# What surface do I get by attaching $g$ handles as well as $k$ crosscaps to a sphere?

I recently found out that there is a classification of compact connected surfaces that says that every such surface (or, $2$-manifold) is homeomorphic to either $S_g$, the sphere with $g \geq 0$ handles, or $N_k$, the sphere with $k \geq 1$ crosscaps.

If my understanding is correct:

- attaching a handle to a sphere means deleting two discs from the sphere and attaching the ends of a cylinder to the boundaries of these holes. I am able to visualize how attaching one handle to a sphere gives a torus, attaching two handles gives a double torus, etc.
- attaching a crosscap to a sphere means deleting a disc and pasting a Möbius strip to the boundary of this hole.

My question is, what happens if I attach a handle *and* a crosscap to a sphere? I presume the result is still a compact connected surface (I don't see how it cannot be if every $S_g$ and $N_k$ is). So, by the classification theorem, this surface must also be homeomorphic to some $S_g$ or $N_k$. How can I find out to what surface it will be homeomorphic to? More generally, if I attach $g$ handles *and* $k$ crosscaps to a sphere, what is the resulting surface homeomorphic to as per the classification theorem?

## 1 answer

My question is, what happens if I attach a handle

anda crosscap to a sphere?

By Dyck's theorem, the connected sum of a torus (sphere with one handle) and a projective plane (sphere with a cross-cap) is the same as the connected sum of three projective planes. So you get the sphere with three cross-caps.

More generally, if I attach

ghandlesand kcrosscaps to a sphere, what is the resulting surface homeomorphic to as per the classification theorem?

You use the same principle, but with more complicated arithmetic. `:-)`

#### 3 comments

I think in the presence of a crosscap, each handle should be convertible to two crosscaps, therefore $g$ handles and $k$ crosscaps with $k\ge 1$ should be equivalent to $2g+k$ crosscaps. But I'm not entirely sure.

Yes, @celtschk .

+1 This is exactly what I needed, thank you for the clear explanation and the reference! :)

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