Boundary tori inject into the circle-bundle group

Each of the two possible bases has nonempty boundary and deformation-retracts to a graph, so ${\pi }_2(B)=0$. The homotopy exact sequence [F2] therefore injects the fiber subgroup ${\pi }_1(S^1)\cong \mathbb{Z}$ into ${\pi }_1(N)$, with image the kernel of $p_{\ast }:{\pi }_1(N)\to {\pi }_1(B)$.

The fundamental group of a pair of pants is free on two generators $a,b$, and its three boundary loops represent $a$, $b$, and $(ab{)}^{-1}$, each of infinite order. The once-punctured Möbius band has free generators $x,c_1$ and boundary loops represented by $c_1$ and $c_1^{-1}{x}^{-2}$; both are nontrivial cyclically reduced words and have infinite order in the free group. Thus every boundary circle of $B$ injects on fundamental groups.

The restriction $T\to \mathrm{\partial }B$ is a circle bundle over a circle, hence induces a surjection ${\pi }_1(T)\to {\pi }_1(\mathrm{\partial }B)$. If $g\in {\pi }_1(T)$ maps trivially to ${\pi }_1(N)$, its projection is a power of the corresponding boundary loop and is trivial in ${\pi }_1(B)$; step 1.2 forces that power to be zero. Therefore $g$ lies in the fiber subgroup, whose inclusion into ${\pi }_1(N)$ is injective by step 1.1, so $g=1$. This proves that ${\pi }_1(T)\to {\pi }_1(N)$ is injective.