I am reading Lectures on algebraic topology by Albrecht Dold. Let $t$ be an additive functor from the category of free abelian groups to the category of abelian groups. In VI 7.3 he writes:
For any pair $(X, A)$ the sequence $$0\rightarrow SA\rightarrow SX\rightarrow S(X,A)\rightarrow 0$$ is exact and splits in every dimension. Since $t$ is applied dimension-wise the sequence $$0\rightarrow tSA \rightarrow tSX \rightarrow tS(X, A) \rightarrow 0 $$ is also exact (and splits in every dimension).
I am not sure what exactly he uses to come to this conclusion. I know that for the tensor product we can argue that it is a left adjoint, hence right exact and free abelian groups are flat, hence the result follows, but in general I don't see why this should hold.
The property we need seems to that $t$ is a right adjoint. Is this true or is this too strong of an assumption?
Notation:
- We have pair $(X,A)$ if $X$ is a topological space and $A$ a subspace.
- $SX$ is the chain complex consisting of free abelian groups $S_nX$ generated by the singular $n$-simplices of $X$.
- $S(X,A)$ is the chain complex $SX/SA$
In general, an additive functor $t$ preserves split short exact sequences (just apply the functor $t$ to the algebraic equations that capture being a split short exact sequence). Thus, each component $t_n$ of the functor $t$ preserves the corresponding split short exact sequence.