Representation of (co)homology classes of $3$-manifolds by embedded surfaces Let $M$ be a closed oriented $3$-manifold. Theorems in algebraic topology allow us to identify
 $$H_2(M) \ \cong \ H^1(M) \ \cong  \ \langle M,S^1\rangle$$
where (co)homology is meant with integer coefficient and the last object is the set of (homotopy classes of) maps $M\to S^1$. So, given a homology class $a \in H_2(M)$, we get a map $f_a : M \to S^1$ (up to homotopy) and for any regular value $t \in S^1$ of this map we have that $f_a^{-1}(t)$ is an embedded oriented surface in $M$. I'm fine with this.
It's believable that if we change the point $t \in S^1$ then we get different surfaces, but I am having difficulty understanding how different these surfaces might be, especially in terms of their genus or Euler characteristic. I think that something happens when we go through a critical value of $f_a$ on $S^1$ and that some $S^1$-valued Morse theory is going on here. I would like to know if this is correct and to see an explicit example of this phenomenon (i.e. topologically different surfaces representing the same homology class in $H_2(M)$ ). I apologize if this is too basic, but I'm not used to think about (co)homology in geometric terms beyond dimension $1$.
 A: I think I can prove:

For any pair of closed orientable surfaces $N_1$ and $N_2$, there is a smooth function $f:S^3\rightarrow S^1$ with $f^{-1}(-1) \cong N_1$ and $f^{-1}(1) \cong N_2$

Here, I'm thinking of $S^1\subseteq\mathbb{C}$ as the unit complex numbers.
The proof will occur via a couple of lemmas.
Lemma 1:  For any two closed orientable surfaces $N_1$ and $N_2$, there is a 3-dimensional manifold-with-boundary $M$ embeddable in $S^3$ with boundary diffeomorphic to the disjoint union of $N_1$ and $N_2$.
Proof: Embed $N_1$ into a very tiny open subset of $\mathbb{R}^3$, small enough that this open subset fits entirely inside a handle of $N_2$.  Take as $M$ all points in between $N_1$ and $N_2$.  Since $M$ is embedded in $\mathbb{R}^3$ and $\mathbb{R}^3$ can easily be embedded in $S^3$, $M$ can be embedded in $S^3$. $\square$
Lemma 2:  With $M$ (in $\mathbb{R}^3$) as above, there is a smooth function $g:\mathbb{R}^3\rightarrow S^1$ for which $-1$ and $1$ are regular values, $g^{-1}(1) = N_2$, $g^{-1}(-1)= N_1$.  Further $g$ is smaller than $-1$ only outside of $N_1$, identically $2$ far enough outside of $N_1$, and bigger than $1$ only inside $N_2$.  
Proof: (Sketch)  Given such an $M$ (still embedded in $\mathbb{R}^3$), consider the function $\tilde{g}:M\rightarrow \mathbb{R}$ with $$\tilde{g}(p) = \frac{d(p,N_1)-d(p,N_2)}{d(p,N_1) + d(p,N_2)}$$ where $d$ is the Euclidean distance (where distance measured from inside is negative).  Since the boundary of $M$ is a disjoint union, the denominator never vanishes, so $\tilde{g}$ defines a continuous (but not necessarily smooth) function.  Direct computation shows that $\tilde{g}^{-1}(1) = N_2$ and $\tilde{g}^{-1}(-1) = N_1$.
Invoking some Riemannian geometry (using the exponential map and compactness of the $N_i$), it's not too hard to see that $\tilde{g}$ is actually smooth near both $N_1$ and $N_2$, and, then, by denseness of of smooth functions, we can approximate $\tilde{g}$ by a smooth function $g$ which agrees with $\tilde{g}$ near $N_1$ and $N_2$.  Further, by attaching cylinders $N_i\times [0,1]$ to each respective boundary, we may assume wlog that $1$ and $-1$ are regular values of $g$ and $g^{-1}(1) = N_2$ and $g^{-1}(-1) = N_1$.
Finally, inside of $N_2$, we can extend $g$ to be smooth and take on values in $(1,2]$, and likewise outside of $N_1$, we can extend $g$ to be smooth and take on values in $[-2,-1)$ with $g(p) = -2$ for all $p$ sufficiently far from $N_1$.  $\square$
From here, the original proposition is easy.  Via stereographics projection, we embed $\mathbb{R}^3$ into $S^3$ and transfer $g$ to a function $f$ defined on $S^3$ with a point $q$ removed.  However, since $g$ is identically $-2$ far enough outside $N_1$, we can extend $f$ to a smooth function on $S^3$ by defining $f(q) = -2$.
