How show that a surface embedded is non-orientable? Let be $M$ a compact $3$-manifold. If $\Sigma$ is a embedded surface in $M$, such that $\Sigma$ is homeomorphic to $\mathbb{RP}^2$.
If $i: \pi_1(\Sigma) \longrightarrow \pi_1(M)$ is not injective, then $\Sigma$ is non-orientable?
Note that $\pi_1(\Sigma) = \mathbb{Z}/2$.
This is utilized in the proof of Proposition 3 in http://arxiv.org/abs/0909.1665.
Thank you!
 A: The claim they're making is not about whether or not $\Sigma$ is orientable (it can't be since it's homeomorphic to $\mathbb{R}P^2$), but whether or not $TM$, when restricted to $\Sigma$ (that is, $i^* TM$) is an orientable vector bundle.
A vector bundle is orientable is the transition maps can be chosen to lie in $Gl^+$ (orientation preserving linear isomorphisms) instead of just in $Gl$ (all linear isomorphisms).
Alternatively, a vector bundle is classified by a (homotopy class of) map $\Sigma\rightarrow BGl$ and orientability is precisely the condition that there is a lift of this map to $BGl^+$.
Alternatively again, (and this is the characterization I'll use), a bundle $\xi$ is orientable iff the first Stiefel-Whitney class $w_1(\xi)\in H^1(\Sigma; \mathbb{Z}/2)$ is $0$.  (I honestly forgot how to prove the equivalence between the three notions, if I ever knew it.  I'm thinking Lawson's book on spin geometry has a proof of the equivalence, but I could be misremembering).
The tangent bundle over $\Sigma$ is not orientable (since $\mathbb{R}P^2$ isn't), but other vector bundles over $\Sigma$ may be.  For example, the trivial bundle (of any rank) is always orientable.
Anyway, to see why their claim is true, note that if $\pi_1(\Sigma)\rightarrow \pi_1(M)$ is not injective, it's actually the $0$ map.  This implies the induced map $H_1(\Sigma;\mathbb{Z}/2)\rightarrow H_1(M;\mathbb{Z}/2)$ is the $0$ map, and this in turn implies the induced map $H^1(M;\mathbb{Z}/2)\rightarrow H^1(\Sigma, \mathbb{Z}/2)$ is the $0$ map.
By naturality of characteristic classes, we have $w_1(i^* TM) = i^*w_1(TM) = 0$ since $i^*$ is the $0$ map, so the bundle $TM$, when restricted to $\Sigma$ is orientable.
Now they use the fact that $i^*TM = T\Sigma \oplus \nu$ where $\nu$ is the normal bundle to argue that $\nu$ is nontrivial.  If it were trivial, it'd be orientable, so $T\Sigma\oplus \nu$ wouldn't be, but $i^*TM$ is, giving a contradiction.
