Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I think I have solved it, but not sure however. The "solution" is in the most recent update.

Currently I'm reading through Hatcher's Algebraic Topology, and I'm stuck on one thing in the text. At page 236, he states that if $M$ is a closed connected $n$-manifold, then if $M$ is not $R$-orientable, the map $H_n(M;R) \rightarrow H_n(M|x;R) $ is injective, with image $\{r \in R | 2r = 0 \}$. This should follow from Lemma 3.27 in the text, where he says that :

Let M be a manifold of dimension n and let $A \subset M$ be a compact subset. Then if $x \rightarrow \alpha_x $ is a section of the covering space $M_R \rightarrow M$, then there is a unique class $\alpha_a \in H_n(M|A;R)$ whose image in $H_n(M|x;R)$ is $\alpha_x$ for all $x \in A$.

He shows that this gives an isomorphism between the set of sections $\gamma_R(M)$ and $H_n(M;R)$. Okay, I agree with that. What I don't follow is why the theorem follows from this, and that a section is uniquely determined by its value at one point.

Any help with explaining why this is, would be appreciated.

Update: Still no real progress. I have, through reading Hatcher, noticed that we can identify $H_n(M|x;R)$ with $H_n(M|X) \otimes R$ (when leaving out $R$, I mean $Z$-coefficients). So, maybe a section decides a subcovering of some sort. Still, that's about it. I've thought about seeing is this covering has some automorphisms or something, but it doesn't quite seem to work.

Yet another update!(Which I think solves it) Okay, I think I might have got it. Let M be a non-orientable manifold. Look at the sections $f:M \rightarrow M_R$, $-f$. A section is determined by its value at an unique point, so the image of these two sections will be $M_r = \{\pm 1 \otimes r\}$, a covering space of M. If $r$ fails to have order $2$, this will be isomorphic to the double oriented covering of M, let's call it $\hat{M}$, which is oriented. So, since $f$ is clearly bijective onto its image, continuous and with a continuous inverse, we have shown that M must be homeomorphic to a submanifold of the oriented manifold $\hat{M}$, but M was assumed not to be orientable, so $r=-r$, that is, it has order 2.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.