Homology of singular chain complex modulo subdivision

Question

Let $S_p(X)$ be the $p$-th singular chain group and $\mathcal S(X)$ be the singular chain complex of a topological space $X$. There is a barycentric subdivision operator (which is also a chain map) $\operatorname{sd}: S_p(X) \to S_p(X)$ as is defined in standard topology textbooks such as Munkres. There is a chain homotopy from $\operatorname{sd}$ to $\operatorname{id}_{\mathcal S(X)}$.

Let $D_p$ be the subgroup of $S_p(X)$ generated by $T - \operatorname{sd}(T)$ such that $T \in S_p(X)$. Then $\mathcal D:= (D_p)$ is a chain subcomplex of $\mathcal S(X)$. I wonder whether the homology of the quotient chain complex $\mathcal S(X)/\mathcal D$ is isomorphic to the homology of $\mathcal S(X)$.

In fact, this is how I used to understand the proof of the excision theorem of singular homology "intuitively". However, now I begin to doubt this intuition.

There are two possible ways I can think of to prove the statement above. One is to show $\mathcal D$ is acyclic and use zig-zag lemma, the other is to construct a chain homotopy between $\mathcal S(X)/\mathcal D$ and $\mathcal S(X)$.

This question has been posted in MSE (link) for two days, but there is no answer.

Answer 1

I believe $\mathcal D$ is not acyclic, unless $X$ is a discrete space, and therefore $S(X)/\mathcal D$ is not quasi-isomorphic to $S(X)$.

In fact, assuming that I did not make a mistake in the claim below, we have the following description of $\mathcal D$: it is zero in degree zero, and is isomorphic to $S(X)$ in positive degrees. It follows that $H_0(\mathcal D)=0$, $H_1(\mathcal D)$ is the huge group $S_1(X)/\partial S_2(X)$, and for $i>1$ $H_i(\mathcal D)\cong H_i(X)$. However, the inclusion $\mathcal D\hookrightarrow S(X)$ induces zero in homology.

It is obvious that $D_0$ is the zero group. For higher degrees, we have the following claim

Claim: for all $n>0$, the operator $T\mapsto T-sd(T)$ defines an injective homomorphism $S_n(X)\to S_n(X)$. It follows that $D_n\cong S_n(X)$ for all $n>0$.

Sketch of proof (hope it is correct, I did not check every detail): Define a partial ordering on the set of singular $n$-simplices of $X$. Given two singular simplices $\sigma, \tau\colon \Delta^n \to X$, we say that $\sigma\le \tau$ if $\sigma=\tau$, or there exists a contracting linear map $h\colon \Delta^n\to \Delta^n$ such that $\sigma=\tau \circ h$. By contracting we mean that there exists a $0\le r<1$ such that $\mbox{dist}(h(x),h(y))\le r\cdot \mbox{dist}(x, y)$. It is not hard to check that this is a partial order. Moreover $\sigma=\sigma\circ h$ for some contracting map only if $\sigma$ is constant.

Now let $0\ne T\in S_n(X)$. We can write $T=\Sigma_{i=1}^m n_i \sigma_i$, where $\sigma_1, \ldots, \sigma_m$ are pairwise distinct singular simplices, and $n_i$ are non-zero integers. If all $\sigma_i$ are constant, then $T-sd(T)=T\ne 0$. Otherwise, among the $\sigma_i$s there exists a non-constant simplex $\sigma_{i_0}$ that is maximal with respect to the order that we defined. The singular chain $T-sd(T)$ is a linear combination of $\sigma_i$s and simplices that are smaller than $\sigma_i$. It is clear that there is no summand that can cancel $\sigma_{i_0}$, so $T-sd(T)\ne 0$.