A few friends and I have been stuck on this old qualifying question for quite some time now...
Let $D$ be the diagonal subspace of $\Bbb S^2 \times \Bbb S^2$. Show that the projection onto the first coordinate from the complement of $D$ gives a homotopy equivalence.
My question is, what exact map do I choose from $\Bbb S^2$ back to $D$ complement? Any map I choose has a hard time sending the origin anywhere nice. And how do I show that a carefully constructed map is actually homotopic to the identity map when composed with the projection?
There is actually a quite canonical map $$i:S^n\to X\setminus D,\quad x\mapsto (x,-x)$$ If $\text{pr}_1:X\setminus D\to S^n$ is the projection onto the first factor, then $\text{pr}_1\circ i$ is the identity on $S^n$. The other map $i\circ\text{pr}_1$ sends $(x,y)$ to $(x,-x)$. One can show that $$(x,y,t)\mapsto\left(x,\frac{ty+(1-t)(-x)}{||ty+(1-t)(-x)||}\right)$$ is a homotopy between $i\text{pr}_1$ and the identity (actually it is a homotopy rel $i(S^n)$ ).