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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.

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up vote 2 down vote accepted

If you were working over $\mathbb Q$ instead of $\mathbb Z$, this would be easy (and the last paragraph of your question suggests that information over $\mathbb Q$ might be useful). The requirement that $f$ (by which I mean its linear extension, as in the question) vanishes on cycles means that it factors as $g\circ\partial$ for some linear map $g$ from the $(n-1)$-dimensional boundaries to $A$. Over a field, we can extend any such $g$ to a linear function on the space of all $(n-1)$-chains in $X$ and even in $Y$. (The extension from $X$ to $Y$ could simply make $g$ send all the new simplices, those not in $X$, to 0; the "field" assumption is involved in getting the extension from the boundaries to all chains on $X$.) Now with $g$ extended in this way, extend $f$ as $g\circ\partial$ on the $n$-chains of $Y$.

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