Figure 7.7. and 7.8 in Loring Tu's An Introduction to manifolds

Question

Figure 7.7 and 7.8 are showed as follows in Loring Tu's An Introduction to manifolds:

However, I cannot understand why $\mathbb RP^2$ cannot be embedded as a submanifold of $\mathbb R^3$. The the orientation of the boundary of Fig7.7 and Fig 7.8 seems not the same, how do they relate to each other? I would appreciate it if someone could explain this for me.

Answer 1

In the first of the four subfigures of Figure 7.8, the author has taken the liberty of breaking the boundary of the disc into four segments instead of just two, and has labelled how to identify those four segments in pairs. But if you compare that with the identifications shown in Figure 7.7, you'll see that what actually happens in both cases is that each point on the circle is identified with the antipodally opposite point of the circle. So the identifications in Figure 7.7 and the first subfigure of 7.8 are really identical, hence both yield $\mathbb RP^2$.

What happens in the next three subfigures of Figure 7.8, though, is that certain additional identifications are taking place in order to get a map into $\mathbb R^3$: besides the antipodally opposite 2-to-1 A,C identification, and the antipodally opposite 2-to-1 B,D identification, the remaining points on the circle are identified 4-to-1. That means that the induced map $\mathbb R P^2 \to \mathbb R^3$ has certain 2-to-1 identifications and hence is not injective.

But that only raises the question: Why isn't there some other continuous function $\mathbb R P^2 \to \mathbb R^3$ which is injective? Well, there isn't any such function. Proving it takes some substantial work in algebraic topology, and it's not immediately obvious how to prove it. But that's really beyond the scope of your question.

Answer 2

I think it is easier to see it if, instead of an equator, you remove a square from $S^2$, identifying the opposite sides (which will have opposite orientations). This is what you see in figure 7.8. The double arrows correspond to a pair of opposite sides of the square and the single ones to the other pair. When you identify them, you end up with the last figure in 7.8, which shows that you necessarily have self intersections (all the points on the vertical segments are double)