Can one deduce the long exact sequence of fibration from the long exact sequence of a pair? Given that the following sequence is exact:
$$\dots \to \pi_{n}(A) \to \pi_{n}(X) \to \pi_{n}(X,A) \to \pi_{n-1}(A) \to\dots$$
I want to show that given a bundle $p:E \to B$ with fiber $F$ if we take $A:=F$ and $X:=E$, then we can obtain the long exact sequence of fibration from the above statement.
The intuition for this comes from my suspicion that  $p^*:\pi_{n}(E,F) \to \pi_{n}(B)$ is actually an isomorphism. This is not rigorous, I only convinced myself with some drawings. The following two points are not so clear:
$1:$ would this be enough to guarantee the exact sequence of fibration?
$2:$ How can we consider induced homomorphisms between relative and absolute homotopy?
I thought that one way to proceed would be showing that the usual map $j^*:\pi_{n}(E) \to \pi_{n}(E,F)$ which just views absolute spheroids as relative ones followed by $p^*$ is essentially the same as just considering $p_{*}:\pi_{n}(E) \to \pi_{n}(B)$ which is the map induced by projection (a commutative diagram.)
Any hints on proving such an isomorphism in general are welcome.
 A: This is exactly how the long exact sequence of a fibration is derived in some books, like Switzer's text (chapter 4). Just as you said, we have the long exact sequence of the pair $(E,F)$ (omitting basepoints):
$$\cdots\xrightarrow{\ \ \ }\pi_n(F)\xrightarrow{\ i_* \ }\pi_n(E)\xrightarrow{\ j_* \ }\pi_n(E,F)\xrightarrow{\ \partial \ }\pi_{n-1}(F)\xrightarrow{\ i_* \ }\pi_{n-1}(E)\xrightarrow{\ \ \ }\cdots$$
Now, your suspicion is right, in fact, given any weak fibration $p:(E,e_0)\to (B,b_0)$ ($p$ only has the HLP with respect to all disks), $B'\subseteq B$ and $E':=p^{-1}(B')$, the induced map
$$p_*:\pi_n(E,E',e_0)\to\pi_n(B,B',b_0)$$
is an isomorphism of groups for $n\geq 2$ and a bijection for $n=1$. This is proven by induction on $n$. In particular, for $B'=\{b_0\}$, $E' = p^{-1}(b_0)=F$, the induced map
$$P_*:\pi_n(E,F,e_0)\to\pi_n(B,\{b_0\},b_0)\cong \pi_n(B,b_0)$$
is an isomorphism for $n\geq 2$ and a bijection for $n=1$. Here $P_*$ is the composition of $p_*$ with the isomorphism at the end. The isomorphism $\pi_n(B,\{b_0\},b_0)\cong \pi_n(B,b_0)$ is how you get from relative to absolute. Hence, we can "cut and paste" with $P_*$ to obtain:
$$\cdots\xrightarrow{\ \ \ }\pi_n(F)\xrightarrow{\ i_* \ }\pi_n(E)\xrightarrow{\ P_*\circ j_* \ }\pi_n(B)\xrightarrow{\ \partial\circ P_*^{-1} \ }\pi_{n-1}(F)\xrightarrow{\ i_* \ }\pi_{n-1}(E)\xrightarrow{\ \ \ }\cdots$$
From here, you can show that $P_*\circ j_* = p_*$, which gives you the usual long exact sequence of a fibration. 
The proof of the isomorphism $p_*:\pi_n(E,E',e_0)\to\pi_n(B,B',b_0)$ takes a little bit of work, so if you'd like a reference, Switzer covers all of the above in chapter 4.
