Degenerate Singular Chains on CW-complexes

Question

look at the end of this post for the origin of my question.

Recall from the book of May, Chapter 16, that a singular $n-$simplex $f$ is said to be degenerate if $ f = s_i (g)$ for some $(n-1)-$simplex $g$, where $s_i (g) (t_0,\dots,t_n) = g(t_0,\dots,t_{i-1},t_i + t_{i+1},t_{i+2},\dots,t_n).$
Given a CW-complex $X$ of dimension $k

Origin of the question: If $M_k$ is the $k-$skeleton of some CW-decomposition of a smooth manifold $M$ with $i_k: M_k \hookrightarrow M$, $f: \Delta_n \rightarrow M_k$ an $n-$simplex with $n>k$ and $\omega \in \Omega^n (M)$ a differential form, I need that $\int_{s{i_k}_*f} \omega = 0,$ where $s$ denotes Lee's smoothing operator. Since the integral of a form over a degenerate simplex is always zero, this leads me to the above question.

In any case, thanks a lot for your help!
Timo

Answer 1

No, this is not the case. To fix ideas, let's say $\Delta^n = \{ (t_0, \dots, t_n) \in \mathbb{R}^{n+1} \mid \sum t_i = 1, t_i \ge 0 \; \forall i \}$.

Take for example $k = 1$ and $X = \Delta^1$ itself. Consider the $2$-simplex $f : \Delta^2 \to \Delta^1$ given by $f(t_0, t_1, t_2) = (t_0^2 + t_1, t_2 + t_0 - t_0^2)$ (note that $0 \le t_0 \le 1$ thus $t_0 - t_0^2 \ge 0$).

Now suppose there were some $g : \Delta^1 \to \Delta^1$ such that $f = s_0 g$, i.e. $$(t_0^2 + t_1, t_2 + t_0 - t_0^2) = g(t_0+t_1, t_2).$$ Then you get a contradiction, because you must have $g(1,0) = g(1+0,0) = f(1,0,0) = (1,0)$ but also $g(1,0) = g(\frac{1}{2}+\frac{1}{2},0) = f(\frac{1}{2}, \frac{1}{2}, 0) = (\frac{3}{4},\frac{1}{4})$. This is absurd. The case $f = s_1 g$ is similar.