$(\Sigma, \Omega)$ is an adjoint pair in $\text{hTop}_*$ proof question From Rotman's Algebraic Topology:
Shouldn't the red underlined function be $H^{\text{#}}: \Sigma X \rightarrow \Omega Y$ since the associate of a function $F : Z \times Y \rightarrow X$ is defined as $F^{\text{#}}:Z \rightarrow \text{Hom}(Y,X)$?

 A: $\Sigma X$ is a quotient space of $X \times I$ by a subspace (which I will call $S$). By the usual properties of quotients, for any space $Z$, there is a one-to-one correspondence between:


*

*Continuous maps $\Sigma X \to Z$

*Continuous maps $f:X \times I \to Z$ such that $f(s) = f(t)$ for all $s,t \in S$.


That is, one of the usual ways to specify a function on $\Sigma X$ is by specifying a function on $X \times I$ that satisfies the relevant properties.
A: There is a confusing abuse of notation here.  I will work in the pointed category.
First, the adjunction.  For $\Sigma X$ we must take the reduced suspension.  So, in $X \times I$ we must collapse not only the top and bottom of the cylinder $X \times \{0,1\}$ but also the baseline $\{x_0\} \times I$.  All this junk becomes the same point (the basepoint) in the quotient $\Sigma X$.  I'll write $[x,t]$ for the equivalence class of $(x,t)$, which shouldn't get too confusing.   The basepoint is $[x_0,t] = [x,0]=[x,1]$ for ANY $t$ or $x$.
Given $F: \Sigma X \to Y$, we get $F^\#: X \to \Omega Y$ via the formula
$$
F^\#(x)(t) = F[x,t].
$$
For instance $F^\#(x)(0)= F[x,0] = F[x_0,0]=y_0$ since $(x,0) \sim (x_0,0)$ by the gluing. So, $F^\#(x)$ is a loop in $Y$ and you can check all the usual adjunction stuff on the level of maps.  Everything works out because of how we define the reduced suspension. It's fun to check it backwards because then you have to define a map on equiv classes and it ends up OK because of our definition of $\Sigma X$ and loops in $Y$.
Now for your question.  Take a based homotopy $H: \Sigma X \times I \to Y$ of maps $F_0, F_1: \Sigma X \to Y$.  So, all this says:
$$
H([x,t],0)= F_0[x,t] \\
H([x,t],1)= F_1[x,t] \\
H([x_0,t],s) = y_0.
$$
The homotopy Rotman is trying to give you is $H^*: X \times I \to \Omega Y$ via
$$
H^*(x,s) = H([x, -],s)
$$
so that $H^*(x,s)(t) = H([x,t],s)$.  You can check that $H([x, -],s)$ defines a loop in $Y$  because, e.g., $H([x,0],s) = H([x_0, 0],s)=y_0$ so that $0$ maps to the basepoint, as required of loops. 
It should now be clear just from the formulas that 
$$
H^*(x,0) = H([x,-],0) = F_0[x,-] = F_0^\#(x)
$$
so that $H^*$ is "right" on the bottom.  Likewise it is right on the top and the sides, so we have Rotman's claim:
$$
H: F_0 \sim F_1 \Rightarrow H^*: F_0^\# \sim F_1^\#.
$$
This justifies the adjunction passing down to homotopy classes of maps.
If you stare at this and the answer of Hurkyl, you can see what his/her point is.  The point of the reduced suspension is that there is a natural bijection between:
$$
\{\text{based maps } f: \Sigma X \to Y\}
$$
and
$$
\{\text{based maps } f: X \times I \to Y  \text{ such that } f(x,0)= f(x,1) = f(x_0, t) =y_0 \text{ for all } x \in X, t \in I\}.
$$
This is absolutely nothing more than the universal mapping property of the quotient space, given those points are exactly the ones that we glue together to create $\Sigma X$ (reduced). 
TL;DR:  Rotman is loosely thinking that $\Sigma X = X \times I$ (it is not) so a homotopy $H: \Sigma X \times I \to Y$ is a map $H: (X \times I) \times I \to Y$ and this has an associate $H^\#: X \times I \to \Omega Y$.  This isn't strictly right but because of the identifications we make in $\Sigma X$ everything still works. 
A: $\Sigma X$ is a quotient of $X \times I'$ ($I'$ is just $I$, but I want to emphasize that these are different copies of $I$), thus $\Sigma X \times I$ is a quotient of $X \times I' \times I$, and Rotman takes the associate with respect to $I'$ rather than with respect to $I$.
This may seem confusing. On the other hand, to get a map to the loop space $\Omega Y$ and not only a map to $\mathrm{Hom}(I,Y)$ (recall that $\Omega Y$ is a subspace of $\mathrm{Hom}(I,Y)$) you have to take the copy of $I$ that "lives in" the suspension $\Sigma$.
