Showing exactness of a short sequence Let $p: E \rightarrow B$ be a fibration with fiber $F:= p^{-1}(b), b\in B$ fixed. If $B$ is path connected,
I have the following question: How can I show that for any space $X$, the sequence $$[X,F] \rightarrow [X,E] \rightarrow [X,B]$$ is exact.
P.S: $\quad i: F\rightarrow E$ is the inclusion map. $\,\,$Also $p_\star:[X,E] \rightarrow [X,B]$ and $i_\star: [X,F] \rightarrow [X,E]$ are induced by $i$ and $p$.
This is all I understood so far. I know that we have to show that $ker 
 \,\,p_\star = Im \,\,i_{\star}$. Also, I know that $p_\star[f] = [p\circ f]$ where $f: X\rightarrow E$. I pretty much know all the definitions here but I havent built that much intuition to get a proof. I have just been stuck on this for couple of days. Any help on this will be highly appreciated.
 A: Follow the definitions. You have a sequence
$$[X,F]\xrightarrow{\ i_* \ }[X,E]\xrightarrow{\ p_* \ }[X,B]$$
which you would like to show is exact, i.e., 
$$\operatorname{im}(i_*)=\ker(p_*) = \{[\phi]\in[X,E]\ |\ p_*[\phi] = [c_b]\},$$ 
where $c_b:X\to B$ is the constant map sending everything to $b$. So, you need to show that $\operatorname{im}(i_*)\subseteq\ker(p_*)$ and $\ker(p_*)\subseteq \operatorname{im}(i_*)$. 
For the first inclusion, let $[g]\in[X,E]$ be an element of $\operatorname{im}(i_*)$. Then, $[g] = i_*[f] = [i\circ f]$ for some $[f]\in[X,F]$. Then, since $(p\circ i)(x)=b$ for all $x\in F$, we have:
$$p_*[g] = p_*[i\circ f] = [p\circ i\circ f] = [c_b].$$ 
This was the easy inclusion. 
For the other inclusion, take a map $[g]\in\ker(p_*)$. Then, $p_*[g]=[c_b]$, or in other words, $p\circ g\simeq c_b$ as maps from $X\to B$. That is, there exists a homotopy
$$H:X\times I\to B,\quad H(x,0)=(p\circ g)(x),\quad H(x,1)=b.$$
This means that the square:
$$\require{AMScd} \begin{CD}
    X @>{g}>> E\\ @V{i_0}VV @VV{p}V\\
    X\times I @>>{H}> B
    \end{CD}$$
commutes, where $i_0(x) = (x,0)$. I think I'll let you take it from here. Apply the lifting property of fibrations, and look at what the new homotopy does at $t=1$.
One minor comment here: path-connectedness of $B$ guarantees that all of the fibers $\{p^{-1}(b)\ |\ b\in B\}$ are homotopy equivalent, and so the choice of fibre doesn't matter. It also guarantees that there is up to homotopy, only one constant map $c_b:X\to B$.
