I want to show that the non-orientable surface of genus $n$ has a 2-sheeted cover by an orientable surface of genus $n-1$. The base cases are easy: $S^2$ covers $\mathbb{RP}^2$ and I worked on a proof that the torus $\mathbb{T}^2$ covers $\mathbb{RP}^2\#\mathbb{RP}^2\approx K$ (Klein bottle).

Here are some of my constraints:

  • Nothing about the relationship between the Euler characteristic and covering maps has been proved, so I can't use this. Nor do I know anything about homology yet. Additionally, I don't yet know how to produce quotient manifolds using the techniques of group actions. The placement of this question in the text I'm using suggests that I don't need those tools. I pretty much only have knowledge of the universal covering space of manifolds.
  • I proved previously that if $M$ and $N$ are connected $n$-manifolds and $M$ has a $k$-sheeted covering space $H$, then $H\#N\#\ldots \#N$ (with $k$ copies of $N$) is a covering space for $M\#N$.

Induction using this fact seems like a good approach. Suppose $R$ is a connected sum of $n$ projective planes, an $T$ is connected sum of $n-1$ tori such that $T$ is a double cover for $R$. I want to show that $T\#\mathbb{T}^2$ is a double cover for $R\#\mathbb{RP}^2$. We know that $T\#\mathbb{RP}^2\#\mathbb{RP}^2 \approx T\#K$ is a double cover for $R\#\mathbb{RP}^2$, but I'm not sure where the go from here. We can't really cover $T\#K$ by $T\#\mathbb{T}^2$ since there are different numbers of sheets whether you're on $K$ or you're on $T$. Also, even if we could then we would end up with a quadruple cover of $R\#\mathbb{RP}^2$.

Also if anyone wants to explain to me how $\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2$ is a double cover of $\mathbb{RP}^2\#\mathbb{RP}^2\#\mathbb{RP}^2$, I'd greatly appreciate it.

This type have been asked before like here and here, but my knowledge of the techniques used in the first link are scant. It's suggested that I don't need the techniques of orbit spaces or quotient manifolds to prove this claim, and I would like to develop a proof without those techniques.

  • $\begingroup$ I believe the first answer to your first link is exactly what you need... You don't have to use the Euler characteristic. You are given a description of how the $n-1$ torus is the double cover of a surface. The only thing left to show is that the latter surface is the sum of $n$ copies of the projective plane... You can do that with induction if you want... $\endgroup$ Commented Apr 9, 2015 at 23:47
  • $\begingroup$ possible duplicate of Covering space of a non-orientable surface $\endgroup$ Commented Apr 9, 2015 at 23:50
  • $\begingroup$ @Theo Please see my prompt again, I clarified what my issue with this problem. $\endgroup$
    – Mnifldz
    Commented Apr 10, 2015 at 0:55

1 Answer 1


Let $kM$ denote the connected sum of $k$ copies of $M$.

As $3\mathbb{RP}^2 = T^2\#\mathbb{RP}^2$, we see that $(2k - 1)\mathbb{RP}^2 = (k-1)T^2\#\mathbb{RP}^2$ and $2k\mathbb{RP}^2 = (k-1)T^2\# 2\mathbb{RP}^2$.

If $n$ is odd, say $n = 2k - 1$, then $n\mathbb{RP}^2 = (2k-1)\mathbb{RP}^2 = (k-1)T^2\#\mathbb{RP}^2$. As $S^2$ double covers $\mathbb{RP}^2$, we see that $2(k - 1)T^2\# S^2 = (2k - 2)T^2 = (n - 1)T^2$ double covers $n\mathbb{RP}^2$ by the second dot point.

If $n$ is even, say $n = 2k$, then $n\mathbb{RP}^2 = 2k\mathbb{RP}^2 = (k-1)T^2\# 2\mathbb{RP}^2$. As $T^2$ double covers $2\mathbb{RP}^2$, we see that $2(k - 1)T^2\# T^2 = (2k - 1)T^2 = (n - 1)T^2$ double covers $n\mathbb{RP}^2$ by the second dot point.

As for why $4\mathbb{RP}^2$ double covers $3\mathbb{RP}^2$, note that $3\mathbb{RP}^2 = 2\mathbb{RP}^2\#\mathbb{RP}^2$ and $S^2$ double covers $\mathbb{RP}^2$, so $4\mathbb{RP}^2\# S^2 = 4\mathbb{RP}^2$ double covers $2\mathbb{RP}^2\#\mathbb{RP}^2 = 3\mathbb{RP}^2$ by the second dot point.

  • 1
    $\begingroup$ Far more concise than I could've anticipated. Thanks for the response. A clean proof is worth the wait. $\endgroup$
    – Mnifldz
    Commented Aug 10, 2017 at 21:48

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