Adjoint theorem for loop and suspension Ref: Davis and Kirk, Lecture Notes on Algebraic Topology
On pp.114 it is the adjoint theorem for the category of Hausdorff compactly generated spaces: for $X,Y,Z$ compactly generated $f(x,y) \mapsto \tilde{f}(x)(y)$ gives a homeomorphism $$ \text{Map}((X \times Y),Z) \longrightarrow \text{Map}(X,\text{Map}(Y,Z)) $$
On pp.137 restricting to pointed maps we obtain its based version which states that there exists the following homeomorphism $$ \text{Map}_*((X \times Y),Z) \longrightarrow \text{Map}_*(X,\text{Map}_*(Y,Z)) $$
On pp.138 substituting in the definition of suspension $\Sigma{X} = S^1 \times X/S^1 \vee X = S^1 \wedge X$ and the definition of based loop space $\Omega{X}$ = Map$_*(S^1,X)$ we obtain the following homeomorphism $$ \text{Map}_*(\Sigma{X},Y) \longrightarrow \text{Map}_*(X,\Omega{Y}) $$ which means that loop and suspension are adjoint
My question pertains to the last step where we substitute in the definition of suspension $\Sigma{X} = S^1 \times X/S^1 \vee X = S^1 \wedge X$.  It is really not a direct product between $S^1$ and $X$ but the theorem seems to suggest that the quotient does not matter.  I don't understand why.
PS: It turns out that the isomorphism that I copied from pp.137 is wrong!  We can't simply add the subscripts to indicate that we are considering basepoint preserving maps.  We should also change $X \times Y$ to $X \wedge Y$.  For reason please refer to Mike's answer below!
 A: One should also have the property that if $\text{Map}(X,A;B)$ denotes the space of maps $X \to B$ that are constant on $A$, $\text{Map}(X,A;B) \cong \text{Map}(X/A,B)$ for a closed subset $A \subset X$. This implies from the product formula that $\text{Map}_*(X \wedge Y, Z) \cong \text{Map}_*(X, \text{Map}_*(Y,Z))$, by (as you say) quotienting: take $A = X \vee Y$. 
To unpack: consider the subset of $\text{Map}_*(X \times Y,Z)$ consisting of maps that are constant (say $z_0$) when restricted to $X \vee Y$ i.e. any map $f$ in that subset satisfies $f(x_0,y) = f(x,y_0) = z_0$, denote by $\tilde{f}$ the image of $f$ under the map $\text{Map}(X \times Y,Z) \to \text{Map}(X,\text{Map}(Y,Z))$, then
(1) $f(x_0,y) = z_0$ implies that $\tilde{f}(x_0)$ is a constant map in $\text{Map}(Y,Z)$ thus $\tilde{f}(x_0)$ is the basepoint of $\text{Map}(Y,Z)$, hence requiring that $\tilde{f}: X \to \text{Map}(Y,Z)$ is basepoint preserving is equivalent to requiring that $f(x_0,y) = z_0$
(2) $f(x,y_0) = z_0$ implies that $\tilde{f}(x)(y_0) = z_0$ $\forall x$, which is exactly the condition that the image of $\tilde{f}: X \to \text{Map}(Y,Z)$ lies within $\text{Map}_*(Y,Z)$---the subset of basepoint preserving maps of $\text{Map}(Y,Z)$
combining the above two conditions we see that maps in $\text{Map}_*(X,\text{Map}_*(Y,Z))$ correspond precisely to those in $\text{Map}_*(X \times Y,Z)$ that are constant when restricted to $X \vee Y$
A: The second display in your question, the one allegedly copied from page 137, is incorrect.  In place of $X\times Y$ it should have $X\land Y$ (see Theorem 6.33).  This correction resolves the suspension problem, because $\Sigma X=S^1\land X$.
