# Obstructions for the lifting problem after a pull-back

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.

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.
• 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 $H^4(S^4;\mathbb{Z}_2) \cong \mathbb{Z}_2$. Jul 25, 2018 at 17:47