What is the fiber of the map $Y\to \Omega SY$ I am trying to write down explicitly the fiber of the map $$Y\to \Omega SY,$$  where $\Omega$ is the loop space functor and $S$ is the suspension functor. 
The map is the adjoint map of identity map $SY \to SY$ via
$$[Y,\Omega SY]_0 \cong [SY,SY]_0.$$
More explicitly, the map sends $y\in Y$ to the element $\gamma\in \Omega SY $ defined by $x\mapsto [(x,y)]$ where $x\in S^1$ and $[(x,y)]$ is the image of $(x,y)\in S^1\times Y$ in $\Omega SY$ through the quotient map.

My motivation is to solve Exercise 177 on Davis-Kirk:

Show that for any $k$-dimensional CW-complex $X$ and for any $(n-1)$-connected space $Y$ ($n\geq 2$) the suspension map 
  $$[X,Y]_0\to [SX,SY]_0$$
  is bijective if $k<2n-1$ and surjective if $k=2n-1$. 

Following the hint, we consider the map $[X,Y]_0\to [Y,\Omega SY]_0$ instead. Convert the map $Y\to \Omega SY$ to a fibration and apply obstruction theory as well as the Freudenthal suspension theorem.
To prove the map is surjective, it is equivalent to consider the lifting problem 
$$\require{AMScd}
\begin{CD}
  @.  Y \\
@. @VVV \\
X @>{}>> \Omega SY
\end{CD}
$$
To solve this, we need the obstruction theory, which requires us to think about the fiber of the map $Y\to \Omega SY$.

P.S. 
The Freudenthal suspension theorem tells us that the suspension homomorphism 
$$S: \pi_k(Y) \to \pi_{k+1}(SY)$$
is an isomorphism if $k<2n-1$ and an epimorphism if $k=2n-1$.
We can use the theorem via
$$\pi_k(\Omega SY)=[S^k,\Omega SY]=[S^{k+1},SY]=\pi_{k+1}(SY)=\pi_k(Y)$$
 A: One can directly see it is injective since if $y$ is not the basepoint the loop $[0,1]\rightarrow \Sigma Y$ given by $x \rightarrow [(x,y)]$ has $[(.5,y)]$ in its image only for that specific y. If $y$ is the basepoint then we get the constant loop, and no other $y$ gives a constant loop since in the image are both $[(.25,y)]$ and $[(.5,y)]$.
There is also a more general categorical observation. If $F:C \rightarrow D$ is a left adjoint and faithful, then if $f^\sharp$ is a monomorphism, $f^\flat$ is as well. In $Top_*$, monomorphisms are injective, basepointed maps. Since $1:\Sigma Y \rightarrow \Sigma Y$ is a monomorphism, its transpose is as well because the suspension functor is faithful.
Here is a proof of the categorical fact:
Suppose $f^\flat h = f^\flat h'$, then we have $(f^\sharp Fh)^\flat = (f^\sharp Fh')^\flat $, so since $\flat$ is an isomorphism we know that $f^\sharp Fh=f^\sharp Fh'$, so $Fh=Fh'$ (since $f^\sharp$ is a monomorphism) which only happens if $h=h'$ (since $F$ is faithful).
