Geometrically reasoning through that $P^{2}$ # $P^{2}$ is the Klein Bottle

Question

Here’s a very vague attempt of mine to geometrically understand why is $P^{2}$ # $P^{2}$ homemorphic to the Klein Bottle. Since I’m currently on my cell phone, let me try and use as many words to explain how I reasoned. The reasoning is vague because my intuition relies on my intuitive idea or orientation while doing geometric transformations.

The first picture shows two copies of $P^{2}$ identified as upper hemisphere with the boundary circle identified appropriately. Carve out two very small hemispherical disks from the top. Now, the proof which gives the polygonal presentation of connected sums argues that a triangle, labelled $a,b,c$ anti-clockwise maps to two coordinate balls in the maps which geometrically realize the manifolds. And it also makes a cutting and pasting argument to identify the images of the boundaries $a,b,c$ and $c^{-1} b^{-1} a^{-1}$, so I’m intuitively arguing that that the circle removed from the first copy is traversed anti-clockwise, and the one in the second copy is traversed clockwise. Now take the second copy, flip it over, and move it up. This flips the orientation of both circles to be anti-clockwise. The boundaries of the circles removed are both homeomorphic to $S^{1}$, so there’s a trivial bijection that allows us to connect both the manifolds. Now, morph the hemispheres so they look like cylinders, giving me the first picture on the second board. Now stretch and twist the resultant copy to get the second picture.

My question is where do I get the Klein Bottle? If the two semicircles in the last are to be identified, then, I’m done. But that isn’t the case, right?

What’s going on? Admittedly this is all vague, but I’m trying to learn how to reason geometrically for basic problems in alg top.

Answer 1

I'm pretty much unable to follow what you are trying to do with that $abc$ triangle, For example, what it means to map that triangle "to two coordinate balls in the maps which geometrically realize the manifolds" is completely unclear. Also, the $w_1$ in your blackboard diagram is undefined.

However, your blackboard photo does have a clue to a clear error in your reasoning. You wrote "Why must $>>$ be identified with $>$?", but in fact $>>$ is NOT to be identified with $>$, and therefore that is NOT how to prove that $P^2 \# P^2$ is homeomorphic to the Klein bottle.

Instead, you need to do a couple more cuts, and then reglue in a different way. On the blackboard photo, the second to last drawing on the right shows a perfect cylinder with two $<<$ arrows on the top circle and two $>$ arrows on the bottom circle. Good so far.

1. Now cut that cylinder into two rectangles, each with one $<<$ arrow on the top side and one $>$ arrow on the bottom side. Carefully label the four vertical sides of those two rectangles with a $>>>$ pair and a $|>$ pair to indicate how the two rectangles should be reglued to form the cylinder.
2. But now, reglue those two rectangles in a different way: glue the $<<$ arrow on one rectangle with the $<<$ arrow on the other rectangle, joining the two into a single polygon.
3. Stare at that polygon for a second, and you will see the standard gluing for the Klein bottle.

This process is a specific example of the proof of the classification of surfaces, such as the proof carried out in Massey's book as suggested in the comment of @William.

And while all this cutting and pasting may seem still somewhat intuitive, in fact it's pretty easy to make it all rigorous by using the universality property of quotient maps.