This is a cross-post from a MSE [question](https://math.stackexchange.com/q/2842085/26141) 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](https://zbmath.org/?q=an:1018.55001).



Obstructions for the lifting problem after a pull-back
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.


