Consider the following situation:
- We have a Hurewicz fibration $p: E \rightarrow B$ with path-connected base $B$ and $(d-1)$-connected fiber $F$ for some $d \geq 1$. In case $d=1$ we require $\pi_1(F)$ to be abelian.
- There is a continuous map $f: X \rightarrow B$ where $X$ is a finite CW complex.
- 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 $\pi_k(F)$.
- For each $k \geq 0$, we write $\rho(f,k): \pi_1(X) \rightarrow Aut(\pi_k(F))$ for the induced action of $\pi_1(X)$ on $\pi_k(F)$. We regard 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.
Under these conditions, there is a well-developed 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 $H^d(X;\pi_{d-1}(F)_{\rho(f,d-1)})$. In case $f$ can be lifted over $X_k$ for some $k \geq d$, the next obstructions lie in $H^{k+1}(X;\pi_k(F)_{\rho(f,k)})$ (but these higher obstructions are not necessarily unique).
Now suppose that $f$ factors through a $d$-dimensional $CW$ complex $Y$: say $u: X \rightarrow Y$ and $v: Y \rightarrow B$ where $u$ is cellular and $vu = f$. The problem of lifting $v$ along $p$ has a similar obstruction theory to above, but for dimension reasons here the primary obstruction $z_v \in H^d(X;\pi_{d-1}(F)_{\rho(v,d-1)})$ is the only obstruction for lifting $v$ along $p$.
Here is my question: suppose $z_v \neq 0$ but $z_f = 0$. In other words (by the naturality of the obstruction classes), $z_v$ lies in the kernel of the induced map $$u^*: H^d(Y;\pi_{d-1}(F)_{\rho(v,d-1)}) \rightarrow H^d(X;\pi_{d-1}(F)_{\rho(f,d-1)}) \, .$$ Does this imply that $f$ can be lifted over the whole $X$ (not just its $d$-skeleton)? My intuition is that because $f = vu$ factors through $v$, the lifting problem for $f$ should not be harder than the lifting problem for $v$. And we know that the only obstruction for lifting $v$ is pulled-back along $u^*$ to zero, hence there should be no obstructions left for lifting $f$.
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.
