Let $\pi: X \to B$ be a fibration with $B$
path-connected CW complex filtered by
$p$-skeleta $B_0 \subset B^1 \subset … \subset B^p \subset …
B^{\dim(B)}=B$. This induces a filtration on $X$ via
$X_p := \pi^{-1}(B^p)$.
By construction the pair $(B, B^p)$ is $p$-connected,
ie $\pi_i(B^p) \to \pi_i(B)$ are isomorphisms for
$i < p$ and $\pi_p(B^p) \to \pi_p(B)$ is surjective;
equivalently the relative homotopy groups
$\pi_i(B, B^p)$ are zero for $i \le p$.
In Allen Hatcher's script on Spectral Sequences (p. 526) is claimed that
since $(B, B^p)$ is $p$-connected, the homotopy lifting property implies that
$(X, X_p)$ is also $p$-connected. Why?
Fibrations are stable under pullbacks, so
since $\pi: X \to B$ is fibration, then the restriction
$\pi: X_p \to B^p$ is fibration too. Fibration satisty
homotopy lifting property Why this implies that $(X, X_p)$ is $p$-connected?
The hard way to do it might be to consider a big
commutative diagram comparing the Puppe
sequences associated to pairs $(X, X_p)$ and $(B, B^p)$ as horizontal
sequences and fibration sequences associated
to $F \to X \to B$ and $F \to X_p \to B^p$ with
isomophic fibers as vertical sequences.
But I think that Hatcher had there a simpler argument
is mind, what alse should be the reason for the brief reminder
on homotopy lifting property above.
Using homotopy lifting property for $\pi: X \to B$
and $p$-connectedness of $(B, B^p)$ I showed that
for every $i \le p$ and every pointed map
$\phi: (D^i, S^{i-1}, s_0) \to (X, X^p, x_0)$ with
$\phi(S^{i-1}) \subset X^p$ there is a
relative homotopy $H: D^i \times [0,1] \to X$
with $H_0= \phi, H_1(D^i) \subset X^p$ and
for all $t \in [0,1]$ we have $H_t(S^{i-1}) \subset X^p$.
Does if suffice to prove that $[\phi]=0$ in
$\pi_i(X, X^p)$? Or should the homotopy above be a
$rel \ S^{i-1}$ homotopy in the sense that
$H_t \vert _{S^{i-1}}$ is independent of $t$?
Equivalently, when is a class $[\varphi] \in \pi_i(X, X^p)$
zero? When there exist just a homotopy $H:D^i \times [0,1] \to X$
with $H_0=\varphi$ and $H_1(D^i) \subset X^p$ or
should this homotopy be a $rel \ S^{i-1}$ homotopy,
ie on $S^{i-1}$ independent of $t$?
Best Answer
Indeed, you have to construct a homotopy $\operatorname{rel}{S^{i-1}}$ but you're almost there. The crux is that you should not only use the homotopy lifting property of $\pi$ with respect to $D^i$ but rather to the pair $(D^i, \partial D^i)$. That $\pi$ indeed has the homotopy lifting property against such a pair is for example described in Hatcher, p. 376.
If you use this and repeat your argument, you will have constructed a homotopy $\operatorname{rel}{S^{i-1}}$.