Fundamental groups of codimension 1 manifold complements

Question

Let $M$ be a smooth manifold of dimension at most $3$ and $S \subset M$ a smoothly embedded compact connected codimension $1$ manifold, separating $M$ into two components, $M_1$ and $M_2$. I wonder now if the following is true:

If $\pi_1(M)$ is finitely generated, so is $\pi_1(M_i)$ for $i=1,2$.

One might be inclined to think that this is a purely algebraic matter, but it is quite possible to find an infinitley generated group $A$ and set up a diagram of groups $A \leftarrow B \rightarrow C$ such that the pushout is finitely generated. Hence, there must be some topological restrictions i do not realize. Any help is appreciated.

Answer 1

This is false in dimension 3. For instance, let $M$ be the Whitehead manifold. Then there exists a compact PL (or smooth as you prefer) submanifold $N$ in $M$ whose complement has infinitely generated fundamental group. This is a special case of the main theorem in

T.W. Tucker, Non-compact 3-manifolds and the missing-boundary problem, Topology 13 (1974), 267-273.

In dimension 2 your conjecture is, of course, true, since surfaces with finitely generated fundamental groups are tame.

Answer 2

I think you should be able to show this using Scott's Core theorem. Pick a compact core $C$ in $M$. Now you might want to distinguish some cases, but you should get somthign like the generation of $\pi_1M_i$ by the image of $\pi_1(C\cap M_i)$ and $\pi_1S$. You should also assume properly embeddedness for this to work.

EDIT: So as pointed out in the comments my answer was just too sloppy to understand so I decided to give a proof.

First: how to you come up with the idea of the proof? So my attempt was to look at graphs of groups or groupoids to avoid issues related to base point, connectedness and so on. And now you just start to look at suitable sets covering the core $C$ and the seperating surface $S$ and partition them more and more (and sometimes less) to get nice generating sets for different parts. You get a graph of groups and which contains the relations among the fundamental groups of the sets. Eventually I got multiple partitions of sets which would do the trick. Here is a particularly easy one, to which I arrived by using the above technique, but which actually only consists of two sets.

So let* $K= C\cup \nu S$ and $K_i = K \cap M_i$, where $\nu S$ is a trivial tubular neighborhood (trivial by seperating property and $M$ is connected) small enough that everything looks like a product there (especially transverse intersections with $C$). Let $k^i:S \hookrightarrow K_i$ and $l^i$ the same map post composed to embed $S$ into $M_i$. Now we see that by Van Kampen (here we are again slightly ignoring connectedness of the $K_1,K_2$**) we have $$ \pi_1K = \pi_1K_1 *_{k_*\pi_1S} \pi_1K_2. $$

So this is the pushout $(\pi_1 k^1,\pi_1 k^2)$ and we have a map from it $$c:\pi_1K \twoheadrightarrow \pi_1M = \pi_1 M_1 *_{l_*\pi_1S} \pi_1M_2.$$ But we have more structure to that pushout map, namely that it arises not only from a map $\pi_1 K_i \to \pi_1M$ but that those maps each factor through $\pi_1M_i$. Hence our maps $\pi_1 K_i \to \pi_1M_i$ generate at least the part $\pi_1M_i - l^i_*\pi_1S$. That is because $c_*\pi_1K = c_*\pi_1K_1 \cup c_*\pi_1K_2$ and these two images (or rather the two factors) are always disjoint except for the image of the intersection. But as we have the surface in there from the beginning (algebraically you could also say that we are amalgamating over a larger subgroup already which is preserved, i.e. onto) that is no problem and we see that we get surjections $$\pi_1K_i \twoheadrightarrow\pi_1M_i.$$

So we showed: picking a modified core for $M_i$ gives us a finitely generated fundamental group.

* You might want to check that this proof won't work with $C$ and $C_i$ and that this is not even true in general.

** So by choosing our $K$ we made sure, that every component of $C-(C\cap S)$ touches $\cap S$, so we have that $S$ cuts this component containing $S$ into two components. So you could only look at the connected part and do the other components seperately eventually. This is why I ignore this issue. And another argument might be to just choose $C$ connected, which makes this obsolete as well.

Answer 3

Let me make this a new answer, as the old one (particularly the comment section) got really messy. I have been on vacation for a really long time --- excuse my late response.

I am still convinced that the answer should be yes for proper embeddedness and that all the wild excuses kind of miss the point, as they construct wild embeddings. Assume the surface $S$ has a normal bundle $V$ diffeomorphic to $S\times [-1,1]$ s.t. the boundary is also properly embedded. Also assume the compact core $C$ touches $S\times [-1,0]$ and $S\times [0,1]$ (if not you can add a compact $0$-codimensional manifold to $C$).

Then, on fundamental group level, $K=C\cup V$ generates $M$ and $K_i=K\cap M_i$ generate $M_i$. You get this by writing the fundamental groups of $K=K_1\cup K_2$ and $M=M_1\cup M_2$ as pushouts and closely looking at the images $\pi_iK_i\to \pi_1M_i (\to \pi_1M)$.

You can get more details on this by looking at my other answer.