Fundamental group of $X = \{(p, q)|p \neq −q\}\subset S^n \times S^n$ This problems appears in Chapter 2, exercise 3 from "A Concise Course in Algebraic Topology J. P. May" book.
Let $X = \{(p, q)|p \neq −q\}\subset S^n \times S^n$. Define a map $f:S^n\to X$ by $f(p) = (p, p)$. Prove that $f$ is a homotopy equivalence.
Attempt:
I need a map $g:X\to S^n$ such that $f\circ g\sim 1_X$ and $g\circ f\sim 1_{S^n}$.
I guess such a map is $g=\pi_1$ the first projection. It satisfies $g\circ f= 1_{S^n}$ but I have problems proving that $f\circ \pi_1\sim 1_X$. I can't even draw what is happening in the case $n=1$ (a subset of torus which I can't imagine).
A nice corollary is: The fundamental group of $X$ is trivial when $n>1$ and is $\mathbb{Z}$ when $n=1$.
 A: Hint: For $(p,q)\in X$, $p\ne -q$. That is, there is a unique geodesic from $p$ to $q$ (the small arc of the great circle containing $p$ and $q$). This can be used to construct a homotopy from $(p,q)$ to $(p,p)$. 
A: Solution 1.
Let$$g: X \to S^n, \text{ }(p, q) \mapsto p.$$We see that $g \circ f = \text{Id}_{S^n}$. Now, let $(p, q) \in X$. There is a well-defined shortest path $\lambda_{p, q}: [0, 1] \to S^n$ starting at $p = \lambda_{p, q}(0)$ and ending at $q = \lambda_{p, q}(1)$, as $p \neq -q$. Define$$H : X \times [0, 1] \to X,\text{ }H(p, q, t) = (p, \lambda_{p, q}(t)).$$Then$$H(p, q, 0) = (p, p) = (f \circ g)(p, q),\text{ }H(p, q, 1) = (p, q) = \text{Id}_X(p, q).$$Moreover, it is easy to see that $H$ is continuous. (This is at least easy to give a handwaving argument for. Basic open sets in $X$ look like direct products of epsilon balls inside $S^n$. These have preimages under $H$ of direct products of epsilon balls in $X$, crossed with $[0, 1]$, which are of course open in the product topology.) Thus, $f \circ g$ is homotopic to $\text{Id}_X$. Hence, $f$ is a homotopy equivalence between $X$ and $S^n$.
Solution 2.
We show that $X$ is the tangent bundle of $S^n$. The way to see this is to visualize the tangent space at a point $p$ as a hyperplane $\mathbb{R}^{n+1}$ with origin at $p$ and then folding the plane over the second $S^n$ factor using stereographic projection. We see that the only points in $S^n \times S^n$ that we miss with this operation are the anti-diagonal points. Now, the zero section under this identification of $X$ with $TS^n$ is the diagonal. Hence, $X$ deformation retracts onto the diagonal. Since $f$ is an isomorphism of $S^n$ with the diagonal, it is a homotopy equivalence between $S^n$ and $X$.
Solution 3.
Here is a way to prove the result using extra machinery but without the topological visualization. Note that both the sphere and $X$ are CW-complexes, and hence, we just need to prove that $f$ induces isomorphisms on homotopy groups. In fact for $n > 1$. Now, using the same $\pi$, as before, it actually suffices to show that $\pi$ induces isomorphisms on homotopy groups. Now, $\pi$ already induces a surjection on homotopy groups, and thus, since the homotopy groups of the sphere are finitely generated, we only need to show that both the sphere and $X$ have the same homotopy groups. This can be done using the long exact sequence of a fibration. $\pi$ is a fibration with contractible fiber $\mathbb{R}^{n+1}$, and hence, the homotopy groups of $X$ are the same as the homotopy groups of the sphere.
A: Since I didn't see anybody writing out this rather simple homotopy explicitly:
Consider $S^n$ as the subspace of $\mathbb R^{n+1}$ defined by $\|x\|=1$. We are looking for a homotopy between $f\circ \pi_1$ and $\operatorname{id}_{X}$, where $f\circ \pi_1$ takes $(p,q)$ to $(p,p)$. Since $p\neq -q$ for $(p,q)\in X$, we have a well defined homotopy
\begin{align*}
H \colon X\times I&\longrightarrow X,\\
\big((p,q),t\big) &\longmapsto \left( p, \frac{(1-t)q+tp}{\|(1-t)q+tp\|}\right).
\end{align*}
Note that since $p\neq -q$ the straight line parametrized by $t\mapsto (1-t)q+tp$ does not cross the origin. Continuity of $H$ is now a just a question of continuity of the norm, of taking quotients of continuous functions, etc.
Effectively we are taking a straight line homotopy in the second $S^n$ and projecting it radially onto the sphere, which yields the geodesics described in the other answers.
