I have a question on Hatcher exercise 3.3.2 which says that
Show that deleting a point from a manifold of dimension greater than $1$ does not affect orientability of the manifold.
There is a post that shows the converse of the title. I want to prove the title. To do this, I consider some small ball $B\subset M$ that contains $p$. I need to choose a generator $\mu_B\in H_n(M,M-B)$ such that $(i^x)_*(\mu_B) = \mu_x$ for each $x\in B$, where $\mu_x$ is a local orientation and $i^x:(M,M-B)\to(M,M-x)$ is an inclusion map, then we're done. We already know that each $x\in B\setminus\{p\}$ has a local orientation $\mu_x$ so somehow, $\mu_B$ can be determined by $\mu_x$ for $x\in B\setminus\{p\}$ and I think dimension (open-closed argument) is used here. But I don't know how to write a detail and I wonder if this hand-waving makes sense.
Best Answer
The orientation cover of $M \setminus \{p\}$ is the orientation cover of $M$ with $2$ points removed. But the connectedness of a manifold of dimension at least $2$ doesn't change when you delete finitely many points.