Why are $ \sum $ and $ \Omega $ adjoint functors? How to establish rigorously that the suspension $ \sum $ and the loop space $ \displaystyle\Omega $ are adjoint functors on the category of pointed spaces ?
Thanks in advance for tour help.  
 A: It's a special case of the currying adjunction. For sets and nice topological spaces, we have
$$\text{Hom}(X\times Y,Z) = \text{Hom}(X,\text{Hom}(Y,Z))$$
An element of $\text{Hom}(X\times Y,Z)$ is a function of two variables $f(x,y).$ You can treat it as a function-valued function of one variable, $g(x)(y),$ which takes one input $x$ and returns a new function $g(x)$ which then takes another input $y$, with value given by the only thing it can be, viz. $g(x)(y) = f(x,y)$. A function-valued function means an element of $\text{Hom}(X,\text{Hom}(Y,Z)).$
This exchange is called currying in computer science contexts, after Haskell Curry. In math contexts it's also called the tensor-hom adjunction.
If $X$, $Y$ and $Z$ are nice topological spaces (where "nice" means weakly Hausdorff and compactly generated, Steenrod's convenient category of topological spaces. These are mild conditions that make sure the structure maps of this adjunction are continuous), then this is a homeomorphism, and so an enriched adjunction between functors in this category of nice topological spaces.
If you pass to the category of pointed spaces, then the adjunction becomes
$$\text{Hom}(X\wedge Y,Z) = \text{Hom}(X,\text{Hom}(Y,Z))$$
because the pointed maps have to send $*\times Y\cup X\times *$ to the zero map, and $X\times Y$ with its equator and prime meridian $*\times Y\cup X\times *$ smashed to a point is just the smash product $X\times Y/(*\times Y\cup X\times *)=X\wedge Y.$
Setting $Y = S^1$ gives you the required adjunction
$$\text{Hom}(\Sigma X,Z) = \text{Hom}(X,\Omega Z)$$
