Dtjytyk

EDIT: I misunderstood the exercise, as pointed out by @Arthur in the comments. I only showed that a manifold does not lose orientability when a point is removed. But we can imagine a one-dimensional manifold gaining orientability when a point is removed. Here orientability would also be affected.

EDIT: In case you are looking for a solution to the exercise, here is a hint. The proof below shows (in painstaking detail!) that a manifold does not lose orientability when a point is removed. As for the second statement, we prove that an orientation on $Msetminus x$ can be extended to an orientation on $M$. There is at most one way to do this, since an orientation on a path-component is uniquely determined by its restriction to any point of the path-component (paving lemma argument). When $M$ has dimension greater than one, this extension is continuous ("locally consistent") -- use the fact $Msetminus x$ is path-connected (to see this without point-set topology, use LES to compute $H_0$), something which is not true in general when $M$ has dimension $1$.

The following is an exercise in Hatcher:

Show that deleting a point from a manifold of dimension greater than 1 does not affect orientability of the manifold.

I do not understand why the dimension must be greater than one. Is there something wrong with my proof for arbitrary dimension (below)? I have tried to be very explicit about maps, hoping to backtrace my mistake.

The heart of my argument is the commutative diagram near the bottom.

Proof. Let $zin M$ where $M$ is an $n$-manifold. Put $M':= Msetminuslbrace zrbrace$. Choose an orientation
$$lambdacolon Mrightarrow M_R$$
(as in Hatcher, s. 3.3). For each $xin M'$, excision gives a canonical isomorphism
$$
varphi_xcolon H_n(M'vert, x)xrightarrowsim H_n(Mvert ,x)
$$
induced by the inclusion of pairs $(M',M'setminus lbrace xrbrace )hookrightarrow (M,Msetminus lbrace xrbrace )$. For each $xin M'$, this allows us to define
$$
mu_x := varphi_x^{-1}(lambda_x).
$$
We claim that $xmapsto mu_x$ defines an orientation of $M'$. Clearly $mu_x$ generates $H_n(M'vert, x)$ for each $xin M'$ (since $varphi_x$ is an isomorphism). Hence we just need to check "local consistency", i.e. continuity of $xmapsto mu_x$.

For each $xin M'$, the fact that $lambda$ is an orientation means we can find $(B,lambda_B)$ with $Bni x$ an open ball of finite radius and $lambda_B$ a generator of $H_n(Mvert, B)$ with
$$lambda_Bvert_x = lambda_x,$$
where I use the restriction bar "$vert$" to indicate the map induced by the inclusion $(M,Msetminus B)hookrightarrow (M,Msetminuslbrace xrbrace )$. Since any ball contained in $B$ satisfies the same consistency condition, we may assume that $Bsubset M'$ and even that $mathrm{cl}, Bsubset M'$.

(For the last assertion, I used that $M$ is Hausdorff to find open neighborhoods $Uni x$ and $Vni y$ with empty intersection. Then $xin Ucap B$ and $z$ does not lie in the closure of $Ucap B$. Finally, open balls form a basis, so we can choose a smaller open ball around $x$ contained in $Ucap B$.)

The fact that $mathrm{cl}, Bsubset M'$ means that the inclusion $(M',M'setminus B)hookrightarrow (M,Msetminus B)$ induces an isomorphism
$$
psicolon H_n(M'vert B)xrightarrowsim H_n(Mvert B).
$$

Define
$$
mu_B := psi^{-1}(lambda_B).
$$
We claim that $(B,mu_B)$ is a "consistent open ball at $x$", i.e. that
$$
mu_Bvert_y = mu_y
$$
for each $y in B$. But this follows from the diagram:
$$
require{AMScd}
begin{CD}
H_n(M'vert, B) @>>> H_n(M'vert, y) \
@Vpsi VV @VVvarphi_x V \
H_n(Mvert, B) @>>> H_n(Mvert, y)
end{CD}
$$
Here the horizontal arrows are the "restriction maps" mentioned above. All arrows in this diagram are isomorphisms (the vertical arrows are excision maps; the horizontal arrows are induced by homotopy equivalences). Furthermore, the diagram commutes since both the clockwise and the counterclockwise directions are induced by the inclusion $(M',M'setminus B)hookrightarrow (M,Msetminuslbrace yrbrace )$. But the clockwise value of $mu_B$ is $varphi_x (mu_Bvert_y)$, whereas the counterclockwise value is $psi (mu_B)vert_y = lambda_Bvert_y = lambda_y$. But
$$
varphi_x (mu_Bvert_y ) = lambda_y Rightarrow mu_Bvert_y = varphi_x^{-1}(lambda_y ) = mu_y,
$$
completing the proof. Q.E.D.