Codimension 1 homology represented by Embedded Submanifold I'm looking for a reference for the following statement:
Given an oriented manifold $M$ and a class $\xi \in H_{n-1}(M)$, there is some embedded oriented submanifold $F^+ \hookrightarrow M$ such that $i_*[F] = \xi$.
In every reference I've been able to find, this is treated like an easy special case, and the proof sketch is as follows, from this MO question:
"Every class in $H_{n−1}(M;\mathbb{Z})$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\rightarrow S^1$ representing the Poincare dual in $H^1(M;\mathbb{Z})=[M,S^1]$ and take the preimage of a point. In dimensions>2 it can be taken connected."
Can someone supply some more details of this proof? In particular, given that the preimage $F$ is a submanifold:


*

*How do we know that this submanifold is orientable?

*How do we know that $i_*[F] = \xi$?

 A: First, do not take the preimage of an arbitrary point. Instead, take the preimage of a regular value $p \in S^1$. The inverse function theorem then guarantees that $N=f^{-1}(p)$ is a submanifold of codimension 1.
Next, show that $N$ has orientable normal bundle. What this means is that $T(M)\mid N$, the restriction to $N$ of the tangent bundle $T(M)$, has a section which is not tangent to $N$. To put it still another way, there is a continuous function $\nu : N \to T(M)$, such that $\nu(x) \in T_x M$ is not tangent to $N$. You can construct $\nu$ locally using the local charts provided by the inverse function theorem, then you can put the local patches together to get $\nu$ on all of $N$ using a partition of unity argument.
Next, since $M$ is orientable and $N$ has orientable normal bundle, it follows that $N$ is orientable.
The statement that $i_*[F]=\xi$ is, literally, one version of the Poincare duality theorem. You can see this approach to the Poincare duality theorem in the book of Bott and Tu.
A: I am assuming your manifolds to be smooth.
Note that $H^1(M) = Hom(H_1(M),Z)$ by the universal coefficient theorem. Note that the latter is the same as $Hom(\pi_1(M),Z)$, since $Z$ is abelian. It is now possible to show (try it out with a CW complex), that such maps are always induced by a continous map $M \to S^1$. One obtains $H^1(M) = [M,S^1]$. Differential topology tells you that every class in the latter group is represented by a smooth map $f:M \to S^1$ (it tells you even more, that every continous map can be approximated by such smooth maps). Sard's theorem tells you that you can choose a regular value $p$ of this map. And finally one obtains by the regular value theorem, that $f^{-1}(p)$ is an orientable smooth properly embedded submanifold in $M$. That's it for the first part. Note that if $M$ has boundary we want to consider the relative homology $H_{n-1}(M,\partial M) \cong H^1(M)$.
Now the last part follows for example from the (not so hard) problem 11-C in Milnors Characteristic Classes. Don't get confused with the strange sign convention there. It states that the Poincaré (similar Lefschetz) Duality Isomorphism maps the Thom class of the normal bundle of an embedded submanifold (or rather the inclusion of the thom class in the homology of the supermanifold) to the homology class which is represented by the submanifold itself, ie. the image of the fundamental class under the map induced by the inclusion. So it suffices to show, that the Thom class of the normal bundle of the submanifold in $H^1(M)$ is precisely the cohomology class with which we have started. But this is obvious since naturality implies that the Thom class is obtained by pulling back the Thom class of the normal bundle of the regular value in $S^1$ under $f$, but this is precisely the generator of $H^1(S^1)$.
Please ask if you need further details. 

I want to add a few remarks about an intuitive approach using intersection theory (this is how I use to think about this problem and its solution). First off we want to choose a regular value $p$ of a map $f$ representing a cohomology class in $H^1(M)$, just as we did in the first paragraph. Let us now recall the important result of intersection theory for this matter: take a submanifold $N$ representing a homology class in $H_{n-1}(M,\partial M)$ (where $\partial =\emptyset$ is possible). This is dual to a class in $H^1(M)=Hom(\pi_1(M),Z)$, which is just a homomorphism $\phi$. One now obtains $\phi$ (and it is uniquely determined), by evaluating intersections of loops intersecting $N$. I.e. $\phi$ is a homomorphism and $N$ defines a homomorphism by sending a smooth loop intersecting $N$ transversal to the signed sum of intersection points (differential topology shows again, that every element in $\pi_1(M)$ can be represented in that way. Furthermore it is far away from trivial to show that this homomorphism does not depend on this choice.). This is kind of duality follows from example from the formula $[A] \cup [B] = [A\cap B]$ where $A,B$ are transversal submanifolds such that they dimensions add to the dimension of $M$. 
Now how you can think of the problem is the following: let $S^1 \to M$ be a smooth embedding transversal to $N$. Then $S^1 \to M \stackrel f \to S^1$ is a loop in $S^1$ intersecting $p$ transversal (where $p$ is representing a homology class in $H_0(S^1) = H^1(S^1)$ since $p$ is a regular value of $f$ and $f^{-1}(p)=N$. By this transversality the evaluation of the loop on $N$ (by evaluation I mean evaluation of the intersection) boils down to the evaluation of the loop on $p$ in $S^1$. But this is easy - it just measures the degree. Hence let $\alpha : S^1 \to M$ be a smooth, transversal to $N$, represantive of $[\alpha] \in H_1(M)$ (by Hurewicz). Let $[N]^*,[p]^*$ denote the dual cohomology class in $H^1$ of each submanifold, which is then just a homomorphism. They evaluate to: $[N]^*([\alpha])=[p]^*([f\alpha])=deg(f\alpha)$. But since $[p]$ is a generator of $H_0(S^1)$ and hence $[p]^*$ is the generator of $H^1(S^1)$, it follows by the previous formula that $[N]^*$ is the image of the pullback of $[p]^*$ under the map $f$. Consequently $[N]^*$ represents the same homomorphism as $f_*$ which is the one we started with. 
A last note to help you: a cohomology class in $H^1(M)$ can be assigned a function $M \to S^1$ in two ways: (1) $f_*$ induces it (2) it is the image of a generator under $f^*$. Please check that those assignments are precisely and trivially the same.
Try to visualize the process from above: you have a large hyperplane which represents a homomorphism of loops. A loop can be homotoped and will always be evaluated the same way. Hence if you can homotope the loop such that it does not intersect the hyperplane, it gets evaluated to zero by the dual cohomology class. But you can simplify the process of evaluating loops on the hyperplane if it is the preimage of a regular value on $S^1$ by pushing the loops forward under this map to $S^1$. That's as easy as it gets and I really hope it helps. It is really fun and important.
