Everyone learns in the first course in algebraic topology that the singular homology of a topological space with a simplicial decomposition is isomorphic to its simplicial homology.
I want to ask if the following is true. Thank you for any help!
Let $p:E\to B$ be a fibration. Suppose $B$ has a simplicial decomposition. For each $n\in\mathbb{Z}_{\ge0}$, let $C_n$ be the free abelian group generated by the set of pairs $(\sigma,\tau)$ where $\sigma:\Delta^n\to E$ is a singular simplex, $\tau:\Delta^n\to B$ is a simplicial simplex, and $\tau=p\circ\sigma$. There is a boundary operator $\partial:C_n\to C_{n-1}$ defined in the usual way, making use of the faces of $\Delta^n$. Clearly $\partial\circ\partial=0$, and $(C_*,\partial)$ is a chain complex.
My question is: Is the homology of $(C_*,\partial)$ isomorphic to the singular homology of $E$?
Remark: If we assume $\sigma,\tau$ are both singular chains, then $(C_*,\partial)$ is nothing but the singular chain complex of $E$.
Update: If $B$ is a finite-dimensional simplicial complex, then for $n>\dim B$, $C_n$ is defined inductively as follows. For $n=\dim B+1$, let $C_{\dim B+1}$ be the free abelian group generated by the set of pairs $(\sigma,\tau)$ where $\sigma:\Delta^n\to E$ is a singular simplex, $\tau:\Delta^n\to B$ is a singular simplex such that $\tau=p\circ\sigma$ and $\partial(\sigma,\tau)\in C_{\dim B}$, where $\partial(\sigma,\tau)$ is defined in the usual way. Define $C_n$ similarly for all larger $n$.
This cannot possibly be true. Indeed, if $B$ is an $n$-dimensional simplicial complex, there are no simplices of dimension $>n$, so necessarily $C_i=0$ for $i>n$. If the answer to your question was positive, this would imply that $H_i(E)=0$ for $i>n$. For a counter-example, we can call upon the Hopf fibration $S^3\rightarrow S^2$.