I am interested in understanding the topology of the space;
$ X = S^2 \times S^2 \setminus Diagonal = S^2 \times S^2 \setminus \{ (x,x): x\in S^2 \} = \{ (x,y)\in S^2 \times S^2 : x \neq y \} $
I considered another example, namely $ Y = S^1 \times S^1 \setminus Diagonal = S^1 \times S^1 \setminus \{ (x,x): x\in S^1 \} $.
Considering the planar representation of the torus as a square under proper identifications, removing the diagonal leaves an annulus: When we remove the diagonal from the square, we end up with two disconnected triangles and shifting one to the top of the other realizes the identification on one side. Then we end up with a rectangle with only one opposite sides to be identified and identifying these two sides gives us an annulus.
Maybe with a similar argument, can we understand the topology of the space $ X = S^2 \times S^2 \setminus Diagonal = S^2 \times S^2 \setminus \{ (x,x): x\in S^2 \} $ ?
Any kind of answer or hint is appreciated.
Consider the projection $X \to S^2$ on the first component. The fiber of this map over $x \in S^2$ is $S^2 \backslash \{x\} \cong \Bbb R^2$. This means that your space is a family of $2$-planes over the sphere. We can see it as a family of vector spaces, with origin given by $\{(a,-a) : a \in S^2\}$. The exact identification is a bit delicate to write down but I believe it's true. In particular this means $X$ retracts onto $S^2$.
To understand $X$ we just need to find its Euler class $e(X)$ of the vector bundle. I won't write down the definition but here is an example : over $S^1$ there are two possibilities : $e(X) = 0$ and $X = S^1 \times \Bbb R$ (i.e the annulus you found) or $e(X) = 1$ and $X$ is a Moebius band.
To find $e(X)$ we need to take a family of vectors in $X$ and see how many zeroes and poles we found. We take then $e(X) = \#zeroes - \#poles$. Now we take a little rotation around the $z$-axis, say $r : S^2 \to S^2$. This gives $s : S^2 \to X, (x,r(x))$. In fact, $s$ is not defined at the north pole and the south pole, giving two poles. There are 4 zeroes so we find $e(X) = 2$ i.e $X = TS^2$.
To recover back your previous answer, we see that $s : S^1 \to S^1 \times S^1, t \mapsto (t, t+\pi/2)$ has no zero or poles, so it means that $e(X) = 0$ and we find indeed an annulus.
Remark : In fact, I believe we should always have $TS^n \cong \{(x,y) \in S^n : x \neq y\}$. A vague argument is that the normal bundle of the diagonal $\Delta \subset X \times X$ is always isomorphic to the tangent bundle of $X$. Not sure how to exactly apply it but it seems a start ... (in fact see the comment by Balarka Sen. I should say I need to think a bit more about it)