Existence of a surface group ensures the existence of a $\pi_1$-injective immersed surface

Question

The question is simple:

For a $3$-manifold $M$, if $\pi_1(M)$ contains a surface group $\Gamma$ (i.e. the fundamental group of some surface) then $M$ contains a $\pi_1$-injective immersed surface $S$.

Existence of $S$ obviously ensures the existence of $\Gamma$ but I don't know if the converse is true. Actually the statement I found is from here which says that

Showing that every hyperbolic $3$-manifold contains an immersed $\pi_1$-injective surface is equivalent to showing that the fundamental group of every hyperbolic $3$-maniofld contains a subgroup isomorphic to the fundamental group of a surface.

Is this true because the assumption $M$ here is hyperbolic? (because it's $K(G,1)$ space?)

Edit: I didn't add a condition that $\pi_1(S) = \Gamma$ by mistake. The actual question is the existence of a $\pi_1$-injective immersed surface with $\pi_1(S) = \Gamma$.

Answer 1

This fact doesn’t need $M$ to be hyperbolic. It just needs one general theorem about 3-manifold topology, namely the Scott core theorem.

Let $N\to M$ be the covering space corresponding to the subgroup $\Gamma$. Since $\pi_1(N)=\Gamma$ is finitely generated, the Scott core theorem guarantees a compact submanifold $N_0\subseteq N$ such that inclusion induces an isomorphism $\pi_1(N_0)\cong\pi_1(N)$; in particular, $\pi_1(N_0)$ is the surface group $\Gamma$.

It is a byproduct of the proof of the Scott core theorem that none of the boundary components of $N_0$ are spheres: the point is that any spherical boundary component bounds a ball in $N$, so can be filled in. Now, note that $N_0$ is not closed (since it is a compact submanifold of the non-compact manifold $N$), so $\partial N_0$ is non-empty. A component $S\subseteq \partial N_0$ is then embedded in $N$, hence immersed in $M$, as required.

Note that the argument does not build $S$ with $\Gamma=\pi_1(S)$. Dehn’s lemma does guarantee that $\pi_1(S)$ is a subgroup of $\Gamma$, and it must be a subgroup of finite index. With a little more work, one can argue that $N_0$ is an interval bundle, and hence find an immersed surface representing $\Gamma$ itself, but this wasn’t required for the question as stated.

Answer 2

Given a $CW$-complex $X$ and a closed surface group $\pi_1(\Sigma,v) < \pi_1(X,x)$, there exists a map $\phi: (\Sigma,v) \to (X,x)$ such that the image of the fundamental group is this subgroup. Take a 1-vertex triangulation of $\Sigma$ with vertex $v$, map the vertex $v$ to the basepoint $x\in X$, then each edge $e$ represents a closed loop and hence an element of $\pi_1(\Sigma)$. Thus we may map this element into $X$ uniquely up to homotopy rel basepoint. The boundary of each triangle of $\Sigma$ is homotopically trivial in $\Sigma$, so its image in $X$ is homotopically trivial, and hence can be filled in with a disk. More generally, this shows that for any group $G <\pi_1(X)$ and two-complex $C$ with $\pi_1(C)=G$, there is a map $C\to X$ inducing the map $G\to \pi_1(X)$ on the level of fundamental group.

If $X$ is a 3-manifold, then it is shown by Schoen and Yau that $\phi:\Sigma\to X$ is homotopic to a minimal area immersion in its homotopy class. However, this proof is overkill (the proof that an area minimizing surface is immersed is due to Osserman and Gulliver). The essential point is a technique of Whitney which classifies maps of $n$-manifolds into $2n-1$ Euclidean space, showing that they may be approximated by immersions. Bing and Papakyriokopoulos applied this technique to maps of disks and spheres into 3-manifolds, and this was generalized by Gabai and Oertel to maps of surfaces. The proof of the main theorem of Oertel explains this argument. (And also provides the figures below.)

Whitney classified singularities of maps, which in the case of maps of surfaces to 3-manifolds looks like:

The singular set includes arcs and curves of double points, triple points, and branch points.

To get rid of branched points, and hence create an immersion, one starts at a branched point and “zips” to decrease the length of the singular arc of double points as in Figure 7. One may push through triple points (decreasing the number of triple points) as in Figure 8 until another branch point is reached as in Figure 10 (this is the same argument employed by Papakyriokopoulos and Bing; it may go back to Dehn but I didn’t check).

There is another local picture (not shown) which looks like a crosscap; in this case the surface is not $\pi_1$-injective since the core of the Mobius strip is homotopically trivial. In the $\pi_1$-injective case (or more generally simple-loop injective), the immersed arc (shown in Figure 10 ) may be surgered away to give a disconnected surface with a sphere component, and the sphere component removed. After finitely many such operations, the branched points are removed and the map is an immersion.