Finding map from Klein bottle to $RP^2$ that induces an epimorphism of fundamental groups.

Question

I want to find a map from the Klein bottle to $RP^2$ that induces an epimorphism of fundamental groups. Since the fundamental group of $RP^2$ is $Z_2$, intuitively it feels like a good start would be to look for any non-trivial map from $K$ to $RP^2$, but frankly even that is difficult for me. After reading quite a bit about the Klein bottle i still have problems even visualizing it, so any help would be appreciated.

Answer 1

Consider the following transformation : take a small disk in $RP^2$, remove it and glue a Moebius band $M$. You get a space $Y$, I claim it is homeomorphic to the Klein bottle.

If $E \subset M$ generates $\pi_1(M)$ (it is the straight line in the picture), you have a projection $M \to D$, sending the boundary of $M$ to the boundary of $D$, which send the complement of $M \ E$ to $D \backslash 0$ with $p(E) = 0$.

You can extends such projection to $K \to RP^2$ and this gives the morphism you were looking for.

Answer 2

Let $X$ and $Y$ be two connected manifolds of dimension $n \geq 2$.

There is a map $\varphi : X\# Y \to X$ given by mapping $Y$ to a disc $D$. By the Seifert van Kampen theorem, $\pi_1(X\# Y) \cong \pi_1(X^{\circ})*_{\pi_1(S^{n-1})}\pi_1(Y^{\circ})$ where $X^{\circ}$ denotes $X$ with an embedded open disc removed; likewise for $Y^{\circ}$. The map $\varphi$ induces a map

$$\varphi_* : \pi_1(X^{\circ})*_{\pi_1(S^{n-1})}\pi_1(Y^{\circ}) \to \pi_1(X^{\circ})*_{\pi_1(S^{n-1})}\pi_1(D) \cong \pi_1(X)$$

which is the identity on the subgroup $\pi_1(X^{\circ})$ and necessarily trivial on $\pi_1(Y^{\circ})$ (because $\pi_1(D)$ is trivial). In particular, $\varphi_*$ is an epimorphism.

As $K = \mathbb{RP}^2\#\mathbb{RP}^2$, we obtain a map $\varphi : K \to \mathbb{RP}^2$ inducing an epimorphism $\varphi_* : \pi_1(K) \to \pi_1(\mathbb{RP}^2)$.

Answer 3

N.H.'s answer is excellent. Here is an alternative way to visualise the map from the Klein bottle to the real projective plane suggested by N.H:

To go from the Klein bottle (left -hand picture) to the real projective plane (right-hand picture), I'm quotienting out the circle marked with the double-headed arrow.

To make contact with N.H.'s explanation: If you take a small open strip around the circle with the double headed arrow, you have a Mobius band. After quotienting out the circle, the Mobius band becomes a small open disk covering the black blob on left and right of the diagram of the real projective plane.

Since you say that it's hard to visualise these spaces, I should probably also explain how to make sense of these pictures.

• The Klein bottle: Start with a square. If you glue the two vertical sides (marked with the double-headed arrows), you get cylinder. If you then glue the two horizontal sides (marked with the single-headed arrows) in the obvious way, you get a torus. To get a Klein bottle instead of a torus, you need to do the gluing in such a way that these horizontal sides come together with "opposite orientation relative to each other" - hence why the single-headed arrows point in opposite directions in my picture of the Klein bottle.

• The real projective plane is the space of all rays through the origin in 3-dimensional space. This is the same as the sphere with opposite points identified. (Imagine a unit sphere sitting in $\mathbb R^3$, and identify each ray with the two points where it intersects the unit sphere...) This in turn is the same as the northern hemisphere, with antipodal points identified on the equator. And this is what I have drawn.