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


  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.

  • $\begingroup$ 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


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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.