# 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

@M.K. the monoidal product is the smash product, which is adjoint to the function space. Smashing with $S^1$ gives you suspension and 'hom'-ing out of $S^1$ gives you the loop-space. See en.wikipedia.org/wiki/Smash_product#Adjoint_relationship – Juan S Sep 24 '12 at 9:38
@M. K. - I'm not sure what you would consider an appropriate answer here. It is the (symmetric) monoidal product in the category of pointed sets (in the same way the tensor product is the monoidal product in the category of (commutative) $R$-modules. It even satisfies a similar universal property - see Rune's answer: mathoverflow.net/questions/105753/… – Juan S Sep 25 '12 at 9:50