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Let $M^3$ be a compact manifold. If $\mathbb{RP}^2$ is embedded in $M$. Suppose, by contradiction, that $i_\sharp: \pi_1(\mathbb{RP}^2) \longrightarrow \pi_1(M)$ is non-injective and that the normal bundle of $\mathbb{RP}^2$ is non-trivial.

Let $\gamma:[0,1] \longrightarrow \mathbb{RP}^2$ be a nontrivial element of $\pi_1(\mathbb{RP}^2)$ and $\nu(t) \in T_{\gamma(t)}M$ ortogonal to $T_{\gamma(t)}\mathbb{RP}^2$, such that, $\nu$ is continuous and $\nu(0)=-\nu(1)$.

Define

$\gamma_{\varepsilon}(t):=\exp_{\gamma(t)}(\varepsilon \sin(\pi t))\nu(t).$

For $\varepsilon\ll1$, $\gamma_{\varepsilon}$ is well-defined and is a smooth closed curve in $M$. Moreover, $\gamma_{\varepsilon}$ intersects $\mathbb{RP}^2$ exactly one point, and the intersection is transversal.

Can anyone explain why $\gamma_{\varepsilon}$ is non-trivial element of $\pi_1(M)$?

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As you wrote, $\gamma_\epsilon$ intersects $\mathbb{RP}^2$ transversally and at one point. It means that $[\gamma_\epsilon]\in H_1(M,\mathbb{Z}_2)$ and $[\mathbb{RP}^2]\in H_2(M,\mathbb{Z}_2)$ have intersection number $1$, so $[\gamma_\epsilon]\neq 0$, and thus $\gamma_\epsilon$ can't be contractible.

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    Thank you @user8268 ! But only now I'm studying differential topology, because I need! Can you guide a book that explains about cohomology and the result above? Thank you again!2012-11-06