If $(X,A)$ is a pair, then $S_k(X,A):=S_k(X)/S_k(A)$ is free on the singular simplicies of $X$ with image not contained in $A$. Why is this so? I tried to give a proof by checking the mapping property of the free abelian group generated by these simplicies, but failed somehow. Is this result a special case of a more general (probably algebraic) proposition?
2026-03-25 18:59:51.1774465191
Relative singular chains basis
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Well, if $S$ is a set which is a disjoint union $B\sqcup A$, then the free Abelian group on $S$ is the direct sum of the free Abelian groups on $B$ and on $A$. $$F(S)=F(B)\oplus F(A)$$ It follows that $F(B)=F(S)/F(A)$
In your case $S$ is the set of all simplices in $X$, $A$ is the set of simplices in $A$, and $B$ the set of simplices whose image is not in $A$.