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Let $X$ be an abstract simplicial complex. Recall that the usual simplicial chain complex for $X$ is defined as follows. Let $C_k(X)$ be the quotient of the free abelian group on formal symbols $[v_0,\ldots,v_k]$ where $\{v_0,\ldots,v_k\}$ is a $k$-simplex of $X$ which identifies $[v_0,\ldots,v_k]$ and $(-1)^{|\sigma|}[v_{\sigma(0)},\ldots,v_{\sigma(k)}]$ for all permutations $\sigma$ of $\{0,\ldots,k\}$. The differentials are defined in the usual way.

My question is what happens when you instead do the following. Let $D_k(X)$ be the free abelian group on formal symbols $[v_0,\ldots,v_k]_D$ where $\{v_0,\ldots,v_k\}$ is a simplex of $X$ of dimension at most $k$ (so possibly there are repetitions among the $v_i$). We can define a differential on $D_{\ast}(X)$ in the usual way.

To summarize : The difference between $D_k(X)$ and $C_k(X)$ are that

  1. In $D_k(X)$, the ordering matters (not just the orientation); for instance, if $v_0$ and $v_1$ are joined by an edge in $X$, then $[v_0,v_1]_D$ is unrelated to $[v_1,v_0]_D$, while in $C_1(X)$ we would have $[v_0,v_1]=-[v_1,v_0]$.

  2. In $D_k(X)$, we allow "degenerate" simplices that have repetitions in their simplices in our chain complex, while this is not allowed in $C_k(X)$.

Does $D_{\ast}(X)$ also compute the homology of $X$? I can verify that it works in degree $0$ (trivial) and $1$ (easy; but this would not work if we did not allow repetitions). It seems similar to stuff I've seen with simplicial sets, but I don't know the literature there enough to extract it from there.

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@Albert: It seems your question is answered positively in theorem 4.3.8 of Spanier's book "Algebraic topology". It states, in your notation, that the natural map $D_\ast X\to C_\ast X$ is a quasi-isomorphism. The proof given by Spanier is a simple application of the method of acyclic models. –  Ricardo Andrade Apr 10 '13 at 22:57
    
By the way, let me remark that your $D_\ast(X)$ (called $\Delta(X)$ in Spanier's book) is simply the chain complex $C_\ast(S(X))$ associated with the simplicial set $S(X)$ naturally associated with the simplicial complex $X$. The $n$-simplices of $S(X)$ are your elements $[v_0,\ldots,v_n]_D$. –  Ricardo Andrade Apr 10 '13 at 22:58
    
You can also find this in Munkres book. –  Benjamin Steinberg Apr 11 '13 at 13:37
    
Indeed, the positive answer is strongly hinted at by the fact that degenerate mappings are allowed in the definition of singular homology. (Hatcher even says that the name "singular" came about because singularities are allowed in the mapping.) –  Russ Woodroofe Oct 31 '13 at 22:13
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