# Fundamental group of quotient of subspace of $\mathbb{R}^3\times\mathbb{R}^3$

This is exercise 12-6 from J. Lee's Introduction to Topological Manifolds.
Let $$E=\{(x,y)\in\mathbb{R}^3\times\mathbb{R}^3:x\neq y\}$$, and define an equivalence relation by $$(x,y)\sim(y,x)$$ for all $$(x,y)\in E$$. Compute the fundamental group of $$E/\sim$$.
This is the first time I'm asked to compute the fundamental group of something that I can't seem to visualise, and I am not sure how to approach it. Seifert-van Kampen does not seem useful for such a space. What I think I want to do is find a group $$G$$ such that $$q:F\to F/G=E/\sim$$ is a covering map for a simply connected $$F$$, so that the fundamental group is isomorphic to $$Aut_q(F)=G$$. But even then, I am not quite sure how to do this. Can this space be visualised somehow? If not, how would one generally approach such a problem where one's visualisation is insufficient, as I guess it is in most worthwhile problems?
I would appreciate some advice on this.

• Maybe, not sure $\Bbb Z_2$-action on the simply-connected space $E$. Commented Sep 21, 2020 at 9:15
• @0-thUser As in, the automorphism group would just be flipping the decks, and $E$ is a double covering of $E/\sim$? Commented Sep 21, 2020 at 10:20

Consider the map $$L : E\rightarrow \mathbb R^3\times \mathbb R^3-\{ 0\}$$ $$(x,y)\rightarrow \left (x+y,y-x\right )$$ This is clearly a homeomorphism. So your space $$E$$ is homotopy equivalent to $$\mathbb R^3-\{ 0\}\simeq \mathbb S^2$$ which is simply connected.

This already gives you the fundamental group to be $$\mathbb Z_2$$ but I shall attempt to identify the homotopy type of this space.

$$E/_\sim$$ is harder to see. Using the isomorphism $$L$$ we see that $$E/_\sim$$ is homeomorphic to $$X=\mathbb R^3\times \mathbb R^3-\{ 0\}/_{(v,x)\sim (v,-x)}$$

[This should remind you of the tangent bundle of $$\mathbb {RP}^2$$ upto a homotopy which I shall try to make precise]

$$\mathbb R^3\times \mathbb R^3-\{ 0\}\xrightarrow{\text{projection }} 0\times \mathbb R^3-\{0 \}$$ is a strong deformatrion retract. Let $$A= 0\times \mathbb R^3-\{0 \} /_{(0,x)\sim (0,-x)}$$

We have the weak retract $$r: X\rightarrow A$$ $$[(v,x)]\mapsto [(0,x)]$$

Consider the homotopy $$H_t : X\rightarrow X$$ $$[(v,x)]\mapsto [(tv,x)]$$ This clearly shows $$r$$ is a strong deformation retract and hence a homotopy equivalnce. So $$E/_\sim \simeq \mathbb {RP}^2$$ and once again we can see that $$\pi_1(\mathbb {RP}^2)=\mathbb Z_2$$

• Very nice solution, thank you! Commented Sep 21, 2020 at 19:21