TITLE: Is there a orientable surface that is topologically isomorphic to a nonorientable one?
QUESTION [10 upvotes]: Is there a surface that is orientable which is topologically homeomorphic to a nonorientable one, or is orientability a topological invariant.

REPLY [6 votes]: 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 $\tilde\phi$ 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 orientation-preserving.  Given the definition, this is clearly invariant under homeomorphism.