This is a cross-post from a MSE question which received no answers. Beware that the notation here is a little different.

Consider the following lifting problem(s):

$\require{AMScd}$ \begin{CD} & & & & E\\ & & & @VV{p}V\\ Y @>{g}>> X @>{f}>> B \end{CD} Let's work with the following assumptions and notations (edited the indexing error pointed out by Aleksandar Milivojevic):

  1. The map $p: E \rightarrow B$ is a Hurewicz fibration with path-connected base $B$ and $(d-2)$-connected fiber $F$ for some $d \geq 2$. In case $d=2$ we require $\pi_1(F)$ to be abelian.
  2. For every $k \geq 0$, the action of the fundamental group $\pi_1(F)$ on the homotopy group $\pi_k(F)$ is trivial. As a consequence, there is a well-defined action [Davis-Kirk, Proposition 6.62] of $\pi_1(B)$ on $\pi_k(F)$.
  3. $X,Y$ are finite CW complexes, $g$ is a cellular map, and $\dim(X) = d < \dim(Y)$ with the same $d$ in (1).
  4. For each $k \geq 0$, write $\rho(f,k): \pi_1(X) \rightarrow Aut(\pi_k(F))$ for the induced action of $\pi_1(X)$ from (2). Consider this as a local coefficient system over $X$, so that there are well-defined cohomology groups $H^*(X;\pi_k(F)_{\rho(f,k)})$ with local coefficients. There is a similar situation with $\rho(fg,k) : \pi_1(Y) \rightarrow Aut(\pi_k(F))$.

Under these conditions, there is a well-understood obstruction theory. Using [Davis-Kirk, Theorem 7.37] and the remarks later: First of all, $f$ can be lifted over the $(d-1)$-skeleton $X_{d-1}$. Second, no matter which lift over the $(d-1)$-skeleton we choose, the obstruction class for lifting it further over the $d$-skeleton is unique. This primary obstruction $z_f$ is an element of $H^d(X;\pi_{d-1}(F)_{\rho(f,d-1)})$. For dimension reasons $z_f$ is the only obstruction for lifting $f$ along $p$.

The problem of lifting $fg$ along $p$ has a similar obstruction theory to above. There is again a primary obstruction $z_{fg} \in H^d(Y;\pi_{d-1}(F)_{\rho(fg,d-1)})$, but now there might be higher obstructions lying in $H^{k+1}(Y;\pi_k(F)_{\rho(fg,k)})$ for $k \geq d$.

Now suppose $z_f \neq 0$ but $z_{fg} = 0$. In other words (by the naturality of the obstruction classes), $z_f$ lies in the kernel of the induced map $$g^*: H^d(X;\pi_{d-1}(F)_{\rho(f,d-1)}) \rightarrow H^d(Y;\pi_{d-1}(F)_{\rho(fg,d-1)}) \, .$$

Does this vanishing imply that $fg$ can be lifted over the whole $Y$ (not just its $d$-skeleton)? If not, can we at least say that there is a unique nonzero higher obstruction class for lifting $fg$?

My intuition is that the lifting problem for $fg$ should not be harder than the lifting problem for $f$.

Davis, James F.; Kirk, Paul, Lecture notes in algebraic topology, Graduate Studies in Mathematics. 35. Providence, RI: AMS, American Mathematical Society. xvi, 367 p. (2001). ZBL1018.55001.


1 Answer 1


First, note that since you are assuming $F$ is $d-1$-connected, the primary obstruction lies in $H^{d+1}$, not $H^d$.

Now, consider the diagram $\require{AMScd}$ \begin{CD} & & & & S^5\\ & & & @VV{p}V\\ S^4 @>{g}>> S^3 @>{id}>> S^3 \end{CD} where $p$ represents the nontrivial element in $\pi_5(S^3) \cong \mathbb{Z}_2$ converted into a fibration, and $g$ represents the nontrivial element in $\pi_4(S^3) \cong \mathbb{Z_2}$. From the long exact sequence in homotopy groups for a fibration, we see that the fiber $F$ of $p$ is simply connected and $\pi_2(F) \cong \mathbb{Z}$. The primary (and only) obstruction to lifting $id$ lies in $H^3(S^3; \mathbb{Z})$ and does not vanish, since otherwise the identity on $S^3$ would factor through $S^5$ and hence be trivial.

Since $H^3(S^4;\mathbb{Z}) = 0$, the "primary obstruction" $z_{id \circ g}$ to lifting $id \circ g$ vanishes. However, there are higher obstructions to lifting this map, that do not vanish. Indeed, if there were a lift, then we would have that the non-trivial map $S^4 \to S^3$ factors through $S^5$, meaning it would be trivial.

  • 2
    $\begingroup$ The obstruction is in $H^4(S^4; \pi_3(F))$, and from the long exact sequence in homotopy for the fibration $F \to S^5 \to S^3$ we see that $\pi_3(F) = \mathbb{Z}_2$. So, the only obstruction to lifting $g$ is the unique nonzero element in $H^4(S^4;\mathbb{Z}_2) \cong \mathbb{Z}_2$. $\endgroup$ Jul 25, 2018 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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