Connected sum of projective plane $\cong$ Klein bottle How can I see that the connected sum $\mathbb{P}^2 \# \mathbb{P}^2$ of the projective plane is homeomorphic to the Klein bottle?
I'm not necessarily looking for an explicit homeomorphism, just an intuitive argument of why this is the case. Can we see it using fundamental polygons?
 A: Here's an answer more in the spirit of the question. All figures should be read from upper left to upper right to lower left to lower right. 
Fig 1: A Klein bottle...gets a yellow circle drawn on it; this splits it into two regions, which we reassemble, at which point they're obviously both Mobius bands:

To see that $P^2$ minus a disk is a Mobius band, look at the following. In the upper left is $P^2$, drawn as a fundamental polygon with sides identified. In the upper right, I've  removed a disk. The boundary of the now-missing disk is drawn at the lower left as a dashed line, and the two remaining parts of the edge of the fundamental polygon are color-coded to make the matching easier to see. In the lower right, I've morphed things a bit, and if you consider the green-followed-by-red as a single edge, you can see that when you glue the left side to the right, you get a M-band. 

A: For some really nice pictures that should make this quite clear to you, look at some of Carlo Sequin's recent papers like this one: 
http://www.cs.berkeley.edu/~sequin/PAPERS/2012_Bridges_Klein.pdf
Let me try a different tack:
We'll look at IMMERSED versions of each of these (embedded Klein bottles in 3-space being kinda scarce). 
The gist of the argument is the following (and you may not need the pictures once you get it):


*

*$P^2 - D^2 = M$: a projective plane minus a disk is a Mobius band. This is easy to see by taking the usual fundamental polygon and cutting out a disk that "hangs over" the edge, so that it splits each of the two gluing edges into two separate pieces. 

*$P^2 \# P^2$ is therefore a union of two Mobius bands along their common boundary. 

*$K$, the Klein bottle, is a union of two Mobius bands along their common boundary. 
I'll show you part 3 in pictures. We start with a schematic picture (i.e., I'm a lousy artist) of the standard picture of a K-bottle. The red ovals are meant to indicate cross-sections.  We also have a piece of very elastic rubber, shown as a  rectangle in the lower right. It's blue on one side, and red on the other: 

Next, we bandsaw this bottle in half, along the plane of symmetry (i.e., the plane perpendicular to our view direction). I've drawn, in dark blue, the resulting boundary. You can check for yourself that the boundary is an immersed circle in the plane. 

Now I'm going to build up the rest of the "far" half of the K-bottle, in steps. I start by gluing a rectangle of rubber as shown, and press the red mid-line "back" away from us, so that the thing depicted as a blue rectangle is really more like half of a paper-towel-tube. Clear? 

Now I extend this to glue up with more of the boundary:

And even more:

But when I reach the left end, to continue I have to fold the rubber sheet forward, so that you see the back side (in red):

I continue extending, and you can see that to finish the job, I need to glue the sheet to itself in a way that matches the blue side with the red...and that makes a Mobius band. 

Obviously, the front half of the K-bottle is another M-band, and they join along this blue middle "slice-curve" (which is a circle). Hence $K = M_1 \cup M_2 / (\partial M_1 \text{~} \partial M_2)$. QED.
A: [Munkres]=Munkres J. R.: Elements of Algebraic Topology, Reading, MA: Addison-Wesley, 1984.
The proof is just a simple trick. Munkres, p.39, Figure 6.9 only gives one part of the proof. The orientation of each $2$-simplex in the figure is the counterclockwise direction; the left figure represents the Klein bottle. The other part of the proof is to establish a homeomorphism between the right figure in Munkres, p.39, Figure 6.9 and the first figure in Munkres, p.38, Figure 6.8 using the last two figures in Munkres, p.38, Figure 6.8. The last two figures in Munkres, p.38, Figure 6.8 represent $P^2\#P^2$. Let us consider its left branch first. It is a $2$-simplex denoted by the path $D\to D\to -C$ using its directed boundary segments rather than the order of its vertices. The minus sign represents the clockwise direction. The starting point and the end point of the path $D\to D$ are the same. $C$ is a closed curve. The same $2$-simplex can also be represented by $D\to -C\to D$. The corresponding $2$-simplex in the right figure of Munkres, p.39, Figure 6.9 is $C\to -B\to C$. Although the starting point and the end point of the path seem to look different, we can relabel vertices to make the path closed so that it may represent a $2$-simplex.
