Fibrations lift p-connectedness Let $p: X \to B$ be a Serre fibration, and suppose that $B^p \subset B$ is a subspace of $B$ such that $(B,B^p)$ is $p$-connected, i.e. $\pi_n(B,B^p)=0 \ \forall n \leq p$ or, equivalently, $\pi_n(B^p) \to \pi_n(B)$ is an isomorphism $\forall n < p$ and is surjective for $n=p$.
If we define $X^p$ to be the pullback of $B^p \to B$ along $p$, i.e. $  X^p:=p^{-1}(B^p)$ then, is it true that $(X,X^p)$ is $p$-connected? Using the lifting property which defines Serre fibrations I managed to show that $\pi_n(X^p) \to \pi_n(X)$ is onto for each $n \leq p$, but concerning injectivity, I can only show that if $n<p$ and $\alpha :S^n \to X^p$ is nullhomotopic in $X$, then there exists a map $\beta:S^n \to X^p$ such that $\alpha \simeq \beta$ in $X^p$, and $p \circ \beta$ is constant.
I'd like $\beta$ to be constant, to obtain that $\pi_n(X^p) \to \pi_n(X)$ is injective.
Any hint? Thanks in advance.
 A: Let me denote by $i : B^p \to B$ and by $j : X^p \to X$ the given inclusions, and by $q : X^p \to B^q$ the base change of $p$.
Start with $\alpha : S^n \to X^p$, nullhomotopic in $X$ and let $s : \mathrm{Cone}(S^n) \to X$ be a nullhomotopy for $\alpha$. The composition $ps : \mathrm{Cone}(S^n) \to B$ is a nullhomotopy for $q \alpha$. Choose a homotopy $h : \mathrm{Cone}(S^n) \times I \to B$ in such a way that $h(x,0) = ps(x)$ and that $h(-,1) : \mathrm{Cone}(S^n) \to B$ factors through $B^p$.
The inclusion $Y := (\mathrm{Cone}(S^n) \times\{0\}) \cup (S^n \times I) \to \mathrm{Cone}(S^n)$ is an acyclic cofibration (pushout product property applied to the cofibration $S^n \to \mathrm{Cone}(S^n)$ and to the acyclic cofibration $\{0\}\to I$). It follows that it has a lift against $p : X \to B$ (here I am considering the map $\tilde{s} : Y \to X$ which restricts to $s$ on $\mathrm{Cone}(S^n) \times \{0\} \to X$ and is constantly equal to $\alpha$ on $S^n \times I$). Let $\tilde{h}\colon \mathrm{Cone}(S^n) \times I \to X$ be a lifting. It follows that $p\tilde{h}(x,1) = h(x,1) \in B^p$, and therefore the universal property of the pullback shows that $\tilde{h}(-,1) : \mathrm{Cone}(S^n) \to X$ factors through $X^p$. Since additionally $\tilde{h}(x,1) = \alpha(x)$ for every $x \in S^n \times{1}$, we obtained the desired nullhomotopy inside $X^p$, thus completing the proof.
A better proof: since $p$ is a Serre fibration, $X^p \simeq B^p \times_B^h X$, where $h$ denotes the homotopy pullback. Then $\mathrm{hofib}(X^p \to X) \simeq \mathrm{hofib}(B^p \to B)$ and now the result follows from the long exact sequence of homotopy groups for fiber sequences.
Edit: I corrected the wrong proof.
