Let $X$ be a CW complex.
Suppose $\gamma\in C_n(X)$ is a cycle which is a sum of maps $\sigma:\Delta^n\to X$ which factor as $\Delta^n\to\mathbb R^{n-1}\to X$. Does it follow that $\gamma$ is null-homologous?
More generally, we could ask the following which also seems likely to be true but hard to prove:
Let $L\subseteq C_\bullet(X)$ be the subcomplex generated by $\sigma$ and $d\sigma$ for all maps $\sigma:\Delta^n\to X$ which factor as $\mathbb R^{n-1}\to X$. Is $L$ acyclic?
First, we may as well allow the manifolds $M^{n-1}$ to have a boundary, in which case we can replace $M^{n-1}$ by a closed neighbourhood of the image of $\Delta^n$ if necessary, and thus assume that $M^{n-1}$ is compact. In this context, the union of the images of all the sets $M^{n-1}$ will be contained in a finite subcomplex of $X$, so we may assume that $X$ itself is finite, of dimension $d$ say. We can argue by induction on $d$. The case $d<n$ is clear, so suppose that $d\geq n$.
Now assume for the moment that the maps $M^{n-1}\to X$ are reasonably regular, and let $K$ denote the union of the images. Then $K$ will have dimension less than $n$ (in the sense of Hausdorff dimension, say), so for each $d$-cell we can choose an interior point that is not in $K$. After removing these points we get a subspace $Y\subset X$ such that $\text{skel}^{d-1}(X)$ is a deformation retract of $Y$, and $\gamma$ is in the image of the map $C_n(Y)\to C_n(X)$. It is now easy to deduce from the inductive hypothesis that $\gamma$ is nullhomologous.
Unfortunately, it is possible for the maps $\Delta^n\to M^{n-1}\to X$ to be very irregular, like space-filling curves. In that case even $\sigma(\text{skel}^1(\Delta^n))$ can have very high dimension. On each individual simplex we can adjust the maps $\Delta^n\to M^{n-1}\to X$ by an arbitrarily small amount to make them smooth. If we can make these adjustments in a suitably compatible way then we can find a chain $\gamma'$ homologous to $\gamma$ where the previous case applies, and thus deduce that $\gamma$ is nullhomologous. However, I will refrain from any further analysis of this situation unless the OP confirms that he cares about it.
