## Natural Equivalence relating Suspension and Loop Space

Theorem:
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.
Proposition
\label{prop13}
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}.

Proposition
\label{prop8}
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.

Suspension
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 $\displaystyle A: [SX, *; Y,y_0]\to [X, x_0; \Omega Y, \omega_0].$