Poincaré–Lefschetz duality for a disk

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.