Let $A=\{(x,y) \in \Bbb S^n \times \Bbb S^n \mid x \ne y \}$. Let $f: \Bbb S^n \to A, \ x \mapsto(x,-x).$ Show that $f$ is a homotopy equivalence. 
Let $A=\{(x,y) \in \Bbb S^n \times \Bbb S^n \mid x \ne y \}$ where $n \ge0$. Let $$f: \Bbb S^n \to A, \ x \mapsto(x,-x).$$ Show that $f$ is a homotopy equivalence.

I tried to dumb this down as much as possible as I couldn't figure out how the points $x \ne y$ look like on the torus. So if we consider $\Bbb S^0 \times \Bbb S^0$ i.e. $\{-1,1\} \times \{-,1,1\}$ we're left with $\{-1,1\}$ when we impose the condition $x \ne y$. Pictorially we're removing two points if we consider $\{-1,1\} \times \{-,1,1\}$ in the plane (we have two choices on what to remove, but I don't think we lose generality if we remove $(-1,0)$ and $(0,1))$.
So now we have a map $f:\{-1,1\} \to A$ such that $x \mapsto (x,-x)$ so $f(1)=(1,-1)$ and $f(-1)=(-1,1)$. I don't see how the homotopy equivalence should be constructed here. Is it a bad idea to consider the case $n = 0$? I feel like I'm leading myself into a bad path with it.
 A: We have to find a homotopy inverse for $f$. It seems obvious that a nice candidate is
$$g : A \to S^n, g(x,y) = x .$$
In fact $g \circ f = id$. The map $r = f \circ g$ is given by
$$r(x,y) = (x,-x) .$$
We have to show $r \simeq id$. Define
$$H : A \times I \to A, H(x,y,t) = \left(x, \frac{(1-t)y -tx}{\lVert (1-t)y) - tx \rVert} \right).$$
To show that this is well-defined we use
Lemma. If $x,y \in S^n$ and $a, b \ge 0$ such that $a x = b y$ and at least one of $a, b \ne 0$, then  $x = y$.
Proof. We have $a = a \cdot 1 = a \lVert x \rVert =  \lVert ax \rVert = \rVert by \rVert = b$, thus $a = b  \ne 0$ which shows $x = y$.
$H$ is wll-defined because

*

*$(1-t)y \ne tx$ for all $t \in I$:
The equation $(1-t)y = tx$ implies $x =y$ due to the Lemma. This is impossible for $(x,y) \in A$.


*$x \ne  \frac{(1-t)y -tx}{\lVert (1-t)y) -x \rVert}$ for all $t \in I$:
The equation $x =  \frac{(1-t)y -tx}{\lVert (1-t)y) -x \rVert}$ means that $\lambda  x = (1-t)y - tx$ for some $\lambda > 0$, i.e. we have $(\lambda + t)x = (1-t)y$. Since $\lambda + t > 0$ , the Lemma shows $x = y$ which is impossible for $(x,y) \in A$.
Clearly $H_0 = id$ and $H_1 = r$.
A: This is not a complete answer. Just idea.
Let $X=S^n\times S^n$ and $A=\{(x,x)|x\in S^n\}$. We want to prove that $X-A\simeq S^n$.
My idea of for a proof:
We can show that there is an homeomorphism $S^n\times S^n-A\cong S^n\times S^n-(\{x_0\}\times S^n)$. We can do that by a linear map moving $y=x$ to $x=0$.
But, then $S^n\times S^n-A\simeq S^n\times S^n-(\{x_0\}\times S^n)\simeq (S^n-\{x_0\})\times S^n\simeq \{*\}\times S^n\simeq S^n.$
A: Let V be a real vector space of dimension at least 2. Then VxV is the direct sum of the diagonal D, the set of points (x,x), and the antidiagonal A, the set of the points (x,-x). You should have no problem showing that the inclusion A-0 into VxV-D is a homotopy equivalence.
Now normalize vectors and get the same result for spheres.
