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?
Best Answer
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.