# Mistakes in Bredon's book “Topology and Geometry”?

I am preparing the notes for a course in Algebraic Topology, so I decided to borrow some of the material from the classical (and wonderful) book by G. Bredon Topology and Geometry.

Looking at the part regarding the orientation of a topological $$n$$-manifold $$M^n$$, at page 341 we find the following well-known result, with its usual proof (Proposition 7.1):

So far, so good. However, after five pages we find what follows:

This makes me confused, for at least two reasons:

Point 1. The Note after the statement of Proposition 7.10 does not make any sense to me. As defined, the symbol $${}_2G$$ denotes the $$2$$-torsion part of the abelian group $$G$$, so if $$G$$ is torsion-free (for instance, if $$G=\mathbf{Z}$$) then $${}_2G=0$$. This is clearly very different from the free-product $$G \ast \mathbf{Z_2}$$ (here $$\ast$$ seems to denote the free-product, see pages 158-159).

Point 2. In Corollary 7.11, take $$A=\{x\}$$ and $$G=\mathbf{Z}$$. Then, when $$M$$ is not orientable one finds $$H_n(M, \, M-\{x\}, \, \mathbf{Z})=0$$, and this contradicts Proposition 7.1, that yields the (correct, as far as I know) result $$H_n(M, \, M-\{x\}, \, \mathbf{Z})= \mathbf{Z}$$.

Question. Are the issues risen in Points 1, 2 above really mistakes in Bredon's book, or perhaps am I missing something trivial?

• I guess $*$ might be a typo, it would rather be some sort of $\otimes$. – Dima Pasechnik Jan 24 '19 at 11:42
• Every point has an orientable neighborhood (say, a ball), hence $M$ is always orientable along $\{x\}$, so corollary 7.11 says that for every manifold $M$ the formula you give holds. – Denis Nardin Jan 24 '19 at 11:59
• Also it seems that Bredon indicates with $\ast$ what I would call $\mathrm{Tor}_1$, so in particular $A\ast \mathbb{Z}/n$ is exactly the $n$-torsion of $A$. – Denis Nardin Jan 24 '19 at 12:07
• Well, at page 158 it also indicate by $*$ the free product, and in a book of 550 pages it is not easy to understand where the same notation indicates two very different things. Now it makes sense, thanks! – Francesco Polizzi Jan 24 '19 at 12:09
• @GeraldEdgar: Bredon died in 2000, and there is no webpage available. On the Springer's webpage there is no errata, either. Actually, on the web I found nothing (well, maybe I did not look well enough). – Francesco Polizzi Jan 24 '19 at 14:51

Star (in older topology texts) often indicate torsion product of abelian groups, that is, $$A * B := \operatorname{Tor}_{\Bbb Z}(A, B)$$. Usually it is clear from the context whether free product or torsion product is meant.

• Thanks. I was not aware of this (old) notation. – Francesco Polizzi Jan 24 '19 at 13:34
• (This notation is also used in Spanier's text, for example.) – Pedro Tamaroff Jan 24 '19 at 13:51
• @FrancescoPolizzi I'm allegedly a specialist in Algebraic Topology, and I didn't know either, so don't feel too bad :) – Denis Nardin Jan 25 '19 at 12:22
• Bredon carefully defines A*B on page 278 in a section called More homological algebra and even says that it is often denoted Tor(A,B). – Nicholas Kuhn Jan 25 '19 at 17:31
• Yes, but again: the book is 550 pages long, one that is just interested in one topic cannot read carefully all of it. Since at page 158 Bredon uses $*$ to indicated to free product, I supposed (maybe ingenuously) that this notation would have been kept in all the book, even because I never met the notation $*$ for $\mathrm{Tor}_1$ before. – Francesco Polizzi Jan 25 '19 at 18:55

I think that you are missing the definition of 'orientable along $$A$$'. I haven't got that book of Bredon to hand, but presumably 'orientable along $$A$$' means that if you move a local orientation of $$M$$ around a closed path that stays in $$A$$ then it will come back to the same local orientation. In particular, in the case when $$A$$ is a single point, then $$M$$ will always be orientable along $$A$$, regardless of whether $$M$$ is orientable or not, so the case that you view as wrong doesn't arise.

I agree with Denis T's interpretation of the notation $$A*B$$.

• Yes, definitely I was confused about the definition of "orientable along $A$". And I was unaware of the old notation $A*B$ for $\mathrm{Tor}_1(A, \, B)$. – Francesco Polizzi Jan 25 '19 at 17:27