restriction is nonzero.

Let $n>0$ be even. The top integral cellular differential in [F4] is multiplication by $2:\mathbb{Z}\to \mathbb{Z}$, which has zero kernel, and there is no cell above it. Thus [F5] gives $H_n(P;\mathbb{Z})=0$. If $P$ were integrally orientable, [F7], applied using compactness and connectedness from steps 1.1–2.1, would make its top restriction image the whole local group $\mathbb{Z}$, contradicting its zero domain. This proves integral nonorientability without assuming it from a picture of transition signs. The mod-two orientation and class of step 2.1 still exist. At $n=0$, $P$ is a point, so the multiplication-by-two argument has no positive differential to apply to; it is integrally orientable.

Assume [F6]. The compact boundaryless manifold and orientation in steps 1.1–2.1 satisfy exactly the hypotheses of [F3]. It gives the stated cap isomorphism, since compact support is ordinary cohomology on compact $P$. Step 3.1 then computes its domain as well as its codomain: both are ${\mathbb{F}}_2$ for $0\le p\le n$ and zero otherwise. Over ${\mathbb{F}}_2$ the isomorphism between one-dimensional spaces sends the unique nonzero class to the unique nonzero class, so this is a concrete complementary-degree pairing without any sign choice. In particular $1\cap [P{]}_2=[P{]}_2$ and the nonzero top cohomology class caps to the point generator of $H_0$.

The assertions include $n=1$, with zero cellular differential in degrees zero and one and cap interchanging the two nonzero degrees. Empty spaces and zero coefficient rings are not inputs here; the nonempty quotient and the specific field ${\mathbb{F}}_2$ were fixed. Degrees outside $0,\dots ,n$ are zero on both sides by step 4.1, including negative degrees. Collapsed or repeated boundary points in the attaching maps are included in [F4]'s quotient construction, and singular degeneracies remain in the coefficient comparison of step 3.1. The paths in step 2.1 have their prescribed endpoints, with the non-antipodal denominator check preventing a singularity. Only step 4.1 invokes AC, precisely the countable chart-neighborhood and local UCT free-module/projection/lift choices of [F3]; compactness, canonical mod-two orientation, fundamental-class existence and the integral nonorientability calculation above are choice-free.

The equator is nonempty, since it contains $((1,0,\dots ,0),0)$, and proper, since it misses the two poles $p_{\pm }=(0,\pm 1)$. It is closed and bounded in Euclidean space, hence compact by [F5]. It is weakly locally contractible. For $n=1$ it consists of two isolated points. For $n>1$, at a point of $E$ some coordinate $u_j$ is nonzero; restrict to its fixed sign and solve $u_j=\pm \sqrt{1-{\sum }_{k\mathrm{\ne }j}{u}_k^2}$. Projection to the remaining coordinates is a homeomorphism onto an open subset of ${\mathbb{R}}^{n-1}$. Inside any prescribed neighborhood a small ball in these coordinates contracts within it by straight segments. Thus all hypotheses of [F1] hold.

Put $U_{\pm }=\{(u,t)\in S^n:\pm t>0\}$. They are disjoint open sets whose union is the complement of $E$. Projection $(u,t)\mapsto u$ identifies each with the open unit ball $B^n$, with inverse $u\mapsto (u,\pm \sqrt{1-\mathrm{\mid }u{\mathrm{\mid }}^2})$. These formulas are continuous inverses. Contract $B^n$ by $u\mapsto (1-s)u$, which stays in the ball for all $0\le s\le 1$. Transporting this contraction gives a contraction of $U_{\pm }$ to $p_{\pm }$. No path joins the two hemispheres: its last coordinate would change from positive to negative and take zero by [F6]. Hence these are exactly the two path components.

By [F2], $H_0(S^n\setminus E;\mathbb{Z})$ is $\mathbb{Z}[p_{+}]\oplus \mathbb{Z}[p_{-}]$. Its augmentation sends $(a,b)$ to $a+b$, so its kernel consists of $(a,-a)$ and is generated by $[p_{+}]-[p_{-}]$. Thus reduced $H_0$ is $\mathbb{Z}$. In every positive degree the singular chain complex splits as the direct sum of the two hemisphere complexes. To see this on generators, a simplex cannot meet both hemispheres: restrict its last-coordinate function to a straight segment between two preimages of opposite signs and apply [F6]. Each simplex therefore lies in exactly one hemisphere, and every face stays there, proving the direct-sum assertion degreewise and for the differential. Kernels and images in this two-summand complex split componentwise. Both hemisphere homology groups vanish in positive degree by [F4] and step 1.2: the point complex has one generator in each degree with boundary alternating between identity and zero. Therefore all positive reduced homology of the complement is zero. Negative reduced groups are zero by the convention in [F1].

With AC supplied by [F3], apply [F1] to the hypotheses verified in step 1.1. It identifies each group computed in step 2.1 with ${\tilde{H}}^{n-i-1}(E;\mathbb{Z})$, proving the displayed formula. In particular this also computes the equator's reduced integral cohomology: it is $\mathbb{Z}$ exactly in degree $n-1$. The duality image of $[p_{+}]-[p_{-}]$ is a generator there; its sign is the one determined by the fixed ambient orientation and the duality construction. No additional sign convention on an independently chosen equator generator is silently asserted.

At $n=1$, the two hemispheres are open semicircles, and the equator is $S^0$; its reduced degree-zero cohomology is the quotient of ${\mathbb{Z}}^2$ by constant pairs, as in [F1], so the same duality conclusion holds. The excluded case $n=0$ would have empty equator and does not satisfy [F1]'s nonempty hypothesis. Neither hemisphere is empty: its specified pole is its contraction endpoint. The contraction at time zero is the identity and at time one is the pole map. Zero homology degrees and zero input classes were accounted for in step 2.1; degenerate simplices stay in their hemisphere just as other simplices do. Every chart and contraction here is explicit, and only [F1]'s atlas/UCT and controlled-extension choices invoke AC.

Its rational cellular chain group $C_n=H_n(X^n,X^{n-1};\mathbb{Q})$ has dimension $c_n$ by [F4]. There are finitely many nonzero groups, since $X$ has finitely many cells. The identity map preserves every skeleton and induces the identity map of each relative group. Its cellular chain trace in degree $n$ is therefore the sum of the $c_n$ diagonal entries equal to one, namely $c_n$, with trace zero if $c_n=0$. By [F5], the induced homology map is the singular homology identity.

Apply [F3] to the identity chain map in step 1.1. Its homology trace sum is $L({\mathrm{i}\mathrm{d}}_X)$ by [F1], and its chain trace sum is ${\sum }_n(-1{)}^n{c}_n=\chi (X)$ by [F2]. This proves the equality over $\mathbb{Q}$ directly; no change-of-coefficients identification with integral ranks is needed.

If $X$ is nonempty and contractible, [F6] gives its rational homology as that of a point. For completeness, the singular chain group of a point is $\mathbb{Q}$ in each nonnegative degree, with its unique simplex as basis. In positive degree $n$ the boundary is multiplication by ${\sum }_{j=0}^n(-1{)}^j$, which is $1$ for even $n$ and $0$ for odd $n$. Thus in every positive degree the kernel equals the image from the next degree; in degree zero the boundary from degree one is zero. Consequently only $H_0=\mathbb{Q}$ survives. The identity therefore has trace $1$ in degree zero and no other nonzero traces, so [F1] gives $L=1$, and step 2.1 gives $\chi =1$. Every $x\in X$ satisfies ${\mathrm{i}\mathrm{d}}_X(x)=x$; nonemptiness supplies a fixed point without any choice family. This is consistent with the nonzero-Lefschetz sufficient condition, but the identity's fixed point is already explicit.

If $X$ is empty, there are no cells or singular simplices, so both sums are zero. If $X$ is a singleton, step 3.1 gives $1$; for a finite discrete space the same degree-zero identity matrix has one diagonal $1$ per point. Missing dimensions and zero cellular groups are covered by step 1.1. Degenerate singular simplices at a point are exactly the generators used in step 3.1 and do not produce unwanted higher homology. The proof is a finite alternating sum, with no infinite endpoint or limiting convention; the boundedness required by [F3] was verified in step 1.1. All basis choices in that theorem are finite, so this example is choice-free.

Apply the choice-free part of [F1] with $n=2$. It makes $P$ a nonempty compact connected boundaryless $2$-manifold and supplies a canonical mod-two fundamental class. In the quotient model a specified point is $x=[1:0:0]$. By [F4], its local integral top-homology group is infinite cyclic and therefore has nonzero generators. The same holds at every point.

The integral cellular chain complex of [F2] is $0⟶\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\stackrel{0}{\to }\mathbb{Z}⟶0,$ with the three nonzero terms in degrees $2,1,0$. Multiplication by two is injective on $\mathbb{Z}$, so the degree-two cycle group is zero. There is no degree-three cell and hence no incoming boundary. Therefore [F3] gives $H_2(P;\mathbb{Z})=\ker (2)\mathrm{/}0=0.$ For clarity the same complex gives $H_1(P;\mathbb{Z})=\mathbb{Z}\mathrm{/}2$ and $H_0(P;\mathbb{Z})=\mathbb{Z}$; the failure is in the required top degree, not in connectedness.

Every homomorphism from the zero group $H_2(P;\mathbb{Z})$ has image zero. In particular restriction at the explicit point $x$ of step 1.1 cannot hit either generator of its infinite cyclic stalk. Thus no global integral class can have the required generator restrictions at all points. If $P$ had an integral orientation, [F5] with the hypotheses in step 1.1 would make that restriction onto, contradicting step 1.2. Hence $P$ is nonorientable and witnesses the failure of an ordinary integral fundamental class without the orientation hypothesis.

Modulo two, the differential $2$ in step 1.2 becomes zero, as established through characteristic coefficient maps in [F2]; the entire top cellular group survives as $H_2(P;{\mathbb{F}}_2)={\mathbb{F}}_2$. The canonical class from step 1.1 restricts to the nonzero mod-two local generator, so cannot be zero. It is therefore the unique nonzero element of this one-dimensional group. This is a change of coefficients, not an integral generator hidden by an orientation convention.

The example uses the positive even dimension two, not the zero-dimensional point ${\mathbb{R}\mathbb{P}}^0$, which is orientable. The zero integral top group in step 1.2 is compared with a nonzero stalk at a specified point, so the failure is not vacuous. The two endpoints of the one-cell attach to the sole zero-cell, giving $d_1=0$, while the top attaching incidence is $2$, not zero; both are covered by [F2]'s full attaching calculation. Degenerate singular chains are included in the comparison [F3]. No top cap isomorphism or field-cohomology dualization is used, and every clause of [F1] and [F2] invoked above is explicitly choice-free.

For any $x,y\in X$, the straight path $\gamma (t)=(1-t)x+ty$ joins them. By [F4], a degree-zero cocycle $\phi$ must satisfy $\phi (y)-\phi (x)=0$, so it is constant. Conversely a constant function has zero endpoint difference on every path, so is a cocycle. There are no degree-zero coboundaries by [F3]. Evaluation at $0$ and the assignment of the constant function with value $a\in R$ are inverse $R$-linear maps between $H^0(X;R)$ and $R$. In particular the constant cocycle $1$ represents a nonzero class because $1\mathrm{\ne }0$.

The homotopy $H(x,t)=(1-t)x$ contracts nonempty $X$ to its origin, so [F2] identifies its homology with that of a point. At a point there is one singular simplex in each degree, and its boundary in degree $k>0$ is multiplication by ${\sum }_{j=0}^k(-1{)}^j$, equal to $1$ for even $k$ and $0$ for odd $k$. Therefore every positive-degree kernel equals the next boundary image, and $H_k(\ast ;R)=0$ for $k>0$, while $H_0(\ast ;R)=R$. Since $n>0$, this proves $H_n(X;R)=0$ and $H_0(X;R)=R$, the latter with the origin's point class as generator.

An isomorphism $H^0(X;R)\to H_n(X;R)$ cannot exist: by step 1.1 its domain contains the nonzero constant class $1$, while by step 1.2 every element of its codomain is zero, so every homomorphism has that nonzero class in its kernel. This is precisely the degree-zero failure of the proposed replacement of $H_c^p$ by $H^p$ in dimension $n$. The straight-path and point-chain computations used no AC.

Assume now [F5]. The space $X$ is the oriented Euclidean coordinate ball of [F1], which therefore supplies $H_c^p(X;R)=0$ for $p\mathrm{\ne }n$ and the stated normalized cap isomorphism in degree $n$. In particular $H_c^0(X;R)=0$ since $n>0$, so the correct degree-zero duality is $0\to 0$, consistent with step 1.2. In degree $n$, evaluation equal to $1$ on the oriented relative class maps to the origin's generator of $H_0$ by [F1]. The AC use is the freeness/projection and free-comparison-lift selections in that lemma's local UCT proof; it is not needed for step 2.1.

The hypotheses exclude both $n=0$ and $R=0$ for a reason: at $n=0$ the space is a point and both ordinary degree-zero groups equal $R$; over the zero ring both groups in the claimed mismatch are zero. The chosen witness space is nonempty, with the specified origin, and $n=1$ is already a counterexample with the same contraction. The contraction has endpoints the identity and the constant map; the straight paths have their stated endpoints. The point-chain calculation includes degenerate simplices in every positive degree. No infinite selection of points, paths or cocycle representatives is used: all paths and the constant cocycles are given by formulas.

First prove the puncturing fact needed twice below. If $W$ is a connected Hausdorff $3$-manifold without boundary and $p\in W$, its coordinate balls are path connected. Thus its path components are open, and connectedness makes $W$ path connected. For two points distinct from $p$, choose a path between them and a small closed coordinate ball about $p$ contained in a chart and avoiding both endpoints. If the path meets a still smaller concentric closed ball, its closed preimage in $[0,1]$ has first and last points, by compactness in [F8]. Their images lie on the boundary sphere, since the endpoints of the whole path lie outside the ball. Replace the intervening segment by a path on that sphere. Such paths are explicit normalized straight segments between non-antipodal points; for antipodal points insert a unit vector off their line and concatenate. The replacement, as well as the earlier and later path portions, misses $p$. If the smaller ball is never met, the original path already misses $p$. Thus $U=W\setminus \{p\}$ is path connected.

Take an open coordinate ball $V\subset W$ centered at $p$. The sets $U,V$ are open, cover $W$, and are path connected by step 1.1 and convex coordinates. Their overlap is a punctured open ball, homeomorphic to punctured ${\mathbb{R}}^3$ by the radial map $x\mapsto x\mathrm{/}(r-\mathrm{\mid }x\mathrm{\mid })$ for a radius-$r$ ball. Hence it is path connected with trivial fundamental group by [F6]. The ball $V$ contracts linearly to any chosen basepoint, so its fundamental group is trivial. By [F5] the inclusion $U\hookrightarrow W$ induces an isomorphism on fundamental groups, initially based in $U\cap V$: the pushout of $G$ and the trivial group over the trivial group is $G$, as the universal property verifies. For any other prescribed basepoint in $U$, choose a path to the overlap. Conjugating loops by that path changes basepoint and commutes with inclusion; the reversed path gives its inverse since a path followed by its reverse contracts by linear retracing of its parameter. Thus inclusion induces the same isomorphism at every basepoint of $U$. This proves both directions of the puncturing comparison, rather than just surjectivity.

In $S^3=\{(z,w)\in {\mathbb{C}}^2:\mathrm{\mid }z{\mathrm{\mid }}^2+\mathrm{\mid }w{\mathrm{\mid }}^2=1\}$ take $X_0=\{\mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}\sqrt{2}\}$. The parametrization $(z,u)\mapsto (z,u\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2})$ identifies it with a closed disk times $S^1$. Its spherical exterior $W_0=\{\mathrm{\mid }z\mathrm{\mid }>1\mathrm{/}\sqrt{2}\}$ has coordinates $(u,w)\mapsto (u\sqrt{1-\mathrm{\mid }w{\mathrm{\mid }}^2},w)$ with $u\in S^1$, $\mathrm{\mid }w\mathrm{\mid }<1\mathrm{/}\sqrt{2}$. Contracting the open disk factor to a chosen small positive real $w_0$ is a deformation retraction onto that circle, so [F7] gives ${\pi }_1(W_0)\cong \mathbb{Z}$. Choose the point at infinity to be $(1,0)\in W_0$ and the loop $\mu (t)=(\sqrt{1-w_0^2}{e}^{2\pi it},w_0)$, which avoids it. This loop is exactly a meridian of $X_0$ pushed into the exterior: varying the $z$ phase goes around the boundary of the disk factor at fixed $w$ phase, and decreasing $\mathrm{\mid }w\mathrm{\mid }$ slightly pushes it outside the torus. It represents $1$ under the retraction calculation. The puncturing result in step 2.1 shows it is a generator of ${\pi }_1(W_0\setminus \{\mathrm{\infty }\})$. Stereographic projection, with the explicit inverse used in [F3], identifies this with the Euclidean exterior $E_0={\mathbb{R}}^3\setminus X_0$. Dilate the compact image of $X_0$ to diameter at most one. This changes none of these groups or meridian markings and supplies the initial torus permitted by [F4].

Perform [F4]'s exact recursion with this initial meridian and a root meridional slice at the chosen $w$ phase. Set $E_n={\mathbb{R}}^3\setminus X_n$. Inductively its fundamental group is free on the terminal meridians, initially the singleton basis $\{\mu \}$ by step 3.1, and it is path connected. To pass to the next level, process its finitely many disjoint slices lexicographically. The old annular generator is the corresponding pushed torus meridian in those slice coordinates, with its previously chosen whisker. [F4] checks the actual two-sided collar, the cap-only intersection with the remainder, and the preservation of old exterior paths. Hence all hypotheses of [F3], including the free-basis hypothesis furnished by this induction, apply. Its new group is free on the retained generators together with the two new child meridians, and the actual inclusion fixes the retained generators and sends the parent to their marked commutator. The same collared cover preserves path connectedness. The next prepared child slices have these same meridians and transported paths, so the induction continues. Composing the finitely many injective homomorphisms from [F3] proves that each ${\pi }_1(E_n)\to {\pi }_1(E_{n+1})$ is injective. Its reduced-word proof covers empty retained alphabet, powers of one generator, inversions, and recorded whisker conjugations; no abelian linking invariant is substituted for it. In particular the fixed loop $\mu \subset E_0$ represents a nonidentity element in every ${\pi }_1(E_n)$.

Let $B={\bigcap }_n{X}_n=f(D)$ as in [F4], and $E={\mathbb{R}}^3\setminus B$. The $E_n$ are increasing open sets with union $E$: failure to belong to the intersection means failure at some finite index. Each contains $E_0$, so their union is path connected by step 4.1; any two points lie together in some $E_n$. If $\mu$ had a based nullhomotopy in $E$, its continuous image from $[0,1{]}^2$ would be compact by [F8]. The open cover by all $E_n$ would have a finite subcover, and its largest index would contain the entire nullhomotopy. This contradicts step 4.1. The same reasoning applies to any proposed contracting disk. Thus ${\pi }_1(E,\mu (0))$ is nontrivial. The first commutator substitution has zero abelianization, which explains why a linking-number obstruction would not establish this conclusion.

Put $\mathrm{\Sigma }=\mathrm{\partial }B=f(\mathrm{\partial }D)$, the embedded two-sphere furnished by [F4], now regarded inside $S^3$. There is a disjoint partition $S^3\setminus \mathrm{\Sigma }=f(\mathrm{int}D)\bigsqcup W,W=S^3\setminus B\mathrm{.}$ Both sets are open and nonempty: [F4] identifies $f(\mathrm{int}D)=\mathrm{int}B$, and $W$ contains infinity. The first is path connected and homeomorphic to an open ball. For the latter assertion, the radial ball parametrization in [F4] maps interiors to interiors by its explicit radial formula. Also $W=E\cup \{\mathrm{\infty }\}$ is path connected: $E$ is path connected by step 5.1, and a coordinate ball about infinity missing the compact $B$ meets $E$ and joins infinity to it. By [F1] there are exactly two components of $S^3\setminus \mathrm{\Sigma }$, so these two nonempty path-connected pieces are precisely them. The open set $W$ is a connected Hausdorff $3$-manifold. Apply step 2.1 with $p=\mathrm{\infty }$ to obtain the inclusion-induced isomorphism ${\pi }_1(E,\mu (0))\cong {\pi }_1(W,\mu (0))$. By step 5.1 this group is nontrivial. An open $3$-ball contracts linearly to any basepoint, so every based loop contracts there; a homeomorphism transfers such a contraction. Therefore $W$ is not an open $3$-ball, proving the failed conclusion for the promised spherical complement.

All finite exteriors and both final components are nonempty; the meridian is nontrivial already at stage zero, and injectivity preserves its nonzero powers. The sphere embedding, including its boundary points, is established by the finite inverse-control proof in [F4], not by a drawing or a limit of injective maps alone. The point at infinity was treated by a proved isomorphism, not by assuming that adding one point preserves fundamental groups. The only AC uses are those of [F2]: compatible recursive map selection and invariance of domain in [F4], and the duality hypothesis in [F1]. The finite group calculations and single compact-nullhomotopy argument require no further choice. This gives the witness and the exact failed ball conclusion while preserving the two-component separation conclusion.

The map $f$ is continuous, since the inverse image of every open set is itself. Its map on every singular chain is the identity: composing a singular simplex with $f$ changes nothing. Consequently its map on each homology group is the identity. By [F2], the only nonzero rational homology groups are the two one-dimensional groups in degrees zero and one. Their identity matrices are each $(1)$, with trace $1$. All other homology endomorphisms are on zero spaces and have trace zero.

Substitution into [F1] gives $L(f)=(-1{)}^0\cdot 1+(-1{)}^1\cdot 1=1-1=0.$ On the other hand $f(x)=x$ for every $x\in S^1$, and $(1,0)\in S^1$ is an explicit fixed point. Thus the premise of the proposed implication holds and its conclusion fails. The fixed-point set is the whole circle.

The witness is nonempty and connected, so the failure is not due to an empty space or a disconnected-component convention. The zero value in step 2.1 is cancellation of two traces equal to one, not the vanishing of the homology groups. The sole one-cell's two endpoints are attached to the same vertex, giving a valid nonregular CW structure; the calculation uses singular homology and includes degenerate singular simplices. No limiting, relative, orientation or homotopy-endpoint choice is involved, and no AC is used. The fixed-point theorem asserts the different implication from nonzero Lefschetz number to a fixed point, so this example does not contradict it.