Is there a surface that is orientable which is topologically homeomorphic to a nonorientable one, or is orientability a topological invariant.

7Orientability can be defined in terms of relative homology groups, see http://www.map.mpimbonn.mpg.de/Orientation_of_manifolds . In particular, it is invariant under homeomorphisms. – archipelago Aug 10 '15 at 20:25
1 Answers
The answer is no: orientability is a topological invariant.
However, your question made me realise that I'd never thought about orientability for topological manifolds! For differentiable manifolds, it's clear that orientability is diffeomorphisminvariant, because we can define it in terms of the determinant of the differential of a transition function.
For general topological manifolds, we have to first define what we mean by an orientation. Suppose we have a continuous map $\phi : U \to V$, where $U$ and $V$ are pathconnected open subsets of Euclidean space $\mathbb{R}^n$. Choose some point $x \in U$. Then $x$ has some smaller neighbourhood $U' \subset U$ such that the boundary of $U'$ is homeomorphic to the $(n1)$sphere: $~\partial U' \cong S^{n1}$.
$\phi$ is a homeomorphism, so we also have $\phi(\partial U') \cong S^{n1}$. Because $\phi$ is invertible, it induces an invertible map on homology groups $\tilde\phi: H_{n1}(\partial U') \to H_{n1}\big(\phi(\partial U')\big)$. We can canonically identify both the domain and codomain of this map with $H_{n1}(S^{n1}) \cong\mathbb{Z}$. So, since $\tilde\phi$ is invertible, it must act as $\pm 1$. If it is $+1$, we say that $\phi$ is orientationpreserving.
The last thing to check is that the above definition is independent of the point $x$ that we choose. Given some other point $x' \in U$, choose some path $\gamma : [0,1] \to U$ such that $\gamma(0) = x$, $\gamma(1) = x'$. Then, using the fact that $U$ is open and $\text{im}(\gamma)$ is compact, we can argue that if we started with $x'$, we would end up with a map $S^{n1} \to S^{n1}$ homotopic to the one we got starting with $x$. Therefore they induce the same action on homology.
A manifold is orientable if we can find an atlas such that all the transition maps are orientationpreserving. Given the definition, this is clearly invariant under homeomorphism.
 4,253
 16
 26

What about spaces that are different as manifolds but different as topological spaces? – PyRulez Aug 10 '15 at 23:41

I'm not quite sure what you mean, but you can only define orientability for a manifold. For a general topological space, it has no meaning. – Rhys Aug 10 '15 at 23:43

Can a topology be given both a orientable manifold structure and a nonorientable one. There are atleast 7, 7spheres, I believe. – PyRulez Aug 10 '15 at 23:52

There is only one *topological* manifold (up to homeomorphism of course) that you can call a '7sphere'. I think you're talking about differentiable structures, and I'm not sure you actually know which question you want to ask. :) – Rhys Aug 11 '15 at 00:00


The definition of orientable that one usually sees is applicable to differentiable manifolds. It's manifestly diffeomorphisminvariant. I thought you were asking whether there is a homeomorphisminvariant analogue. There is, as demonstrated in my answer. Where has this process gone wrong...? – Rhys Aug 11 '15 at 00:16

I was thinking a topological space could be a manifold in a different ways. I'm a topology noob. – PyRulez Aug 11 '15 at 00:41