How can this function have an associate function?

Question

From Rotman's Algebraic Topology:

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$(\Sigma, \Omega)$ is an adjoint pair of functors in $\text{hTop}_*$, where $\Sigma$ is the reduced suspension and $\Omega$ is the Loop space operator.

PROOF:

If $X$ and $Y$ are pointed spaces, define $\tau_{XY}=[\Sigma X, Y] \rightarrow [X, \Omega Y]$ by $[F] \rightarrow [F^\text{#}]$ where $F^\text{#}$ means the associate function.

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How can this function have an associate function? Associate functions are defined as followed:

If $F : Z \times Y \rightarrow X$ is a function, then its associate is the function $F^\text{#}:Z \rightarrow \text{Hom}(Y,X)$ defined by $F^\text{#}(z)=F_z$ (where $F_z: y \rightarrow F(z,y)$)

But $\Sigma X = X \times I / (X \times \dot I \cup \{x_0\} \times I)$ (where $x_0$ is the basepoint of $X$) is not of this form and therefore an associate function should not be defined here.

Can someone explain to me what's going on here?

Answer 1

A map from $\Sigma X$ to $Y$ induces a map from $X\times I$ to $Y$ (by composing with the projection from $X\times I$ to $\Sigma X$), and that has an associate: a map from $X$ to $\Omega Y$.

Answer 2

There is a very close relationship between $\Sigma X$ and $X \times I$. There is a similarly close relationship between $\Omega Y$ and $Y^I$. These relationships are so close that it's often not worth commenting when passing back and forth between them: the reader is expected to fill in those minor details.

The proof gives the formula for the isomorphism in terms of the adjunction $[X \times I, Y] \to [X, Y^I]$. You, the reader, are expected to fill in both of the aforementioned relationships to turn this into a function $[\Sigma X, Y] \to [X, \Omega Y]$ (and you're expected to check the result is well-defined).