For a non-orientable surface, we can replace a handle by two cross-caps. Can we do the opposite i.e replace any two cross-caps by a handle?

Any help is appreciated!!


2 Answers 2


This is Dyck's Theorem.

You need either to have a third crosscap available, or be careful to use a crossed handle (i.e. a Klein bottle). Otherwise you'd be able to convert a sphere with two crosscaps (which is non-orientable) into a torus (which is orientable).

  • $\begingroup$ Yes, we need an odd number of cross-caps. Thank you for the answer. $\endgroup$
    – yaa09d
    Nov 8, 2011 at 6:44
  • 3
    $\begingroup$ Even will do too, as long as there are at least 3 to begin with. $\endgroup$ Nov 8, 2011 at 14:23

The short answer is, yes, you can do that. Basically you can show that $$\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2\cong \mathbb{T}\#\mathbb{RP}^2.$$ This is a homeomorphism, hence reversible. This relates to your question since $\mathbb{RP}^2$ is a cross-cap, $\mathbb{T}$ is a torus (handle in your terminology) and every nonorientable compact surface has $\mathbb{RP}^2$ as a connect-summand.

In fact, usually one uses this opposite direction to prove the classification of surfaces. Using various techniques you show a surface must be a connect sum of some number of copies of $\mathbb{T}$ and $\mathbb{RP}^2$, and then uses the fact that the presence of a single $\mathbb{RP}^2$ allows you to convert all $\mathbb{T}$'s into pairs of cross-caps, showing every compact nonorientable surface is a connect-sum of some number of cross-caps.

  • $\begingroup$ Thank you! Actually, I was trying to see if the opposite holds i.e I have a non-orientable surface with odd number of cross-caps and I want to convert all but one of them into handles. But since we have the homeomorphism you gave above, it is obevious this works. Thanks again. $\endgroup$
    – yaa09d
    Nov 8, 2011 at 6:50

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