Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The wikipedia page on Borel-Moore homology claims to give several definitions of it, all of which are supposed to coincide for those spaces $X$ which are homotopy equivalent to a finite CW complex and which admit a sufficiently nice embedding into a smooth manifold.

My question concerns one of the definitions on that page: it is asserted that if $X$ is a space and $\bar X$ is some compactification, then there is a definition of Borel-Moore homology given by $$ H_*^{\text{BM}}(X) := H_*(\bar X,\bar X \setminus X) . $$ The entry additionally requires that the pair $(\bar X,X)$ should be a CW pair for this to work.

It then goes on to assert that $\bar X = X^+$, the one-point compactification, will suffice for this.

Frankly, I don't see how $(X^+,X)$ will form a CW pair in most instances. For example suppose $X = \Bbb Z$ is the set of integers which is considered as a discrete space. Then in this instance $(X^+,X)$ certainly fails to be a CW pair.

Another example: $X = S^1 \setminus \ast$ with one point compactification $S^1$. Then $(S^1,S^1 \setminus \ast)$ is not a CW pair either.

My Questions:

(1) For what class of spaces $X$ does $H_*(X^+,+)$ coincide with definition of Borel-Moore homology given by locally finite chains?

(2) Does the wikipedia entry contain a mistaken assumption? Perhaps we do not need to assume that $(X^+,X)$ can be given the structure of a CW pair?

Note: if $X = \Bbb Z$ and if we use the above as a definition of Borel-Moore homology then $ H_0^{\text{BM}}(X) $ is a free abelian group whose generators are given by the underlying set of $\pi_0(X) = \Bbb Z$. This is clearly the wrong answer: it should be the countably infinite cartesian product of copies of the integers indexed over $X$ instead (using, say, the definition of Borel-Moore homology given by locally finite chains).

Another Question:

(3) Is there a definition the above kind (using compactifications) which will work (i.e., coincide with the locally finite chain definition) for $X = \Bbb Z$?

(I suspect not, since ordinary singular homology in degree zero is always free abelian.)

Incidentally, later in the page it lists the main variance property: Borel-Moore homology is supposed to be covariant with respect to proper maps. The page gives a proof using the above definition. But since the above definition doesn't work in general, I don't see how this is really supposed to be a proof.

share|improve this question
What is your exact question concerning BM-homology? –  André Henriques Jul 29 '13 at 23:50
@JohnKlein, if you are looking for a good treatment of locally finite singular or cellular homology, then obviously this Wikipedia page is not a very good source. I recommend instead taking a look at Chapters 3, 4 and Appendix A of Ranicki's Ends of Complexes (maths.ed.ac.uk/~aar/books/ends.pdf). –  Igor Khavkine Jul 30 '13 at 0:42
Thanks Igor. I'm familiar with (Hughes' and) Ranicki's work. But they don't give a comparison of the various definitions. They are content to give two definitions: one the locally finite singular theory and the other a cellular version. –  John Klein Jul 30 '13 at 0:43
@Ricardo and Scott: I did what you suggested. –  John Klein Jul 30 '13 at 1:07
Although $(S^1,S^1\setminus\ast)$ is not a CW pair, it has the homotopy type of one: triangulate $S^1$ on the complement of a tiny neighborhood around the missing point, and then add the vertex-at-infinity. Everything works out in an up-to-homotopy way, so perhaps the only issue is requiring $(X^+,X)$ to have the homotopy type of a CW pair. Perhaps a general scheme would involve cofibrant replacements of $X \hookrightarrow X^+$. –  Vidit Nanda Jul 30 '13 at 11:33

1 Answer 1

up vote 6 down vote accepted

In Chriss & Ginzburg, it's asserted that the Borel-Moore homology is given by $H_*(\bar X, \bar X\setminus X)$ if $\bar X$ is the one-point compactification or any compactification such that the inclusion is cellular. They very pointedly do not assume that the inclusion of $X$ into $X^+$ is cellular. They reference Bredon but don't cite a particular chapter.

share|improve this answer
Ben: I guess you mean that $\bar X \setminus X \to \bar X$ should be cellular. That's a reasonable condition, provided a cell structure on $\bar X$ is given. It seems then that the wikipedia entry has an error then. –  John Klein Jul 30 '13 at 12:53
Ben: I just checked Chriss & Ginzburg. It is clear now that the wikipedia entry has a mistake. –  John Klein Jul 30 '13 at 13:26
I took the liberty to correct the wikipedia page. –  Ricardo Andrade Jul 30 '13 at 22:56
Thanks Ricardo. –  John Klein Jul 31 '13 at 1:44
Another reference (with Google Books preview available) is "Representation Theories and Algebraic Geometry" books.google.pl/books?id=Q0Nyon7-qDMC , pp. 129-130 –  Michał Kukieła Aug 16 '13 at 15:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.