$\pi_1(\text{P}^2(\mathbb{R}))$ and multiplication by $2$

Question

I want to calculate the fundamental group of the real projective plane $\text{P}^2(\mathbb{R})$ using SVK theorem.

To this end, I choose to model $\text{P}^2(\mathbb{R})\simeq D^2/{\sim} $ as the unit disk $\{x:\|x\|\leq 1\}$ in $\mathbb{R}^2$ quotiented by identifying antipodal points lying on the boundary.

I take

• $A= \text{P}^2(\mathbb{R})-\{y\} $
• $B= \text{P}^2(\mathbb{R})-\partial$, where $\partial=\{x:\|x\|=1\}$
• $A\cap B$

which are all path-connected.

Now, fix a point $x_0 \in A\cap B.$

$A$ can be rectracted by deformation to $S^1$, so that $A \approx S^1$ and $\pi_1(S^1)\simeq \mathbb{Z}.$ The retraction $r_A:A \to S^1$ induces an isomorphism $\pi_1(A,x_0)\simeq\pi_1(S^1,r_A(x_0)) \simeq \mathbb{Z}$ which is given by $[\lambda]_A \mapsto [r_A \circ \lambda]_{S^1}$ for every loop $\lambda$ in $A.$

If I call $c$ the loop corresponding to $1 \in \mathbb{Z}$ under the isomorphism, I have the equality $\pi_1(S^1,r_A(x_0)) = \langle c | \emptyset \rangle $; the deformation, giving a path $h :t \mapsto h(t)=H(x_0,t)$ from $x_0$ to $r(x_0),$ also gives a presentation $\pi_1(A, x_0) = \langle hch^{-1},\emptyset \rangle $, where we can see now the generator as a loop with endpoints $x_0$ instead of $r(x_0).$

On the other hand, $B$ can be contracted to $\{x_0\},$ so $$\pi_1(B,x_0) \simeq \pi_1(\{x_0\},x_0)=\{x_0\} \simeq \{\text{id}\}= \langle \emptyset | \emptyset \rangle.$$

Finally, choosing another circle $S^1_{x_0}$ passing through $x_0$, I retract $A \cap B$ to it so that $$\pi_1(A\cap B, x_0)\simeq \pi_1(S^1_{x_0},x_0)= \langle \gamma| \emptyset \rangle.$$

The inclusion $A \cap B \subset B$ induces a morphism $b_*:\pi_1(A\cap B,x_0) \to \pi_1(B,x_0)$ which can only be the trivial map sending everything to the constant path at $x_0.$

Next, the inclusion $A \cap B \subset A$ induces a morphism of groups $a_*:\pi_1(A\cap B,x_0) \to \pi_1(A,x_0)$ given by $[\ell]_{A\cap B} \mapsto [\ell]_A$ for every loop $\ell$ in $A \cap B$ with endpoints $x_0.$

I want to understand how to prove that the map $a_*$ as defined above has to be the multiplication by two $- \cdot 2: \mathbb{Z} \to \mathbb{Z}$

Answer 1

The basic idea is as follows, I think I will carry out a proof similar to your, so bear with me.

As you did, consider the projective plane $X$ and take a point $x_0$ in it. Then $U = X\smallsetminus x_0$ deformation retracts into the sphere.

Take a small ball $V$ around $x_0$, so that $V\cap U$ also deformation retracts into a sphere.

Now for $V\cap U$, you will not have identified any boundary points, but in $U$, at the boundary sphere, you will identify them. This has the following consequence that you can form a commutative diagram

$$\require{AMScd} \begin{CD} U\cap V @>{\pi}>> S^1\\ @VVV @VVV \\ U @>{\pi}>>S^1 \end{CD}$$

where the vertical map is of degree $2$. Essentially, it will send a generating loop in $U\cap V$ that winds once around the boundary to one that will wind twice around the boundary of $U$, since in there you will have identified antipodal points.

Add. If you want to be more precise, note that the generating loop for $U$ can be taken to be a loop in the unit disk that draws a half moon, going from $-1$ to $1$ in an almost straight line missing the origin and then through the arc.This makes it easy to see that the generating loop for $U\cap V$ will represent twice the previous loop in $U$.

Answer 2

Let $i:S^1\to D^2$ be the inclusion of the boundary and $p:D^2\to P^2(\mathbb R)$ the canonical projection.

In particular, $p\circ i: S^1\to P^2(\mathbb R)$ factors through $\partial$ (and thus through $A$, but the inclusion $\partial \to A$ is a homotopy equivalence); let's call $\alpha :S^1\to\partial$ the map we get.

We know that $\partial \cong S^1$, so what map is $\alpha_* : \mathbb Z\to \mathbb Z$ ?

Well you have the following commutative diagram :

$\require{AMScd}\begin{CD}S^1@>>> S^2 \\ @VVV @VVV \\ \partial @>>> P^2(\mathbb R)\end{CD}$

where the map $\partial \to P^2(\mathbb R)$ is the inclusion. If we identify $\partial \cong S^1$, the map $S^1\to S^1$ is simply $z\mapsto z^2$: that's an explicit computation you can make. Perhaps it's easier to actually define $\partial$ that way, and check that you get the same thing.

I think that might be the main point which wasn't clear to you, so if it still isn't, don't hesitate to tell me.

In particular, $\alpha_*=$ multiplication by $2$.

But also, $i$ is homotopic to the inclusion of $S^1$ at a smaller circle in $D^2$, and therefore $p\circ i$ is homotopic to a homeomorphism $S^1\to S^1_{x_0}$.

So you have the following homotopy commutative diagram:

$\begin{CD} S^1 @>>> S^1_{x_0} @>\simeq>> A\cap B \\ @V=VV & & @VVV \\ S^1 @>>> \partial @>\simeq>> A \end{CD}$

Taking $\pi_1$, since $\pi_1(S^1)\to \pi_1(S^1_{x_0})$ is an isomorphism and $\pi_1(S^1)\to \pi_1(\partial)$ is multiplication by $2$, we finally get that $\pi_1(A\cap B)\to \pi_1(A)$ is multiplication by $2$.

(technically you might have to worry about basepoints: there are at least two ways of dealing with this here: 1- note that all the fundamental groups involved are abelian, and so it doesn't change anything; or 2- do the same reasoning but with fundamental groupoids, and in the end patch things up)

Answer 3

The morphism $a_*$ takes a loop $[\ell]_{A\cap B}=\gamma^n \in \pi_1(A\cap B,x_0)$ and sends it to the corresponding loop in $\pi_1(A,x_0),$ which, since the map is induced by the inclusion, is just $[\ell]_A$, (i.e. $\ell$ modulo homotopy in $A$).

Now we see $[\ell]_A$ inside $\pi_1(S^1,r_A(x_0))$ through the isomorphism $[\ell]_A \mapsto [r_A \circ \ell]_{S^1}$, and we have $[r_A \circ \ell]_{S^1}=c^{2n},$ because on the boundary antipodal points get identified, and so we go twice around the external circle as $[r_A \circ \ell (1/2)]=[-r(x_0)]=[r(x_0)]$ (here the equivalence classes are of points in $S^1$).

Pulling back to $\pi_1(A,x_0)$ we get $[\ell]_A=[hc^{2n}h^{-1}=(hch^{-1})^{2n}]_A$ and we conclude.