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Suppose we have a pullback of topological spaces (CW-complexes) $B\rightarrow A\leftarrow C$ which I will denote by $D$.

Assumptions

  1. The induced map $D\rightarrow C$ is a trivial fibration
  2. The map $f:B\rightarrow A$ has weakly contractible fibers i.e., for any $a\in A$ we have $f^{-1}(a)\simeq \ast$
  3. The induced map $D\rightarrow B$ induces an isomorphism in homology.
  4. The map $C\rightarrow A$ induces a surjective map in homology.
  5. We can assume that $A$ is simply connected.

Question What can we say about the map $B\rightarrow A$? Does it induce an isomorphism in homology?

Edit: The spaces $C,B$ are connected.

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  • $\begingroup$ Doesn't the isomorphism on homology follow already from assumption 2 thanks to the Vietoris Begle theorem? $\endgroup$ – Vidit Nanda Oct 14 '15 at 21:50
  • $\begingroup$ No, the Vietoris-Begle theorem only holds for proper maps. $\endgroup$ – Johannes Ebert Oct 14 '15 at 23:37
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Counterexample:

Let $A=[0,1]$ with the usual topology, $B=C=[0,1]^{\delta}$ (this means the discrete topology). The maps $f:B \to A$ and $g:C \to A$ are the identity. Then $f^{-1} (t) = *$ for all $t \in A$, but $f$ is not a homology isomorphism, and $g$ is surjective in homology. Furthermore, the pullback $D$ is the diagonal in $B \times C$, hence it is also $[0,1]^{\delta}$. The two maps $D \to B$ and $D \to C$ are homeomorphisms, hence acyclic fibrations and homology equivalences.

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  • $\begingroup$ Thank you, I forgot to say that the spaces B and C are connected. $\endgroup$ – Ilias A. Oct 15 '15 at 5:41
  • $\begingroup$ @Ilias: Having B, C connected does not help. Just take the suspension of Johannes' example. Then B, C, D are connected with uncountable fundamental group and uncountable H_1, but A is still contractible. And if you want B, C to be simply connected, then just suspend once more. $\endgroup$ – Sebastian Goette Oct 17 '15 at 9:06

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