# Is there a orientable surface that is topologically isomorphic to a nonorientable one?

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

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 diffeomorphism-invariant, 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 path-connected 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 $(n-1)$-sphere: $~\partial U' \cong S^{n-1}$.

$\phi$ is a homeomorphism, so we also have $\phi(\partial U') \cong S^{n-1}$. Because $\phi$ is invertible, it induces an invertible map on homology groups $\tilde\phi: H_{n-1}(\partial U') \to H_{n-1}\big(\phi(\partial U')\big)$. We can canonically identify both the domain and codomain of this map with $H_{n-1}(S^{n-1}) \cong\mathbb{Z}$. So, since $\phi$ is invertible, it must act as $\pm 1$. If it is $+1$, we say that $\phi$ is orientation-preserving.

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 $S^{n-1} \to S^{n-1}$ 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 orientable. Given the definition, this is clearly invariant under homeomorphism.

• 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, 7-spheres, 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 '7-sphere'. 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 0:00
• Oh, nvm.${}{}{}$ – PyRulez Aug 11 '15 at 0:05