Proving the existence of some deformation retract I am trying to find out if the set   $  S ^n \times S^n \setminus \left\lbrace (x,x) \mid x \in S^n \right\rbrace$  deformation retracts onto the subspace $\left\lbrace (x,-x) \mid x \in S^n \right\rbrace$ .
I've been trying to find the homotopy to no avail. 
Any help is greatly appreciated.
 A: Hint: For any pair of distinct points x, y on the sphere there exists a unique great circle passing through these points and this circle depends continuously on x, y. Now, slide y along this circle towards -x. 
A: This is not a complete answer (sorry), but instead a suggestion for a new approach to finding your homotopy.  Instead of the standard coordinates for $\mathbb{R}^n$, try to use the following coordinate system.
For any $(x_1, x_2, \cdots, x_n) \in \mathbb{R}^n$ the following provides a change of variable to $(\rho, \theta_1, \theta_2, \cdots, \theta_{n-1})$:
\begin{eqnarray*}
x_1 &=& \rho \sin(\theta_{n-1}) \sin(\theta_{n-2}) \cdots \sin(\theta_2) \sin(\theta_1) \\
x_2 &=& \rho \sin(\theta_{n-1}) \sin(\theta_{n-2}) \cdots \sin(\theta_2) \cos(\theta_1) \\
&\vdots& \\
x_i &=& \rho \sin(\theta_{n-1}) \sin(\theta_{n-2}) \cdots \sin(\theta_{i-1}) \sin(\theta_i) \cos(\theta_{i-1}) \\
&\vdots& \\
x_n &=& \rho \cos(\theta_{n-1})
\end{eqnarray*}
Where $\rho \in \mathbb{R}^+ \cup \{0\}$, $0 \leq \theta_1 < 2\pi$, and $0 \leq \theta_i \leq \pi$ provided $2 \leq i \leq n-1$.
It can be shown that this system is sufficient to generate the 1-sphere $\mathbb{S}^n$ for all $n \in \mathbb{Z}^+$, although the details of a proof are a bit complicated.  For your purposes it is enough to let $\rho = 1$.
This is a generalization of polar coordinates that I came up with in order to answer a similar question about retracts in the $n$-sphere.  Essentially, the problem I had faced was how to keep the map in $\mathbb{S}^n$.  This provided just such a way.  I hope this helps, if even just a little.
