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

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

## 1 Answer

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).