Let $X$ be a (finite) simplicial complex and let $f$ be a map from the set of its $n$-Simplices to a abelian group $A$, with the property, that every cycle maps to $0$ (extending $f$ linearly).

Let $Y$ be the cone of $X$. Is it possible to extend this map to a map from the $n$-simplices of $Y$ to $A$ with the same property ?

Let $m$ be the number of $n$-simplices in $X$ and $n$ be the number of $n-1$-simplices of $X$. We have to find a value for each new $n$-simplex in $Y$, so we get a linear system of equations in $n$ variables. Furthermore the boundary of each new $n+1$-simplex must be mapped to $0$ (the property I am looking for). So we get $m$ equations.

Here is one example, where $X=\partial \Delta^2,n=1$. Given $a,b,c$ with $a+b+c=0$.

We have to find $x,y,z$ satisfying $a=z-y,b=x-z,c=y-x$. Then one can begin with the guess $y=0$. The first equation yields $z=a$ and the last equation gives $x=-c$. Inserting this into the second equation, we get $b=-c-a$, which is also fulfilled by assumption. So the claim is true for this special choice of $X$. I am looking for a proof for every $X$.

I am writing a computer program, that computes the simplicial homology with coefficients in a $\mathbb{Q}$ of a simplicial complex. This is, where my motivation came from.