# Quail: Cl. mnfld. story

I did this around 1/5/09.  Here $M$ is always an $n$-manifold. Here orientation always refers with respect to some ring $R$.  Some generalities$H_n(M|A):= H_n(M,M - A)$ is shorthand for local homology.  For $B \subset A \subset X$ there’s an inclusion $(X, X - A) \to (X, X - B)$ giving natural maps $H_n(M|A) \to H_n(M|B)$.

For me one of the big motivations for this story is the following result.

If $M$ is closed and connected then $M$ is compact. This is an easy corollary of

Orientation result:

a) $M$ orientable implies $H_n(M) \to H_n(M|x)$ is iso.

b) $M$ non orientable implies $H_n(M) \to H_n(M|x)$ is injective with image $\{r | 2r = 0\}$

c) $H_i(M) = 0$ for $i > n$.

pf of compact: take a generator $g \in H_n(M)$ and define an orientation via $x \mapsto g_x$ where $g_x \in H_n(M|x)$. Now $g_x \ne 0$ only for $x$ in the closure of the image of some cycle $\Delta^n \to X$ representing $g$.

If $g \in H_i(M)$ misses a point $x$ then it is contained in the complement $M - x$, hence represents zero in $H_n(M|x)$.

By the result $g_x$ is always nonzero; on the other hand the prev. comment says it can only be nonzero on the image of cycle which has compact image, hence $g \colon \Delta^n \to M$ is surjective and $M$ is compact.

The orientation result is a special case of

Orientation Lemma:

A) $A \subset M$ with $A$ compact, and $\sigma \colon M \to M_R$ then $\exists \alpha \in H_n(M|A)$ s.t. $\sigma(x) = \alpha_x$ for $x \in A$.

B) $H_i(M|A) = 0$ for $i> n$.

proof: Step 1, Reduce to the case $M = \mathbb{R}^n$.

For $A \subset M$ arb. right $A = A_1 \cup ...\cup A_m$ where $A_i = A \cap U_i$ and $\mathbb{R}^n \cong U_i \subset M$. (sim. to cov. spaces, van kampfen, schemes).  Now use an induction argument.

The case $n = 1$ is covered by the reduction hypothesis.  If it holds for $A_1 \cup ... \cup A_{m-1}$, then adding $A_m$, have that it holds for $A_1 \cap A_{m}\bigcup ... \bigcup A_{m-1} \cap A_m$ .

The induction argument is finished because the result for $A,B, A\cap B$ implies it for $A\cup B$ via the Mayer Veitoris sequence

$0 \to H_n(A \cup B) \to \begin{smallmatrix} H_n(A) \\ \oplus \\ H_n(B) \end{smallmatrix} \to H_n(A\cap B) \to ...$

B) now follows easily via the l.e.s by things being squeezed by zeros. As for A), basically get classes $\alpha_A, \alpha_B, \alpha_{A \cap B}$ and the l.e.s shows they glue.  As for uniqueness the idea is $\alpha_x = o$ for all $x$ implies $\alpha = 0$, this is proved via it holds for $A,B, etc.$ and the first map in l.e.s is injective.

Step 2, $M = \mathbb{R}^n$, and reduce to the case $A$ is convex.  In fact if it holds for convex things it holds for a countable union of convex things via a similar argument as step 1.

Assuming it holds for convex things, the idea is to treat an arb. $\alpha \in H_n(M|A)$ as if its coming from slightly bigger $K$ which is a union of convex things.

$\alpha \colon \Delta^n \to M$ with $\delta \alpha \in M - A$, then the boundary is compact and it has positive distance from $A$ and so each point $x \in A$ has a closed ball nbd not intersection $\partial \alpha$.  The union of these balls is $K$.  The key idea, is by construction the relative cycle $\alpha \colon \Delta^n \to M = \mathbb{R}^n$ and it lifts $\alpha \in H_n(M|A)$.

B) can now be proved by taking $\alpha$ to be generator of $H_i(M|K)$, it lifts to zero so it must have been zero to begin with.  As for A), uniqueness uses $\alpha_x = 0$ for all $x \in A$ means its zero for all $x \in K$, this is proved via a commutative triangle having legs

$H_n(M|K) \to H_n(M|B) \to H_n(M|x)$

$H_n(M|K) \to H_n(M|x)$

where $B$ is any ball $B \subset K$.  For existence just take the answer for any ball $B \supset K$.

Step 3, $M = \mathbb{R}^n$ and $A$ convex.

Here the result is evident via $H_n(M|A) \cong H_n(M|pt)$.  B) That the higher homology is zero is clear.  As for A), uniqueness is clear for similar argument as before.  Existence is just taking a generator of $H_n(M|pt) \cong H_{n-1}(S^{n-1}) = \mathbb{Z}$.