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

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.

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 $$n$$.

My question is: Is the homology of $$(C_*,\partial)$$ isomorphic to the singular homology of $$E$$?

• I think I know how to prove it: your complex includes in the usual singular complex for E; filter both by the skeletal filtration on the base; an inductive proof shows that the inclusion induces an isomorphism on exact couples, which gives an isomorphism on spectral sequences; finally, noting that the larger filtered complex defines the Serre spectral sequence, we obtain the result. Nov 5, 2022 at 13:46
• But I'm guessing that the proof of the Serre spectral sequence actually uses a result like this one, so I'm hoping you get an answer that is short and sweet! Nov 5, 2022 at 13:47
• Cross-posted on MSE. Nov 5, 2022 at 16:00

My question is: Is the homology of $$(C_*,\partial)$$ isomorphic to the singular homology of E?

Yes. Observe that the singular complex functor sends Serre fibrations to Kan fibrations. Thus, the map $$\def\Sing{\mathop{\rm Sing}} \Sing p\colon \Sing E\to \Sing B$$ is a Kan fibration.

Denote by $$T$$ the underlying simplicial set of the triangulation of $$B$$ so that we have a homeomorphism $$|T|→B$$ and an adjoint simplicial weak equivalence $$t\colon T→\Sing B$$.

Now the chain complex $$C$$ is simply the simplicial chains of the pullback of the span of simplicial sets $$T→\Sing B←\Sing E.$$

The base change of $$\Sing p$$ along $$t$$ yields a map $$P→T$$.

One leg of the above span is a Kan fibration, therefore the pullback computes the homotopy pullback. The map $$T→\Sing B$$ is a weak equivalence, hence so is its base change $$P\to \Sing E$$.

Thus, the map $$P→T$$ is weakly equivalent to $$\Sing p$$, so the chain complex of $$P$$ is quasi-isomorphic to the chain complex of $$\Sing E$$. The chain complex of $$P$$ is precisely $$C$$.

• In case you aren't familiar with the construction, \DeclareMathOperator\Sing{Sing} has substantially the same result as \def\Sing{\mathop{\rm Sing}}. Nov 5, 2022 at 15:40
• @LSpice: If it has the same result, then what is your point? Nov 5, 2022 at 17:04
• Re, increased knowledge, if you didn't already know, and hopefully no harm done, if you did already know. Nov 5, 2022 at 18:14