Let $M$ a compact, connected $m$-manifold. Assume that $H_m(M;\mathbb{Z})\cong \mathbb{Z}.$ I want to show that then $M$ is orientable.
My ideas:
Let $o_M$ one of the two generators of $H_m(M;\mathbb{Z})\cong \mathbb{Z}.$ If I could show that for any $x\in M$ the restriction map $(\rho_{x,M})_{\ast}\colon H_m(M;\mathbb{Z})\rightarrow H_m(M,M\setminus x;\mathbb{Z})$ is an isomorphism I would be done, since then $((\rho_{x,M})_{\ast}(o_M) \vert x\in M)$ would be a family of coherent generators. I convinced myself that $(\rho_{x,M})_{\ast}\colon H_m(M;\mathbb{Z})\rightarrow H_m(M,M\setminus x;\mathbb{Z})$ is injective for any $x\in M$ using the LES and Poincaré duality. But why is each restriction surjective? Can't it be multiplication by some $n\neq 0,1,-1$? If so it would have to be multiplication by the same $n$ for all $x$ by the argument given in the answer to my question Global orientation class of $M$ implies top homology is $\mathbb{Z}$. But why can't that be?
My notes argue like that:
For surjectivity consider $(\rho_{x,M})_{\ast}o_M)\vert x\in M)=(k_xo_x)\vert x\in M )$ for some collection $k_x$. But $k_x$ is locally constant on $M$ and thus constant, and if there is a coherent choice $(ko_x\vert x\in M)$ it follows that $(o_x \vert x\in M)$ is a coherent choice of orientation.
I included the typos, maybe how I am fixing them in my head is not how they should be fixed. Anyway, I do not understand that passage.
EDIT:
- Mikhail Katz claims in the comments that my question is equivalent to asking why the map $H_m(S_m)\rightarrow H_m(S^m,S^m\setminus x)$ is surjective. Why is that equivalent to my question?
- I would also be happy with a reference.
Since you are using Poincare duality already, I think it is fair to quote the following theorem that is usually stated in preparation for Poincare duality:
(Orientation Theorem): Let $M$ be compact manifold. Then the map
$ j : H_n(M;R) \to \Gamma(M, \omicron_M \otimes R) $
is an isomorphism, the latter $R$ module referring to the set of sections of the covering map $p: \omicron_M \otimes R \to M$ [look for example from Haynes Miller lecture notes in AT: 31.13 https://math.mit.edu/~hrm/papers/lectures-905-906.pdf].
Also, consider the following composite map
$\tau: \omicron_M = \bigsqcup_{x \in M} H_n(M|x) \to^{\sqcup_x \overline{\phi}_x} \bigsqcup_{x \in M} \mathbb{Z}/\{ \pm 1\} \to \mathbb{Z}/\{ \pm 1\} $
where $\overline{\phi}_x$ is the composite $H_n(M|x) \to^{\cong} \mathbb{Z} \to \mathbb{Z}/\{ \pm 1\} $. $\tau$ is continuous with respect to the topology on $\omicron_M$ defined as e.g. the above lecture notes.
With this in mind:
By your assumption and the above theorem, $ \Gamma(M, \omicron_M \otimes \mathbb{Z}) \cong H_n(M; \mathbb{Z}) \cong \mathbb{Z}$. Hence there is a non zero section, i.e. a continuous map $s: M \to \omicron_M \otimes \mathbb{Z}$. As it is non zero, $\exists x_0 \in M, k \in \mathbb{Z}\setminus \{ 0 \} : s(x_0) \in \tau^{-1}(\{ \pm k \})$.
As composite $\tau \circ s: M \to \mathbb{Z}/\{\pm 1\}$ is continuous, but since $M$ is connected and $\mathbb{Z}/\{ \pm 1\}$ is discrete, it must be constant. Hence $\forall x \in M : s(x) \in \tau^{-1}(\{ \pm k \})$, and $s: M \to \tau^{-1}(\{ \pm k \} ) $. One can see that
$ f: (\omicron_M \otimes \mathbb{Z})^{\times} = \tau^{-1}(\{ \pm 1\}) \to \tau^{-1}(\{ \pm k \}), \alpha \mapsto k \cdot \alpha $
is a homeomorphism and since it preserves the fibres, an isomorphism of the two fibre bundles. Thus the composite $f^{-1}\circ s : M \to \omicron_M$ is a section, showing that $M$ is orientable.
If you find mistakes in the argumentation let me know, I am also just now learning this subject.
EDIT
You can consider the commutative diagram of $R$ modules, for arbitrary $x \in M$: $\require{AMScd}$ \begin{CD} H_n(M) @>{j_{\emptyset, x}}>> H_n(M, M\setminus x)\\ @V{j}VV @A{ev_x}AA \\ \Gamma(M, \omicron_M) @>{id}>> \Gamma(M, \omicron_M) \end{CD} where $j$ is defined as
$\forall \alpha \in H_n(M): \quad j(\alpha) : M \to \omicron_M, x \mapsto j(\alpha)(x) = j_{\emptyset, x}(\alpha) $.
One can check that $j(\alpha)$ is continuous and is therefore well defined. The evaluation map evaluates a given section at a point $x \in M$.
Now if you take it as a fact that the top map $j_{\emptyset, x}$ is injective, it follows that $j$ must also be injective (btw, the map $ev_x$ is also injective, stemming from the fact that $M$ is connected). In particular there is a non-zero section in $\Gamma(M, \omicron_M)$, since by assumption $H_n(M) \neq 0 $. Given this non-zero section, you can again apply the proof above, without referring to the Orientation Theorem.
If you find something even more elementary let me know.