Chain carried by subcomplex (Munkres' notation): What does it mean? In Munkres' book "Elements of Algebraic Topology", page 31, there is a definition:

We shall say that a chain $c$ is carried by a subcomplex $L$ of $K$ if $c$ has value 0 on every simplex that is not in $L$.

I am not quite sure what does this mean. A chain $c$, in my understanding is simply a linear combination of simplices, how can it take a value 0 (behaving like a function)?
My guess is that if $c=m_1\sigma_1+\dots+m_k\sigma_k$ is carried by $L$, then all the $\sigma_1,\dots,\sigma_k$ belong to $L$? Is this what Munkres is trying to say?
Thanks for any enlightenment.
 A: Munkres does define chains as functions. See page 27, he defines a $p$-chain as:

For a simplicial complex $K$, a $p$-chain $c$ is a function from the set of oriented $p$-simplices of $K$ to the integers such that $c(\sigma)=-c(\sigma')$ for opposing orientations and $c(\sigma)=0$ for all but finitely many $p$-simplices $\sigma$.

Further he defines and elementary chain to be $0$ on all but a particular $\sigma$ on which the function takes the value $1,-1$ respecting orientations. 
In the proof of Lemma 5.1 Munkres states that one may write a $p$-chain call it $\gamma$ as a sum of elementary chains. That is $$ \gamma(x) = \sum n_i c_i(x) $$ where each $c_i$ is an elementary chain corresponding to some $\sigma_i$. Thus we get $\gamma(\sigma_i)=n_i$. The statement that a subcomplex $L$ carries a chain can then be interpreted as for each simplex $\sigma$ not in $L$ no elementary chain $c_i$ in the sum corresponds to $\sigma$. So we have $c_i(\sigma) = 0 \, \, \forall i$ and thus $\gamma(\sigma) = 0$.
Your guess is basically the correct idea but I wished to show that the chain is indeed a function.
