Natural Equivalence relating Suspension and Loop Space

If (X,x_0), (Y,y_0), (Z,z_0)\in\mathscr{PT}, X, Z Hausdorff and Z locally compact, then there is a natural equivalence \displaystyle A: [Z\wedge X, *; Y,y_0]\to [X, x_0; (Y,y_0)^{(Z,z_0)}, f_0] defined by A[f]=[\hat{f}], where if f:Z\wedge X\to Y is a map then \hat{f}: X\to Y^Z is given by (\hat{f}(x))(z)=f[z,x].

We need the following two propositions in order to prove the theorem.
The exponential function E: Y^{Z\times X}\to (Y^Z)^X induces a continuous function \displaystyle E: (Y,y_0)^{(Z\times X, Z\vee X)}\to ((Y,y_0)^{(Z,z_0)}, f_0)^{(X,x_0)} which is a homeomorphism if Z and X are Hausdorff and Z is locally compact\footnote{every point of Z has a compact neighborhood}.

If \alpha is an equivalence relation on a topological space X and F:X\times I\to Y is a homotopy such that each stage F_t factors through X/\alpha, i.e.\ x\alpha x'\implies F_t(x)=F_t(x'), then F induces a homotopy F':(X/\alpha)\times I\to Y such that F'\circ (p_\alpha\times 1)=F.

Proof of Theorem
i) A is surjective: Let f': (X,x_0)\to ((Y,y_0)^{(Z,z_0)},f_0). From Proposition \ref{prop13} we have that E: (Y,y_0)^{(Z\times X, Z\vee X)}\to ((Y,y_0)^{(Z,z_0)},f_0)^{(X,x_0)} is a homeomorphism. Hence the function \bar{f}: (Z\times X, Z\vee X)\to (Y,y_0) defined by \bar{f}(z,x)=(f'(x))(z) is continuous since (Ef'(x))(z)=f'(z,x) and thus \bar{f}=E^{-1}f'. By the universal property of the quotient, \bar{f} defines a map f:(Z\wedge X, *)\to (Y,y_0) such that f[z,x]=\bar{f}(z,x)=(f'(x))(z). Thus \hat{f}=f', so that A[f]=[f'].

ii) A is injective: Suppose f,g: (Z\wedge X, *)\to (Y,y_0) are two maps such that A[f]=A[g], i.e.\ \hat{f}\simeq\hat{g}. Let H': X\times I\to (Y,y_0)^{(Z,z_0)} be the homotopy rel x_0. By Proposition \ref{prop13} the function \bar{H}: Z\times X\times I\to Y defined by \bar{H}(z,x,t)=(H'(x,t))(z) is continuous. This is because \bar{H}(z,x,t)=(E\bar{H}(x,t))(z) so that E\bar{H}=H', thus \bar{H}=E^{-1}H' where E is a homeomorphism. For each t\in I we have \bar{H}((Z\vee X)\times\{t\})=y_0. This is because if (z,x)\in Z\vee X, then z=z_0 or x=x_0. If z=z_0, then (H'(x,t))(z_0)=y_0. If x=x_0, (H'(x_0,t))(z)=y_0 as H' is the homotopy rel x_0. Then by Proposition \ref{prop8} there is a homotopy H:(Z\wedge X)\times I\to Y rel * such that H([z,x],t)=\bar{H}(z,x,t)=(H'(x,t))(z). Thus H_0([z,x])=(H_0'(x))(z)=(\hat{f}(x))(z)=f[z,x] and similarly H_1([z,x])=(H_1'(x))(z)=(\hat{g}(x))(z)=g[z,x]. Thus [f]=[g] via the homotopy H.

Loop space
If (Y,y_0)\in\mathscr{PT}, we define the loop space (\Omega Y, \omega_0)\in\mathscr{PT} of Y to be the function space \displaystyle \Omega Y=(Y,y_0)^{(S^1,s_0)} with the constant loop \omega_0 (\omega_0(s)=y_0 for all s\in S^1) as base point.

If (X,x_0)\in\mathscr{PT}, we define the suspension (SX,*)\in\mathscr{PT} of X to be the smash product (S^1\wedge X, *) of X with the 1-sphere.

Corollary (Natural Equivalence relating SX and \Omega Y)
If (X,x_0), (Y,y_0)\in\mathscr{PT} and X is Hausdorff, then there is a natural equivalence \displaystyle A: [SX, *; Y,y_0]\to [X, x_0; \Omega Y, \omega_0].

Author: mathtuition88

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