Loop space suspension/adjunction

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 $$ \langle \Sigma X,Y \rangle \cong \langle X, \Omega(Y) \rangle \,\,\, ? $$

Solution 1:

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).