I'm trying to prove this using purely topological arguments, no differential geometry as I haven't been exposed to it.

I've been playing around with definitions a bit and here's what I have so far.

Let $M$ be an orientable manifold. Let $N$ be its covering space. Then we have an orientation function $\mu : M \rightarrow \{\pm 1\}$ that satisfies:

$\forall x \in M, \exists U \cong D^n$ an open neighborhood of $x$ and $\mu_u$ a generator of $H_n(M, M-U) \cong \mathbb{Z}$ such that $\forall y \in U$ $H_n(M, M - {y}) \longleftarrow H_n(M, M - U) \longrightarrow H_n(M, M-{x})$ where each map is an isomorphism.

Now since $N$ is the covering space of $M$ we know that there is a function $p: N \rightarrow M$ surjective and continuous s.t. $\forall x \in U$ open, $p^-1(U)$ is a disjoint union of open sets $v_i \in N$ s.t. $p(v_i)$ is homeomorphic to $U$.

My conjecture is that $\mu \circ p$ is an orientation function on $N$ satisfying the compatibility conditions i.e. $\forall x \in N, \exists U \cong D^n$ an open neighborhood of $x$ and $\mu_u$ a generator of $H_n(N, N-U) \cong \mathbb{Z}$ such that $\forall y \in U$ $H_n(N, N - {y}) \longleftarrow H_n(N, N - U) \longrightarrow H_n(N, N-{x})$ where each map is an isomorphism.

I need some help moving forward from here. Is it perhaps true because for $x \in N$ we can choose a neighborhood $V$ s.t. $V$ is one of the open sets which maps homeomorphically onto some neighborhood $U$ of $x' \in M$ and we know that the compatibility conditions are true for any $y' \in U$ so it must be true for any $y \in V$?

  • $\begingroup$ Do you know about orientation coverings or about orientation reversing curves? $\endgroup$ Dec 8, 2014 at 8:27

2 Answers 2


You can use orientation coverings or orientation reversing curves (together with the observation, that every curve in the covering comes from a curve in your manifold) to prove your statement in one line. If you want to prove it by definition, as you are conjecturing correctly, $\mu$ gives rise to $\hat \mu$ on $\hat M$ by post-composing (here $\hat M \to M$ is the covering. (note that in nyour definition you forgot to mention the important fact that you need preferred generators on local homology i.e. you need that the maps, induced by inclusions map $\mu_U \mapsto \mu_x$).

To finish your proof consider the diagram: $$ \begin{array}{c} H_n(\hat M, \hat M-U) &\to& H_n(\hat M, \hat M - \{x\})\\ \downarrow && \downarrow\\ H_n(M,M-\pi(U)) &\to & H_n(M,M-\{\pi(x)\}), \end{array} $$ where $U$ is chosen such that $\pi|U$ is a homeomorphism and the lower horizontal map is as desired and sends preferred generator to preferred generators. Since such diagrams commute, you certainly obtain your $\hat \mu_x =\mu_{\pi(x)}$.

  • 1
    $\begingroup$ Can you elaborate on how to use the orientation covering to prove in one line? $\endgroup$
    – nekodesu
    Apr 12, 2018 at 17:32
  • 1
    $\begingroup$ How is the left vertical map defined? I think $\pi$ does not map $\hat{M}-U$ to $M-\pi(U)$. $\endgroup$
    – blancket
    Apr 25, 2021 at 9:58

Using local homology seems like overkill to me. How about just using atlases, something like this.

Let $d$ be the dimension of $M$, and let $B^d \subset \mathbb{R}^d$ be the open unit ball. Choose an atlas of covering maps $\phi_i : U_i \to B^d$ for $M$ such that every overlap map $$\phi_{ji} : \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j) $$ preserves orientation. We may assume that each $U_i$ is an evenly covered open set with respect to the covering map $q : N \to M$.

Now define an atlas for $N$, with charts of the form $\psi_{i,k} : V_{i,k} \to B^d$ so that the $V_{i,k}$ is one of the components of the set $q^{-1}(U_i)$, and $\psi_{i,k}$ is the composition $$V_{i,k} \xrightarrow{q} U_i \xrightarrow{\phi_i} B^d $$ From this construction it follows almost immediately that the overlap maps of the atlas of $N$ preserve orientation.

  • $\begingroup$ Local homology is being used here in the definition of orientability, I believe. $\endgroup$ Dec 8, 2014 at 0:57
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    $\begingroup$ @MattS: I thought that might be the case, but the OP did not specify. Even if it were, I might suggest as a preliminary step proving that the two notions of orientation are equivalent, and then using this atlas argument. $\endgroup$
    – Lee Mosher
    Dec 8, 2014 at 15:16
  • $\begingroup$ why two notions of orientation are equivalent? $\endgroup$
    – Saikat
    Oct 5, 2015 at 20:46

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