3

Let $C_n(X)$ be the free Abelian group generated by all the singular $n$-simplices of the topological space $X$. I'm reading Bredon's Topology and Geometry and there's a statement which says:

The group $C_n(X,A) = C_n(X)/C_n(A)$ is free Abelian. This gives a splitting $C_*(X,A) \rightarrow C_*(X)$. However, this splitting is not a chain map and so it does not induce a map in homology.

By splitting, does it mean $C_*(X) \cong C_*(A) \oplus C_*(X,A)$? What does it mean for a splitting to be a chain map? (I know the definition of a chain map, but for a splitting?) Also, we do have an exact homology long sequence of the pair $(X,A)$, so what is the induced map in homology in the statement referring to?

pamplemousse
  • 223
  • 1
    I think he means it splits pointwise, but not as chain complexes. That is, for all $n$ you get $C_n(X) \cong C_n(A) \oplus C_n(X, A)$ in $\textbf{Ab}$, but when you look at $C_*(X)$, it's a chain complex, but I suspect it doesn't split in the category of chain complexes. – Robert Cardona Dec 10 '18 at 01:14
  • ^This is essentially right. You can split it termwise, but there's no way to make sure all those individual splittings assemble together by some miracle to give a chain map. So, the splitting won't pass down to homology. – Randall Dec 10 '18 at 01:43
  • 1
    If it split in the category of chain complexes then you would have $H_(X) = H_(A) \oplus H_*(X,A)$. You can check in examples that this often is not true. –  Dec 10 '18 at 03:36
  • 1
    The splitting, for Bredon, is the collection of maps $j: C_n(X,A) \to C_n(X)$ which induce the direct sum decomposition, and he is saying that this family of maps is not a chain map. It does not induce a map in homology, so there is in general no map $H_n(X,A) \to H_n(X)$, and in general $H_n(X) \not \cong H_n(A) \oplus H_n(X,A)$. – John Palmieri Dec 10 '18 at 04:39

0 Answers0