Suppose $X$ a CW complex, Y its subcomplex, $T$ is any parameter space.
If we have map: $Y \to X \to X/Y$, the first map is inclusion and the second map is collapsing map, if we apply all spaces with mapping functor obtaining:
$Map(X/Y, T) \to Map(X, T) \to Map(Y, T)$.
I wonder if this sequence of mappings a fibre bundle?
I'm reading the Geometry of Four Manifolds by Donaldson, he states a special case for $X$ a simply-connected 4 manifold, $Y$ its $2$-cells, $T$ the classfying space $BG$ and he only consider a connected components of $Map(S, BG)$, here $S$ can be $X, Y, X/Y$.
Thanks for you answer!
Now I believe this general case is not true.
Suppose we have pair of space $(S^n, D^{n+1})$ and $T$ is some space with nontrivial $n$th homotopy group.
Now $Map(S^n, T)$ consists of all nontrivial equivalent classes represent the nontrivial homotopy group of $T$ but $Map(D^{n+1}, T)$ only consists of the map which homotopy to trivial map, so for this case it fails to be a fibre bundle.
Maybe the fibre mentioned in this book is just the preimage of a fix based point.