# Proof that a map from an orientable surface to a non-orientable surface has even degree.

For a smooth map $f:M\to N$ from an orientable closed surface $M$ to a non-orientable closed surface $N$, we define its parity (also called modulo 2 degree, and denoted $\deg_2(f)$) as the parity of the number of preimages of any regular value of $f$.

By a geometric construction, I am able to convince myself that in fact $f$ is even, but I suspect that there exists an argument of algebraic topology to show this more easily.

Avoiding the construction, I can say the following:

• If there exists an odd map $f:M\to N$, then there exists an odd map $gf:M\to\mathbb P^2(\mathbb R)$, since for any non-orientable $N$ we can construct an odd map $g:N\to\mathbb P^2(\mathbb R)$ by collapsing to a point the complement of a tubular neighbourhood of an orientation reversing loop. So we can assume that $N=\mathbb P^2$, which has odd Euler characteristic. This is interesting because $M$ has even characteristic.

• If $f$ is a local homeomorphism, then it's a covering map, so it factors via the orientation covering of $N$, which is even. So in this case it's easy to show that $f$ is even.

Of course, it would also be good to know what happens in greater dimensions. I think that a similar construction shows again that the map must be even.

And although it really isn't necessary for the construction, I would like to know what happens when you collapse to a point the border of a compact manifold-with-border. When can I say that I get a topological manifold? In this case I collapsed a tubular neighbourhood of an orientation-reversing curve. In even dimensions I obtain the projective space, but in odd dimensions I have no idea. I can't obtain the projective space because I can't obtain an orientable manifold.

• I don't have time to write a full answer, but look at the induced map $f^\ast :H^\ast(\mathbb{R}P^2,\mathbb{Z}_2) = \mathbb{Z}_2[x]/x^3 \rightarrow H^\ast(M_g) = \mathbb{Z}_2[y_1,...y_g, z_1,..., z_g]/I$ for appropriate ideal $I$. Note that $I$ contains the squares of the $y_i$ and $z_j$. Then $f^\ast(x) = \sum a_i y_i + \sum b_j z_j$, so $f^\ast(x^2) = f^\ast(x)^2 = \sum a_i^2 y_i^2 + \sum b_j^2 z_j^2 + 2\sum a_i b_j y_i z_j = 0 (mod\text{ }I)$ since everything is mod $2$. – Jason DeVito Sep 26 '14 at 12:56

## 1 Answer

I think that you can prove it in the following way. Every non-orientable manifold $Y$ has an orientable double cover $\tilde{Y}$. It should be possible to prove that if you have a map from an orientable manifold $X$ to $Y$, then this lifts to the orientable double cover, and so the degree of the map $f : X \to Y$ must be even.

Recall that a map $f : X \to Y$ lifts to $\tilde{Y}$ if and only if we have that $$f_*\pi_1X \subset p_*\pi_1\tilde{Y}$$ where $p : \tilde{Y} \to Y$ is the covering map. This subgroup consists of exactly those loops in $Y$ which do not reverse orientation.

So it would suffice to show that if you have a path in an orientable space $X$, then its image in $Y$ cannot reverse orientation. However, this would be true as long as you can represent the path $\gamma$ by one which is a local isomorphism (i.e. the map restricted to a neighbourhood of $\gamma$ is a covering map), which should be true.

• It is not that easy: I can show examples of maps from orientable manifolds to non-orientable manifolds that don't lift. But of course, they are even (in fact, non-surjective). Example: Project the torus to $S^1$, and send $S^1$ to an orientation-reversing loop. – Marcos Cossarini Sep 26 '14 at 12:41
• Also, one cannot in general modify a loop $\gamma$ in the domain of $f$ so that it avoids the singularities of $f$. For example, we can perturb the identity $1_{S^1}$ to obtain a singular map $g$. Then, the map $f=1_{S^1}\times g$ has a wrinkle that can't be avoided by curves homotopic to $\{p\}\times 1_{S^1}$. – Marcos Cossarini Sep 26 '14 at 12:50
• I think that I was a little too focused on the case of algebraic varieties over $\mathbb{C}$, where the singularities are always in high enough codimension that it works. So I suppose this isn't really what you're looking for... – Simon Rose Sep 26 '14 at 13:01