$[X,F] \to [X,E] \to [X,B] $ is exact sequence of pointed sets

Question

how to show:

if $F \to E \to B $ is a fibration then for any space $X$ the sequence $[X,F] \to [X,E] \to [X,B] $ is exact sequence of pointed sets.

any hints, thanx.

Answer 1

From your comments I think you're misunderstanding the question. $[X,F] \to [X,E] \to [X,B]$ is exact, but $1 \to [X,F] \to [X,E] \to [X,B] \to 1$ is not. That's the whole point of the long exact sequence of homotopy groups for a fibration (where $X = S^n$). So if you've proved that the kernel of the second map is equal to the image of the first, you're done.

For example there is a fibration sequence $\mathbb{Z} \to \mathbb{R} \to S^1$ given by the exponential map $\theta \mapsto e^{i\theta}$. It's clear that $[X,F] \to [X,E]$ will not necessarily be injective, since $E = \mathbb{R}$ is contractible but $\mathbb{Z}$ is not. Similarly $[X,\mathbb{R}] \to [X,S^1]$ will not necessarily be surjective.