Let be $X,Y$ Hausdorff spaces. I denote with $\langle \cdot, \cdot \rangle$ the basepoint preserving homotopy classes of maps, with $\Sigma$ the reduced suspesion and with $\Omega$ the loop space. How can I prove that $$ \langle \Sigma X,Y \rangle \cong \langle X, \Omega(Y) \rangle \,\,\, ? $$
Algebraic Topology – Loop Space Suspension and Adjunction
algebraic-topologytopological-k-theory
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Quite accidentally I've stumbled on a result that basically answers my question.
Proposition A.16 in the appendix of Hatcher is the following:
The natural bijection $$\newcommand\Maps{\operatorname{Maps}}\Maps(X,\Maps(Y,Z)) \simeq \Maps(X\times Y,Z)$$ is a homeomorphism assuming $X$ is Hausdorff and $Y$ is locally compact Hausdorff.
Sketch of proof:
The key point is that $X\times Y$ and $X$ are Hausdorff spaces, so compact subsets of these spaces will be compact Hausdorff, and thus normal. We can use this normality to prove the following two lemmas, which combine to give the result.
Let $M(K,U)$ denote a compact-open subbasic set. Then
- $M(A\times B, U)$ is a subbasis for $\Maps(X\times Y,Z)$, with $A$ compact in $X$, $B$ compact in $Y$, $U$ open in $Z$.
- If $X$ is Hausdorff, then for any space $Q$, $M(A,V)$ forms a subbasis for $\Maps(X,Q)$ as $V$ ranges over a subbasis for $Q$, and $A$ ranges over all compact sets in $X$.
Then applying the second result to $Q=\Maps(Y,Z)$, with subbasis $V=M(B,U)$, we find that $M(A,M(B,U))$ is a subbasis for $\Maps(X,\Maps(Y,Z))$.
Since $M(A\times B,U)\leftrightarrow M(A,M(B,U))$ under the natural bijection, this proves that the natural bijection is in fact a homeomorphism. $\blacksquare$
This then answers my question, at least to a point where I am satisfied. (I would still be interested in a counterexample to the conclusion of the proposition when the assumption that $X$ is Hausdorff is dropped.)
To get the desired result, we note that when $X$ is Hausdorff, $Y$ is locally compact Hausdorff, the homeomorphism induces a natural homeomorphism $$\Maps_*(X,\Maps_*(Y,Z)) \simeq \Maps_*(X\wedge Y, Z)$$ using that $\Maps$ applied to a quotient map in the contravariant variable or an embedding in the covariant variable gives an embedding.
Then taking $Y=S^1$, we get that the adjunction $$\Maps_*(\Sigma X, Z)\simeq \Maps_*(X,\Omega Z) $$ is a homeomorphism whenever $X$ is Hausdorff.
This is a general fact about sequential colimits (or even more generally colimits over posets). $\newcommand\colim{\operatorname{colim}}$
Let $A : \mathbb{N}\to \mathcal{C}$ be some sequential diagram. $\mathbb{N}$ has an endomorphism $s$ defined by $s(n)=n+1$. There is a corresponding natural transformation $\sigma : A \to s^*A$, which at $n$ is the map $f_n : A_n\to A_{n+1}$ which is $A(n\le n+1)$, the transition map in the sequential diagram.
In fact, more generally, let $t : \mathbb{N}\to \mathbb{N}$ be any nondecreasing map, then $t$ gives an endomorphism of $\mathbb{N}$ as a category, so we can consider the pullback $t^*A$. If $t$ additionally satisfies $t(n)\ge n$ for all $n$, then there is a natural transformation $\tau: A\to t^*A$ which at $n$ is the unique map $A_n\to A_{t(n)}$ in the diagram $A$.
Now if we assume that $t$ is cofinal (for all $n\in \mathbb{N}$, there exists $N\in\mathbb{N}$ such that $t(N)\ge n$), the following is true.
Proposition. $\colim A$ exists if and only if $t^*\colim A$ exists. In this case if $t(n)\ge n$ for all $n$, so $\tau$ exists, then $\tau$ induces a morphism $\tilde{\tau}:\colim A\to \colim t^* A$ on passing to colimits which is an isomorphism.
Proof.
It suffices to give a natural isomorphism between $\newcommand\cocone{\operatorname{co-cone}}\cocone_A$ and $\cocone_{t^*A}$, the functors that the colimits represent. Given some cocone to $A$, $(X, i_n: A_n\to X)$, we get a cocone to $t^*A$ by taking $(X, i_{t(n)}: A_{t(n)}\to X)$. However, it turns out that we can also go backwards, since $t$ is cofinal. If we are given $(Y,j_m : A_{t(m)}\to Y)$, we can define $i_n : A_n\to Y$ by taking $N$ large enough that $t(N)\ge n$, and then defining $i_n$ to be the composite $A_n \to A_{t(N)} \xrightarrow{j_N} Y$. This is well defined regardless of the choice of $N$ because $\mathbb{N}$ is a poset and $(Y,j_m)$ are a cocone to $t^* A$. This gives us a cocone $(Y, i_n)$ to $A$. These operations are inverse to each other, so we have a natural isomorphism as desired.
Finally we need to show that when $t(n)\ge n$ for all $n$, and if $\colim A$ exists then $\tau$ induces an isomorphism on the colimits. First let's fix some notation. Let $i_n : A_n\to \colim A$ be the inclusions for the colimit of $A$, let $j_n : A_{t(n)}\to \colim t^*A$ be the inclusions of the colimit of $t^*A$, and let $\tau_n: A_n\to A_{t(n)}$ be the components of $\tau$. Then we observe that $\tilde{\tau}$ is defined by applying the universal property of the colimit $\colim A$ to the cocone $(\colim t^*A, j_n \circ \tau_n)$. But we know (up to isomorphism) what the $j_n$ are. By the first half of this proof the $j_n$ are $i_{t(n)}$, so $$j_n\circ \tau_n = i_{t(n)}\circ \tau_n = i_n,$$ since $\tau_n : A_n\to A_{t(n)}$ is a map in the diagram $A$ and $i$ is a cocone to $A$. Thus after potentially composing with an isomorphism $(\colim t^*A, j_n)\to (\colim A, i_{t(n)})$, $\tilde{\tau}$ is actually given by the colimiting cone itself, so we have that for some isomorphism $\phi$, $\phi\circ \tilde{\tau} = 1_{\colim A}$, and thus $\tilde{\tau}$ is an isomorphism itself (namely $\phi^{-1}$). $\blacksquare$
Edited to fix the final argument
Build a modified version of the stable homotopy group colimit sequence as follows.
Fix $a$, let $P_{2n} = \pi_{n+a}(X_a)$, let $P_{2n+1} = \pi_{n+a}(\Omega\Sigma X_a)$, let the maps $f_{2n} : P_{2n}\to P_{2n+1}$ be $\pi_{n+a}(\eta_{X_a})$, and let the maps $f_{2n+1} : P_{2n+1}\to P_{2n+2}$ be given by the composite of $\pi_{1+n+a}$ applied to the structure map $\Sigma X_a\to X_{a+1}$ with the isomorphism $\pi_{n+a}(\Omega\Sigma X_a)\cong \pi_{n+1+a}(\Sigma X_a)$.
Let $d: \mathbb{N}\to \mathbb{N}$ be given by $d(n) =2n$, and let $\delta : P\to d^*P$ be the corresponding natural transformation. Note that $d^*P$ is precisely the sequence that usually defines the stable homotopy groups.
Now consider the $\sigma$ natural transformation, $\sigma_P$, for this sequence. If we restrict to even $n$, then $\sigma_P$ corresponds to $\pi_{\bullet+a}(\eta)$. In other words $d^*\sigma_P = \pi_{\bullet + a}(\eta)$. We have a commutative diagram $$ \require{AMScd} \begin{CD} P @>\sigma_P>> s^* P\\ @V\delta VV @V\delta VV \\ d^*P @>>\pi_{\bullet+a}(\eta)> d^*s^*P.\\ \end{CD} $$ Termwise this commutative diagram is $$ \require{AMScd} \begin{CD} P_n @>>> P_{n+1}\\ @VVV @VVV \\ P_{2n}=\pi_{n+a}(X_a) @>>\pi_{n+a}(\eta_{X_a})> P_{2n+1}=\pi_{n+a}(\Omega \Sigma X_a).\\ \end{CD} $$
Now if we take colimits of this diagram of functors, three of the maps, $\sigma_P$ and both $\delta$s induce isomorphisms on the colimit. Therefore the last map does as well.
Best Answer
Let $(X,x_0)$ be a pointed topological space. Note that:
$\Sigma X=\left(X\times [-1,1]\right)/\left((x,1)\equiv (x_0,t)\equiv (x,-1)\text{ for all }x\in X,t\in [-1,1]\right)$ ($\equiv$ denotes the equivalence relation by which we are taking the quotient)
$\Omega X=\{f:[0,1]\to X:f(0)=x_0=f(1)\}$.
The bijection:
(1) $\phi:\left\langle \Sigma X,Y \right\rangle\cong \left\langle X,\Omega Y\right\rangle$
is defined by the rule:
(2) $\left(\phi(f)(x)\right)(t)=f(x,t)$.
Exercise 1: Prove that $\phi$ is indeed a well-defined bijection. Explicitly write down the inverse $\psi:\left\langle X,\Omega Y\right\rangle\to \left\langle \Sigma X,Y\right\rangle$ of $\phi$ as a rule similar to (2).
I hope this helps! I think this is really one of those results that, while straightforward to prove, is best understood on one's own. So, I recommend staring at (2) for a while until you can really see that it defines a map $\phi$ as in (1). (Also, it's most satisfying to see geometrically how one could arrive at (2); look at the suspension of the circle, for example.) Once you're comfortable that (1) gives you a well-defined map, it would probably be good to do Exercise 1 by explicitly finding an inverse for $\phi$ (from which, of course, it follows that $\phi$ is a bijection).