Brief survey of duality in manifolds

Let $X$ be a topological manifold.

Theorem: Let $x$ be a point in $X$. There is a small open neighborhood $U$ of $x$ such that for every $x'$ in $U$, there is a homeomorphism of $X$ with itself carrying $x'$ to $x$.

Actually, we can say something substantially stronger, but it's a bit more subtle and harder to digest.
Theorem: Let $x$ be a point in $X$. There is a small open neighborhood $U$ of $x$, such that for each $x'\in U$, there exists a homeomorphism $f_{x'}$ from $X$ to itself, carrying $x'$ to $x$; moreover, this family of homeomorphisms can be chosen such that the disjoint union $f= \coprod_{x'\in U}f_{x'} : U\times X\to U\times X$ is a homeomorphism of $U\times X$ with itself, carrying $(x', y)$ to $(x',f_{x'}(y))$, and in particular sending $(x',x')$ to $(x',x)$. The proof is an elegant geometrical construction.

Let $\delta(U)\subset U\times X$ be the set of ordered pairs $(x,x)$ with $x\in U$. Then the homeomorphism $f : U\times X\cong U\times X$ is an automorphism of the trivial fiber bundle $\pi_1: U\times X\to U$, where $\pi_1(x,y)=x$, and $f$ carries $\delta(U)$ homeomorphically onto $x\times U$. We can thus say that $f$ is an isomorphism of the fiber bundle pair $(U\times X,U\times X-\delta (U))$ with $(U\times X,U\times (X-x))$, where both bundles are fibered over $U$.

Corollary: If $X$ is a connected manifold, then the group of homeomorphisms of $X$ with itself acts transitively on the points of $X$. Proof: Let $x,x'$ be any two points. Choose a path between them; cover it with finitely many opens where such an open $U$ exists. and apply compactness.

Corollary: If $X$ is connected, the homeomorphism type of the pair $(X,X-x)$ is independent of $x$.

Corollary: If $X$ is connected, the projection map $(X\times X, X\times X - \delta(X))\to X$ onto the first coordinate is a fiber bundle projection. Choose any basepoint $x_0\in X$; then the fiber of the bundle is $(X,X-x_0)$, which by the previous corollary is well-defined independently of the choice of basepoint; and by the main theorem from earlier, every point $x\in X$ admits an open neighborhood $U$ around which the fiber bundle trivializes, as $(U\times X,U\times X-\delta(U))\cong (U\times X, U\times (X-x))\cong (U\times X, U\times (X-x_0))$.

We call this fiber bundle the homology tangent bundle.

In cohomology theory there is a natural map (for any spaces $(X,A),(Y,B)$, and for any $n,q$) called the slant product map, taking the form
\begin{equation}
\label{eq:1}
H^n((X,A)\times (Y,B);R)\otimes H_q(Y,B;G)\to H^{n-q}(X,A;G)
\end{equation}
where $R$ is the ground ring, and $G$ is a module of $R$. Here, $(X,A)\times (Y,B)$ is a shorthand we introduce for the pair $(X\times Y, A\times Y\cup X\times B)$.

We will say that $X$ is oriented over the ring $R$ if there exists an element $u\in H^n(X\times X,X\times X-\delta(X);R)$ (called an orientation of $X$) which restricts in each fiber $H^n(x\times X, x\times (X-x);R)$ to a generator of that module (which is $0$ in dimensions other than $n$, and isomorphic to $R$ in dimension $n$, by an easy argument by excision and the LES in cohomology.)

Let $X$ be an oriented manifold, and let $u$ be an orientation. We fix both of these for the time being.

Let $A, B$ be a pair of subsets of $X$, $B\subset A$. Then $A\times (X-A)\subset X\times X-\delta(X)$, and so is $B\times (X-B)\subset X\times X-\delta(X)$, so $(X-B,X-A)\times (A,B)$ is a subspace of $(X\times X,X\times X-\delta(U)$. Let $u'$ denote the restriction of $u$ to this subspace.

The slant product map gives a map
\begin{equation}
\label{eq:2}
H^n((X-B,X-A)\times (A,B);R)\otimes H_q(X-B,X-A;G)\to H^{n-q}(A,B)
\end{equation}
Fixing the first argument to be $u'$ gives us a map
\begin{equation}
\label{eq:3}
\gamma_u : H_q(X-B,X-A;G)\to H^{n-q}(A,B)
\end{equation}
Notice that both $H_q(X-B,X-A;G)$ and $H^{n-q}(A,B)$ are both contravariantly functorial with respect to the pair $(A,B)$ under inclusion, i.e. if $A\subset A', B\subset B'$ there are natural maps $H^{n-q}(A',B')\to H^{n-q}(A,B)$ and $H_q(X-B',X-A')\to H_q(X-B,X-A)$.

Theorem: The maps $\gamma_u$ are natural with respect to the pair $(A,B)$.

Observation: Fix $(A,B)$ to be any pair. Let $\mathcal{S}$ be the directed set of all pairs of open sets $(U,V)$ with $V\subset U$ and $B\subset V, A\subset U$. Taking colimits of $H^{n-q}(X-V,X-U;G)$ and $H_q(U,V;G)$ over all such pairs $U,V$ gives a map between the colimits
\begin{equation}
\label{eq:4}
\overline{H}_q(X-B,X-A) \to \overline{H}^{n-q}(A,B)
\end{equation}
It is not hard to show that on the left hand side, the colimit is isomorphic to the true homology, as homology is compactly supported. So this map reduces to
\begin{equation}
\label{eq:5}
H_q(X-B,X-A;G)\to \overline{H}^{n-q}(A,B;G)
\end{equation}
Duality Theorem: If $(A,B)$ are compact, this map is an isomorphism.

If $X$ is itself compact, taking $A=X$ and $B =\emptyset$, it is not hard to see that $\overline{H}^{n-q}(X;G)\cong H^{n-q}(X;G)$, so this specializes to Poincare Duality:
\begin{equation}
\label{eq:6}
H_q(X;G)\cong H^{n-q}(X;G)
\end{equation}
It can be shown that if $X$ is $\mathbb{R}^n$, and $A$ is homeomorphic to a simplicial complex, $\overline{H}^{n-q}(A;G)\cong H^{n-q}(A;G)$. Therefore, applying the duality theorem in the case $(A,\emptyset)$, we get $H_q(\mathbb{R}^n;\mathbb{R}^n-A;G)\cong H^{n-q}(A;G)$. Applying the LES of the pair $(\mathbb{R}^n,\mathbb{R}^n-A)$ and using that $\mathbb{R}^n$ is contractible, we see that $H_q(\mathbb{R}^n,\mathbb{R}^n-A;G)\cong H_{q-1}(\mathbb{R}^n-A;G)$. Therefore we get Alexander Duality:
\begin{equation}
  \label{eq:7}
  H_{q-1}(\mathbb{R}^n-A;G)\cong H^{n-q}(A;G)
\end{equation}