TITLE: Simplicial Cohomology With Compact Support
QUESTION [5 upvotes]: This is from Hatcher's discussion on page 242, section 3.3.
$X$ a locally compact $\Delta$-complex, which is equivalent to saying that each point has a neighborhood which meets only finitely many simplices.
$\Delta_{C}^i(X;G)$ is the subgroup of the simplicial cochain group $\Delta^i(X;G)$ consisting of the cochains that are compactly supported, which is equivalent to taking nonzero values on a compact subset of $X$.
He says that for $\varphi \in \Delta^i(X;G)$, the coboundary $\delta \varphi$ can have a nonzero value only on those $(i+1)$-simplices having a face on which $\varphi$ is nonzero and...

... there are only finitely many such simplices by the local compactness assumption.
I'm dubious about this last statement. Why can't we just use the fact that $\varphi$ has compact support? The faces of a $n$-simplex are $(n-1)$-simplices, so if there are infinitely many $(i+1)$-simplices where $\varphi$ has nonzero values, then $\varphi$ would not have had compact support.
Is there something wrong with my argument? If so, how does one use the local compactness assumption?

REPLY [2 votes]: The problem is that $\delta\varphi$ can have larger support than $\varphi$.  For $\delta\varphi$ to be nonzero on a simplex $\sigma$, that means $\varphi$ is nonzero on $\partial\sigma$.  So as long as $\sigma$ has any boundary face in the support of $\varphi$, $\sigma$ may be in the support of $\delta\varphi$, even if the interior of $\sigma$ is not in the support of $\varphi$.
However, the local compactness assumption ensures that for each simplex in the support of $\varphi$, there are only finitely many higher-dimensional simplices which have it as a boundary face.  Thus if $\varphi$ has compact support, $\delta\varphi$ will as well (though its support may be larger!).