Let $M$ be a connected closed orientable smooth $n$-manifold and $\nu \in H_{n-1}(M, \mathbb{Z})$ a non-trivial codimension-one homology class. It is known that $\nu$ can be represented by an embedded submanifold $N \subseteq M$ (this is nicely explained in the answer to this question on math stackexchange).
I'm curious when is it true that $\nu$ can be realised by an embedded compact submanifold that is $\pi_1$-injective on each component? Are there obvious obstructions to this being possible?
I will explain why the general answer to this question is negative in dimensions $\geq 4$. The next remark expresses the fundamental obstruction. We say a simplicial tree $T$ equipped with an action of a group $G$ (a so-called $G$-tree) dominates another $G$-tree $T’$ if there is a $G$-equivariant map $T\to T’$.
Remark: If $\phi\in H^1(M^n,\mathbb{Z})$ is Poincaré dual to an embedded submanifold $N^{n-1}\subseteq M^n$ then the corresponding action on the real line is dominated by the Bass—Serre tree of the splitting of $\pi_1(M)$ obtained by cutting $M$ along $N$.
This can be applied along with the following lemma; it must be well known but I don’t know a reference, so I will give a proof.
Lemma: If the kernel $K$ of $\phi:G\to\mathbb{Z}$ is finitely generated then the corresponding action on a line is not properly dominated by any other minimal action.
Proof: Let $T$ be a minimal $G$-tree that dominates $\phi$. We claim that $K$ fixes the whole of $T$.
First, we claim that $K$ fixes some point of $T$. If not then there is a minimal $K$-invariant subtree $T_K$. On the one hand, $K$ acts cofinitely on $T_K$ because $K$ is finitely generated. On the other hand, $K$ is $G$-invariant because $K$ is normal in $G$. By the $G$-minimality of $T$, it follows that $T=T_K$ and so $K$ acts cofinitely on $T$. Let $\Gamma_K=K\backslash T$ be the quotient graph of groups. Any $g\in G$ now acts as a covering transformation on $\Gamma_K$, so acts by graph automorphisms on the underlying graph. Since $\Gamma_K$ is finite, a large enough power $g^n$ fixes an edge, whence $g^n\in K$. But this is a contradiction when applied to any $g\in G$ with $\phi(g)=1$.
Therefore, $K$ fixes a vertex in $T$. Now, again because $K$ is normal in $G$, the fixed subtree $\mathrm{Fix}(K)$ is a $G$-invariant subtree, so $K$ must indeed fix the whole of $T$ by minimality, as claimed.
Thus, the kernel of the action of $G$ on $T$ is contained in $K$, so $T$ is the line given by $\phi$. QED
Combined with the remark, the lemma gives many examples. The simplest occurs in dimension 4, using the fact that $\mathrm{cd}_\mathbb{Q}(\pi_1(N^3))\leq 3$ for any closed 3-manifold
Example: Let $M^4$ be any closed oriented 4-manifold with $\pi_1(M)\cong\mathbb{Z}^5$. Then any $\phi\in H^1(M)$ has $\ker\phi\cong\mathbb{Z}^4$. Since $\mathbb{Z}^4$ is finitely generated, by the remark and the lemma it follows that if the dual of $\phi$ is represented by a closed manifold $N^3$ then $\pi_1(N)\cong\mathbb{Z}^4$, a contradiction.
In fact, the same argument works whenever $\pi_1(M^4)$ is a rational Poincaré duality group of dimension $\geq 5$.
For $\mathrm{dim}(M)\geq 5$, we can also construct counter examples, using @IanAgol’s observation in comments.
Example: Let $M^5$ haver $\pi_1(M)\cong F_2\times F_2$, with $F_2$ the free group of rank 2. Let $\phi:F_2\times F_2\to\mathbb{Z}$ send every generator to 1. Then $K=\ker\phi$ is finitely generated but not finitely presented. By the remark, if the dual of $\phi$ is represented by closed 4-manifold $N$ then $\pi_1(N)=K$. But $K$ is not finitely presented, so this is a contradiction.
More generally, @MoisheKohan’s answer appears to prove that this necessary condition is also sufficient. This is a very nice and natural group-theoretic question, that I haven’t seen addressed before:
Question: Let $G$ be a finitely presented group. Which $\phi:G\to\mathbb{Z}$ are dominated by a splitting over a finitely presented subgroup?
The statement is clear in the case $n=2$. In the case $n=3$ the statement is a consequence of the Loop Theorem and, I think, is due to Waldhausen's 1968 paper "On irreducible 3 -manifolds which are sufficiently large." You can find some form of it in Hempel's book "3-manifolds" (Lemma 6.6) and, in a cleaner form, in Hatcher's "Notes on basic 3-manifold topology," Lemma 3.6. Note that one does not need $M$ to be closed in this case. I do not know what happens in dimension 4, I suspect the statement is false. In dimensions $n\ge 5$ it is true and the proof is somewhat similar to the 3-dimensional case. Instead of the Loop Theorem one uses transversality:
First of all, since $M$ is compact, the kernel of $\pi_1(N,x)\to \pi_1(M,x)$ is the normal closure of a finite subset of $\pi_1(N,x)$ for each $x\in N$ (note that $N$ can be disconnected). The submanifold $N$ is 2-sided, let $\nu(N)=[-1,1]\times N$ denote its tubular neighborhood in $M$. If each component of $\partial \nu( N)$ is $\pi_1$-injective in $M_0:=M-int(\nu(N))$, then (by the Seifert-Van Kampen theorem)
$N$ is $\pi_1$-injective in $M$ and there is nothing to prove. Again, by compactness, we obtain a finite collection of simple loops
$c_i$ in $\partial M_0$ ($i=1,...,k$) which normally generate kernels of the homomorphisms $\pi_1(\partial M_0,x)\to \pi_1(M)$.
Without loss of generality, we may assume that the loops $c_i$ are pairwise disjoint. We will modify $N$ by killing off the loops $c_i$.
Then, by transversality, each $c_i$ bounds smooth embedded 2-disk $D_i$ in $M_0$ such that $D_i\cap \partial N_0=c_i$ and the disks are pairwise disjoint. Next, attach pairwise disjoint 2-handles $H_i$ in $M_0$ to $\nu(N)$ (and smooth the corners!) so that $D_i$ is the core of $H_i$. The boundary of $W=\nu(N)\cup (H_1\cup...\cup H_k)$ consists of two parts: One of these is $N$ and the other is a submanifold $N'$ homologous to $N$. By the construction, $N'$ is $\pi_1$-injective in $M$.
This argument breaks down in dimension $4$ since we cannot guarantee the existence of embedded and pairwise disjoint disks $D_i$.
Edit. There is a problem with my proof in dimension $\ge 5$ since the image subgroup of $\pi_1(M)$ need not be finitely presentable.
