# What does it mean for a (non-smooth) homeomorphism between oriented smooth manifolds to be orientation preserving?

The definitions I know of orientability of manifolds are in terms of tangent spaces. However, for example in this answer there is mention of orientation preserving homeomophisms (between orientable complex manifolds).

My first thought would be something like instead of considering a basis of tangent vectors (or their determinant), to integrate these to consider a set of short curves emanating from a point on the manifold (e.g. by mapping curves in a coordinate chart), and considering the ordering of their image under the homeomorphism. It is not clear to me that this would work however, since a curve through a point could get folded in such a way that one end of it would give you one ordering, but the other a different one, or so it seems. Probably there could be other kinds of pathological behaviour as well.

In any case my questions are

1. How can we define if a homeomorphism between oriented smooth manifolds is orientation-preserving?
2. More generally, can we speak of orientability of topological manifolds without relying on any differentiable structure? Or even for more general topological spaces?

Thanks!

• For 1. I think we can make a smooth function arbitrarily close to a given continuous function.. For 2. I doubt we can define orientability in those generalities.. – Berci Sep 19 '18 at 8:35

Orientation can be defined on manifolds using singular homology theory. The easiest case is where $M$ is a compact $n$-manifold. Then $H_n(M;\Bbb Z)$ is isomorphic to $\Bbb Z$ when $M$ is orientable, and is zero when $M$ is non-orientable. An orientation of $M$ is a choice of a generator of $H_n(M;\Bbb Z)$. A homeomorphism $f:M\to M'$ is orientation-preserving if $f^*$ maps your favoured generator of $H_n(M;\Bbb Z)$ to your favoured generator of $H_n(M';\Bbb Z)$.

More generally one can use local homology groups; these are relative homology groups $H_n(M,M-\{x\};\Bbb Z)$ where $x\in M$. These determine whether or not a map between manifolds is orientation-preserving at a given point.

For details, see texts in algebraic topology, for instance Hatcher's.

• Indeed $H_n(M,M-\{x\};\Bbb Z)\cong \mathbb Z$ and there are two possible isomorphisms to $\mathbb Z$. The choice of such an isomorphism is the same as an orientation at the point $x$. If one can make this choice consistently, the manifold is orientable. – Cheerful Parsnip Sep 19 '18 at 17:52

Lord Shark the Unknown has completely answered the question, but let me draw your attention to the concept of a "microbundle" introduced by Milnor. See for example https://en.wikipedia.org/wiki/Microbundle and

Switzer, Robert M. Algebraic topology - homotopy and homology. Springer, 2017.

As you certainly know, an orientation of a smooth manifold can be defined as an orientation of its tangent bundle. Topological manifolds $M$ do not have a tangent bundle, but you can consider the triple $E = (M,\Delta,p)$, where $\Delta : M \to M \times M$ is the diagonal map and $p : M \times M \to M$ is the projection onto the first coordinate. Around $\Delta(M)$ the space $M \times M$ looks like a vector bundle. In fact, let $p \in M$ and $U$ be an open neighborhood homeomorphic to $\mathbb{R}^n$. Then $U \times U$ is an open neighborhood of $\Delta(p)$ which is homeomorphic to $U \times \mathbb{R}^n$, and the restriction of $p$ to $U \times U$ corresponds to the projection $U \times \mathbb{R}^n \to U$. This means that $(M,\Delta,p)$ is a microbundle.

Orientations of microbundles can be defined homologically. See again Lord Shark the Unknown's answer. Intuitively, each microbundle with fiber $\mathbb{R}^n$ admits a sphere bundle with fiber $S^{n-1}$ around the zero section of the base space. An orientation of such a sphere bundle is a "compatible family" of orientations of the fibers $S_p \approx S^{n-1}$, and this is given by a family of generators of $H_{n-1}(S_p)$. For a topological manifold you easily see that $H_{n-1}(S_p) \approx H_n(M,M \setminus p)$.

• I had never heard of this, thanks! – doetoe Sep 19 '18 at 16:31