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Let $(M,g)$ be a compact riemannian 3-manifold and $\Sigma \subset M$ an embedded compact surface homeomorphic to the projective plane. Consider the application $i_\#:\pi_1(\Sigma)\to \pi_1(M)$ given by $i_{\#}([\alpha])=[i(\alpha)]$, where $i$ denotes the inclusion application of $\Sigma$ into $M$. Why $i_{\#}$ not injective implies $TM\Big|_{\Sigma}$ orientable?

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  • $\begingroup$ This follows from examining the first Stiefel-Whitney class. If you don't know about those, be more direct: if this bundle were not orientable, there would be some loop you could walk around on which it changes its orientation. But all loops in $\Sigma$ are null-homotopic in $M$, so... $\endgroup$ – user98602 Jul 15 '15 at 16:46
  • $\begingroup$ @MikeMiller can you point me some reference? $\endgroup$ – Lawen Jul 16 '15 at 16:22
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Let $j$ be a smooth embedding of $M$ into $\Bbb R^N$ for some large enough $N$. Now the map $j$ induces a map from $ \hat{j}: M \to \operatorname{Gr}_3(\Bbb R^N)$ where $\operatorname{Gr}_3(\Bbb R^N)$ is the space of 3-planes in $\Bbb R^n$ (see Grassmanian) and $\hat j (x)=\operatorname{Image}(dj(x))$ (the image of the derivative of $j$ at $x$). The universal cover of $\operatorname{Gr}_3(\Bbb R^N)$ is the oriented Grassmanian $\widetilde{\operatorname{Gr}_3(\Bbb R^N)}$, the space of oriented 3-planes in $\Bbb R^N$.

Now consider $\hat j \circ i$. If this map lifts to $\widetilde{\operatorname{Gr}_3(\Bbb R^N)}$, then we can pullback the orientation on each oriented 3-plane in $\Bbb R^N$ to get an orientation on $TM|_{\Sigma}$. For the map on $\pi_1$, we have $(\hat j \circ i)_*=\hat j_* \circ i_*=0$. So the $\hat j \circ i$ does lift to $\widetilde{\operatorname{Gr}_3(\Bbb R^N)}$ and $TM|_{\Sigma}$ is oriented.

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