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I want to know the proof of $[\Sigma X, K] = [X,\Omega K]$ which is called "adjoint relation "

Here $\Sigma X = SX/\{x_0\}\times I$, and $\Omega K$ is a space of loops in $K$ at chosen base point.

And $[Y,Z]$ is a set of homotopy classes of a map $Y\to Z$.

Thanks in advance.

[Refer]

Same question is already discussed :

What is the easiest way to see $\langle \Sigma X, Y \rangle\cong \langle X,\Omega Y\rangle $

But this article provides a rough idea, or hom tensor adjunction argument, which is algebraic.

[Partial Explanation]

By definition of $[\Sigma X, K]$, we have $[\Sigma X, K] \subset [X,\Omega K]$ and for $f\in [\Sigma X, K]$, $f$ gives $f \in [X,\Omega K] $ such that $f(x_0)$ is a constant loop.

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HK Lee
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1 Answers1

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A proof is provided in my answer:

Loop space suspension/adjunction

I'm not sure I follow your partial explanation: the homotopy class of the map $f\in [X,\Omega K]$ sending all $x\in X$ to the constant loop at the basepoint of $K$ corresponds to the homotopy class of the constant map in $[\Sigma X,K]$ under the loop space/suspension adjunction. The restriction of a map $\Sigma X\to K$ to $X$ doesn't even define a map $X\to \Omega K$. (Why?) So, we don't have an inclusion $[\Sigma X,K]\subseteq [X,\Omega K]$ as you describe.

I hope this helps!

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Amitesh Datta
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