# "singular homology = simplicial homology" relative to a fibration

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 $$n$$.

• Really? There are, what I required is that $\tau$ is a singular chain but the boundary of $\tau$ is a simplicial chain.
– Yeah
Nov 6, 2022 at 13:30
• Sorry, my reading comprehension failed. I'll think about it. Nov 6, 2022 at 13:51
• 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. Nov 6, 2022 at 15:09

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 $$S^3\rightarrow S^2$$.
• Dear Thorgott, thank you for your objection. I should also have defined $C_n$ for $n>\dim B$. Please see my update.