at.algebraic-topologydifferential-topology
Does anyone have a reference for:
The Euler-class for an open non-compact
manifold possibly with twisted coefficients (if the
group action on the manifold does not preserve
orientation) and/or for a compactification
e.g. the one point compactification
jim
The answer to the question as it is stated is that there is probably no "largest" class of spaces for which the Euler characteristic makes sense.
The answer also depends on where you would like the Euler characteristic to take values. Here is the tautological answer (admittedly not a very exciting one): if you have a category $C$ of spaces closed under taking cones and cylinders, then there is the universal Euler characteristic for that category: just take the free abelian group $K(C)$ that has a generator $[X]$ for each $X\in C$ and quotient it by the span of $[X]+[Cone(f)]-[Y]$ for all $X,Y\in C$ and any morphism $f:X\to Y$ in $C$. The Euler characteristic of any $X$ in $C$ is set to be $[X]$. (There may be variations and/or generalizations of this approach.)
The group $K(C)$ is complicated in general but for some choices of $C$ it has interesting quotients. This can happen e.g. when $C$ admits a good "cohomology-like" functor. For example if $C$ is the category of spaces with finitely generated integral homology groups then $K(C)$ maps to $\mathbf{Z}$ and this gives the usual Euler characteristic. If one takes $C$ to be formed by spaces that admit a finite cover with finitely generated integral homology groups (typical examples are the classifying spaces of $SL_2(\mathbf{Z})$ and more generally of mapping class groups), then $K(C)$ does not map to $\mathbf{Z}$ any more, but it maps to $\mathbf{Q}$. This gives the rational Euler characteristic.
Finally, let me address the last remark by Dmitri. For some categories the group $K(C)$ maps to $\mathbf{Z}$ in several different ways. Let us take e.g. $C$ to be the category formed by spaces whose one-point compactification is a finite CW-complex (with proper maps as morphisms). Then there are (at least) two characteristics; one is obtained using the ordinary cohomology and another one comes from the Borel-Moore homology. On complex algebraic varieties both agree. But the Borel-Moore Euler characteristic of an open $n$-ball is $(-1)^n$.
Here is the answer to the second question: suppose $Y$ is a locally closed subspace of a locally compact space $X$ such that $X,Y,\bar Y,\bar Y\setminus Y, X\setminus\bar Y$ and $X\setminus Y$ are of the form "a finite CW-complex minus a point". Then $\chi(Y)+\chi(X\setminus Y)=\chi(X)$ where $\chi$ is the Euler characteristic computed using the Borel-Moore homology.
The case when $Y$ is closed follows from the Borel-Moore homology long exact sequence. In general we can write $\chi(X)=\chi(X\setminus\bar Y)+\chi(\bar Y)=\chi(X\setminus\bar Y)+\chi(\bar Y\setminus Y)+\chi(Y)$. In the last sum the sum of the first two terms gives $\chi(X\setminus Y)$ since $X\setminus\bar Y$ is open in $X\setminus Y$.
A connected $n$-manifold $M \neq \emptyset$ is compact iff $H^n (M;\mathbb{Z}) \neq 0$.
EDIT: my old reference to Bredon, Topology and Geometry, page 346 ff, does not completely settle the issue, as George pointed out. So let me give a proof here. For compact $M$, Corollary 7.14 of the cited book gives the answer.
If $M$ is not compact, then Corollary 7.12 shows that $H_n (M;\mathbb{Z})=0$. By Corollary 7.13, $H_{n-1}(M;\mathbb{Z})$ is torsionfree. That is not by itself enough to conclude that $H^n (M,\mathbb{Z})=0$, since for example it could potentially happen that $H_{n-1}(M;\mathbb{Z})=\mathbb{Q}$, leading to a horrendous Ext-term. But such pathologies do not happen.
By Poincare duality, $$H^n(M;\mathbb{Z}) \cong H_0^{lf}(M;\mathbb{Z}_w)$$ (locally finite homology twisted by the orientation character). Any class in $H_0^{lf}(M;\mathbb{Z}_w)$ is represented by a locally finite collection of points, each labelled with an element of the fibre of $\mathbb{Z}_w$.
In other words, for every class $x \in H_0^{lf}(M;\mathbb{Z}_w)$, there is an at most countable discrete space $S$, a proper map $f:S \to M$ and $y \in H_0^{lf} (S,f^{\ast} \mathbb{Z}_w)= H_0^{lf}(S,\mathbb{Z})$ such that $f_\ast (y)=x$.
To show that $x=0$, I show that each proper map $f: S \to M$ can be extended to a proper map $h: S \times [0,\infty) \to M$. This settles the claim as $H_0^{lf}(S \times [0,\infty))=0$.
For each point $s \in S$, there is an 'exit path', i.e. a proper map $q_s: [0, \infty) \to M$, $q(0)=f(s)$. I wish to choose $h:S \times [0,\infty) \to M$ as the disjoint union of all the $q_s$'s, but without further care, $h$ can fail to be proper. We have to arrange the exit paths in such a way that each compact region $K \subset M$ is only hit by finitely many $q_s$'s.
To overcome this problem, write $M= \bigcup_{n\geq 0} N_n$ as the ascending union of compact codimension $0$ submanifolds with boundary. After replacing $N_n$ with $M_n$, the union of $N_n$ with all relatively compact connected components of $M-N_n$, the following is true: $M= \bigcup_{n \geq 0} M_n$, $M-M_n$ has no relatively compact connected component. The construction of the exhaustion guarantees that if $f(s) \in M_n-M_{n-1}$, $q_s$ can be chosen to stay in $M-M_{n-1}$. Hence the required finiteness is guaranteed.
Best Answer
For the untwisted case see Dold's "Lectures on algebraic topology" section VIII.11. If $N$ is a oriented topological submanifold of an oriented manifold $M$ of codimension $k$, then one looks at the Thom class in $H^k(M, M-N)$ and then restricts it to $H^k(N)$ to get the Euler class. Compactness of the submanifold $N$ is never needed.