I'm trying to justify the following claim I've encountered in a paper but am having no luck. Here's the setup.

Let $F$ be a connected, essential surface (possibly with boundary) in a compact, connected, orientable, irreducible 3-manifold $M$ (possibly with boundary), and suppose that there is a component $C\subset M-F$ such that $\pi_1(C)\to \pi_1(M)$ is an isomorphism.

Claim: $F$ separates $M$. In particular, if $F$ is non-separating, then $\pi_1(M-F) \to \pi_1(M)$ is not surjective. In fact, even $H_1(M-F) \to H_1(M)$ is not surjective.

Any insight would be greatly appreciated!

Note: By essential surface, I mean a bicollared surface $F\subset M$ whose components $F_i$ all satisfy:

  • $F_i \not\cong S^2$
  • $F_i$ is not boundary-parallel
  • $\pi_1(F_i) \to \pi_1(M)$ is injective.

The solution is just a Mayer-Vietoris argument! We can cut away most of the assumptions and prove the following:

Let $M$ be a connected manifold and $F\subset M$ a connected, bicollared hypersurface. If $F$ is non-separating, then $H_1(M-F) \to H_1(M)$ is not surjective.

Let the open sets for Mayer-Vietoris be $M-F$ and the bicollar neighborhood of $F$. The intersection deformation retracts onto the ends of the bicollar, which we identify with two copies of $F$. Mayer-Vietoris gives $$ H_1(F)\oplus H_1(M-F) \xrightarrow{f} H_1(M) \xrightarrow{\partial} H_0(F)\oplus H_0(F) \xrightarrow{g} H_0(F)\oplus H_0(M-F). $$

Since $F$ is non-separating, we have $H_0(M-F) =\mathbb Z$. Thus, writing $$g: \mathbb Z^2 \to \mathbb Z^2 \\ (1,0) \mapsto (1,1) \\ (0,1) \mapsto (1,1)$$ we see $\ker g = \operatorname{im}\partial \neq 0$. On the other hand, if $H_1(M-F) \to H_1(M)$ is surjective, so is $f$. But then $\ker \partial = \operatorname{im}f = H_1(M) \implies \operatorname{im}\partial = 0$, a contradiction.


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