# Fundamental group of $M$ has no subgroup of index $2\Rightarrow M$ is orientable

Let $M$ be a connected smooth manifold such that, for every $p\in M$, the fundamental group $\pi_1(M,p)$ has no subgroup of index $2$. Prove that $M$ is orientable.

Here's what I know: there is a smooth, orientable manifold $\widetilde{M}$ and a covering map $\pi:\widetilde{M}\to M$ whose fibers have cardinality $2$, and such that $M$ is orientable $\Leftrightarrow\widetilde{M}$ is disconnected.

Supposing by contradiction that $M$ is not orientable, $\widetilde{M}$ must be connected. I suppose the "subgroup of index $2$" part has to do with the fact that the fibers have cardinality $2$, and probably $\widetilde{M}$ being connected also plays a role.

I don't know much about covering spaces except for basic definitions, so I'm pretty much stuck.

• Do you know how covering spaces of $M$ are related to the fundamental group? – Eric Wofsey Sep 9 '18 at 21:22
• See my comments on math.stackexchange.com/questions/2906719/… – Lord Shark the Unknown Sep 9 '18 at 21:29
• @EricWofsey, according to Lord Shark's comment, the fact that $\widetilde{M}$ is connected means that there is a correspondence between $2$-fold coverings and subgroups of index $2$, but I don't know how to see this. – rmdmc89 Sep 9 '18 at 22:23

Assume $M$ is not orientable. We then have that the orientable double cover $\widetilde{M}$ of $M$ is path-connected.
For a path-connected covering $f: \widetilde{M} \to M$, it is well-known that the number of elements on a fiber is given by the index of $f_{\#}(\pi_1(\widetilde{M}) )$ on $\pi_1(M)$*. It follows that $f_{\#}(\pi_1(\widetilde{M}) )$ is a subgroup of index $2$ (since we have a double cover), a contradiction.
*Let's prove this assertion. For this, note that $$f_{\#}(\pi_1(\widetilde{M},p) )=\{[c]\mid c \text{ is a loop in } f(p) \text{ which lifts to a loop in } p\}.$$ Thus, $f_{\#}(\pi_1(\widetilde{M},p) )$ is precisely the isotropy of the point $p$ with respect to the action of $\pi_1(M,f(p))$ on the fiber which contains $p$, which takes $\alpha \in \pi_1(M,f(p))$ and $p'$ in such fiber and takes them to where the lift of $\alpha$ which begins on $p'$ ends.
You can verify that this is indeed a group action. It is also transitive, and this is where path-connectedness is important. Now, it is an easy exercise to verify that $$\pi_1(M,f(p))/f_{\#}(\pi_1(\widetilde{M},p) ) \to F_{f(p)}$$ $$[c] \mapsto [c] \cdot p$$ is well-defined and is a bijection, where $F_{f(p)}$ is the fiber that contains $p$ (this is an algebraic exercise).
Consider and atlas $(U_i,f_i)_{i\in I}$ of $M$, we can suppose that $U_i\cap U_j$ is connected and therefore the sign of the Jacobian $Jac(f_i\circ f_j^{-1})$ is constant (it takes its values in $\{-1,1\}$).Defines the $\mathbb{Z}/2$-bundle over $M$ whose coordinate changes are $g_{ij}:U_i\cap U_j\rightarrow sign(Jac(f_i\circ f_j^{-1}))$. This bundle is flat, (in fact it is a covering of $M$) thus define by a morphism $f:\pi_1(M)\rightarrow\mathbb{Z}/2$, (we can also says that since the bundle is a $2$-covering, its fundamental group is a subgroup of index $2$ of $\pi_1(M)$) the condition implies that $f$ is trivial and $Jac(f_i\circ f_j^{-1})$ is constant, we deduce that $M$ is orientable.