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

Let $X$ and $Y$ be topological spaces. Let $\langle X,Y\rangle$ denote the homotopy classes of maps from $X$ and $Y$. The reduced suspension $\Sigma(-)$ has the adjoint $\Omega(-)$. In other words, we have $$\langle \Sigma X, Y \rangle\cong \langle X,\Omega Y\rangle$$ for all $X$ and $Y$.

I am always confused with on which side I should put $\Sigma$. What is the easiest or intuitive way to think of this isomorphism? Are there a good way to memorize this this formula?

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I think of the suspension as being made-up of a bunch of "loops" that crash through $X$.  So a map out of $\Sigma X$ to $Y$ is a map from $X$ to $\Omega Y$. – Ryan Budney Sep 24 '12 at 6:37
I agree with Ryan. Given a map $f:\Sigma X \rightarrow Y$, one can get a loop $f_{x}:\{x\}\times I\rightarrow Y$ for each $x\in X$, where $I$ is the unit interval appearing in the definition of reduced suspension. – M. K. Sep 24 '12 at 8:00

## 1 Answer

The easiest way is to think about the adjunctions you do know. I like to think about the Hom-Tensor adjunction. That is all the above is (in the category of pointed spaces). Let me know if you need me to elaborate.

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I would like to see how tensor appears in this context. – M. K. Sep 24 '12 at 8:01

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– Juan S Sep 24 '12 at 9:38
My problem is why the smash product plays the role of the tensor product in the category of pointed spaces. – M. K. Sep 25 '12 at 7:24

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– Juan S Sep 25 '12 at 9:50
Thanks, Juan. I will think about it more. – M. K. Sep 26 '12 at 3:12
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