I am confused about one exercise in my homework.
$N$ (a submanifold of $M$) is said to be $2$-sided if there exists a neighborhood of $N$ homeomorphic to $N × [0, 1]$, and non-separating if $M$\ $N$ is connected. (1) If $N ⊂ M$ is $2$-sided and non-separating, show that there exists a map $M → S^1$ which is injective on cohomology. (2) the fundamental class $[N]$ generates a free factor in $H_{n−1}(M)$ if $N$ is non-separating.
Suppose $N$ is $2$-sided and non-separating. Let's parameterize $S^1$ by the quotient map $[0,1]\to S^1$ sending $0$ and $1$ to the same point. Define the quotient map $f:M\to S^1$ by sending $(x,t)\in N\times [0,1]$ to $t$ and by sending everything in $M\setminus(N\times [0,1])$ to $0$ (which can be thought of as collapsing the complement of the regular neighborhood, then projecting onto the normal axis).
The induced map would be $f^*:H^*(S^1)\to H^*(M)$. It's injective on $H^0$ since $M$ is nonempty. For injectivity of $H^1(S^1)\to H^1(M)$, we just want to see that $f^*[S^1]$ is a free factor of $H^1(M)$. By the Universal Coefficient Theorem, $H^1(M)$ has no torsion, so it suffices to show that $f^*[S^1]$ is nonzero. Since $N$ is non-separating, in $M$ there is a closed loop $\gamma$ from one side of $N$ to the other, intersecting $N$ at only one point. The corresponding loop $f\circ\gamma$ in $S^1$ has degree $\pm 1$, hence $f_*:H_1(M)\to H_1(S^1)$ is surjective, and by naturality of the Universal Coefficient Theorem along with left exactness of the dualization functor, $f^*:H^1(S^1)\to H^1(M)$ is injective.
Really, $f$ is representing a first cohomology class of $M$ (via $$H^1(M)\cong \hom(H_1(M),\mathbb{Z})\cong\hom(\pi_1(M),\mathbb{Z})\cong\langle M,S^1\rangle$$ where $\langle M,S^1\rangle$ is homotopy classes of maps $M\to S^1$, since $S^1$ is a $K(\mathbb{Z},1)$), and the question was whether the class is nontrivial. This is the Poincaré dual of $N$, and $\gamma$ was a loop for which $N\cdot \gamma\neq 0$, which is to say $N$ and $\gamma$ have nonzero intersection number, and hence both are homologically nontrivial.
That's part (2): $[N]$ is the Poincaré dual of $f$ as a cohomology class.