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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 $$\pi_k(\Omega SY)=[S^k,\Omega SY]=[S^{k+1},SY]=\pi_{k+1}(SY)=\pi_k(Y)$$

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  • $\begingroup$ Do you intend this to be the reduced suspension? $\endgroup$ – Connor Malin Apr 26 at 11:39
  • $\begingroup$ It seems to me this map is injective. $\endgroup$ – Connor Malin Apr 26 at 11:42
  • $\begingroup$ @ConnorMalin You should give an official answer. In fact, the map $\iota$ is given by $\iota(y)(z) = [z,y] \in SY = S^1 \wedge Y$. $\endgroup$ – Paul Frost Apr 26 at 14:46
  • $\begingroup$ I guess I was not confident I was interpreting the question correctly. I will add an answer. $\endgroup$ – Connor Malin Apr 26 at 14:57
  • $\begingroup$ @ConnorMalin yes, it is the reduced suspension $\endgroup$ – Aolong Li Apr 26 at 21:46
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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).

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