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 $n$.

  • $\begingroup$ Really? There are, what I required is that $\tau$ is a singular chain but the boundary of $\tau$ is a simplicial chain. $\endgroup$
    – Yeah
    Nov 6, 2022 at 13:30
  • $\begingroup$ Sorry, my reading comprehension failed. I'll think about it. $\endgroup$
    – Thorgott
    Nov 6, 2022 at 13:51
  • $\begingroup$ As a remark, consider fibrations of the form $p\colon M\times N\rightarrow N$. Then, your definition (in degrees up to $\dim(N)$) is just $C_i=C_i^{sing}(M)\otimes C_i^{simp}(N)$. General theory, on the other hand, dictates that the chain complex computing the homology of $M\times N$ is $C_{\bullet}^{simp}(M)\otimes C_{\bullet}^{sing}(N)$, the tensor product of chain complexes. This is distinct from the level-wise tensor product of chain complexes, which is just not sensible from the perspective of homological algebra. This leads me to suspect your hypothesis should also fail in lower degrees. $\endgroup$
    – Thorgott
    Nov 6, 2022 at 15:09

1 Answer 1


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 $S^3\rightarrow S^2$.

  • $\begingroup$ Dear Thorgott, thank you for your objection. I should also have defined $C_n$ for $n>\dim B$. Please see my update. $\endgroup$
    – Yeah
    Nov 6, 2022 at 8:24

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