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I would like to understand how to prove that the connected sum $\mathbb{R}P^2 \# T^2$ of the projective plane with a torus is homeomoprhic to $\mathbb{R}P^2 \# \mathbb{R}P^2 \# \mathbb{R}P^2$.

I got as far as showing that it must be equivalent to a connected sum of projective planes, how can I argue though that I need precisely three projective planes ?

Thanks for your help!

(P.S. not a homework exercise, this is for me to understand the classification of surfaces).

harlekin
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  • You can do this by hand, by making successive modifications to a gluing diagram for one surface until you have a diagram for the other. http://www.ornl.gov/sci/ortep/topology/topo5.gif – Sammy Black Apr 11 '13 at 20:16
  • Another answer: http://math.stackexchange.com/questions/2124520/prove-that-mathbbrp2-k-and-mathbbrp2-mathbbt2-are-ho/2126276#2126276 – rgm Feb 02 '17 at 19:50

2 Answers2

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The following picture will help you to understand this problem intuitively. (If you can understand why $\mathbb{R}\mathrm{P}^2 \# \mathbb{R}\mathrm{P}^2 = K$, where $K$ stands for the Klein bottle.)

The figure in the upper right corner is $K \setminus \text{disk} $. It may take some time to think why it looks like this.

T^2 # \mathbb{R}\mathrm{P}^2 = \mathbb{R}\mathrm{P}^2 # \mathbb{R}\mathrm{P}^2 # \mathbb{R}\mathrm{P}^2

Source of the picture: link

Fang Hung-chien
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Calculate the Euler characteristic. This, combined with the fact that the resulting surface is non-orientable gives you the complete set of invariants, enough to single out the $\mathbb{R}P^2 \# \mathbb{R}P^2 \# \mathbb{R}P^2$.

xyzzyz
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    I believe that you're "putting the cart before the horse" here. The OP's intent is to understand the classification of surfaces, one important step of which is to establish the stated homeomorphism. – Sammy Black Apr 11 '13 at 20:19
  • Is the statement that a handle is equivalent to two cross-caps true in the case of surfaces with boundary? – QGravity Jun 03 '19 at 02:56