End of 3-manifold

Question

Let $M$ be an irreducible, orientable, open 3-manifold with finitely generated fundamental group, this gives us a Scott's Core $C\hookrightarrow M$ so that the inclusion is an homotopy equivalence and such that the ends of $M$ are in 1-1 correspondence with the components of $M\setminus int(C)$.

Let $\mathcal E$ be an end and $\partial_{\mathcal E} C$ be the boundary component facing it. I want to show that the inclusion $\partial_{\mathcal E} C\hookrightarrow Z$, for $Z$ the component of $M\setminus int(C)$ containing $\mathcal E$, is an isomorphism on $H_1$.

I can get surjectivity but I fail to get injectivity.

Answer 1

$H_i(M,C)=0$ by assumption that the inclusion of $C$ is a homotopy equivalence. Now use excision on the interior of $C$ to see that $H_i(M \ \text{int }C,\partial C)=0$. But $(M \setminus \text{int } C,C)$ is the disjoint union of the spaces you're interested, and (relative) homology is additive under disjoint union, hence all of your groups are zero.

(If one is pedantic they say "You need the set you're excising to have closure contained in $C$! But just take a tubular neighborhood of the boundary etc.)

Answer 2

This is an answer to the follow-up question you asked in the comments to Mike's answer.

Theorem. Suppose that $C\subset M$ is a Scott compact core of an open irreducible 3-manifold $M$. Then for each component $E\subset M- int(C)$ (informally known as an "end" of $M$), for the (necessarily connected) boundary surface $S=\partial E$, the inclusion $i: S\to E$ induces an isomorphism $i_*$ of fundamental groups.

Proof.

1. First, let us check the harder part, the injectivity of $i_*$. (It is harder since it uses some nontrivial results of 3-dimensional topology.) I will work in the PL category. Suppose that $i_*$ is not injective. Then, by the Loop Theorem, there exists a simple loop $L\subset S$ bounding a properly embedded disk $D_1\subset E$ such that $L$ does not bound a disk in $S$. However, the inclusion $C\to M$ is $\pi_1$-injective. Therefore, the loop $L$ as above, is null-homotopic in $C$. By the Dehn Lemma, since $L$ is simple, it bounds a properly embedded disk $D_2\subset C$. (This is a historic accident that one of the results I use is called a theorem and the other is called a lemma, both are hard, with the first correct proof due to Papakyriakopoulos.) Since $M$ is irreducible, the 2-sphere $D_1\cup D_2$ bounds a 3-ball $B\subset M$. Let $S':= S\cap B$. This subsurface of $S$ has exactly one boundary component, namely, $L$. I claim that $S'$ is is the 2-disk, which will contradict the fact that $L$ does not bound a disk in $S$. Suppose not. Then the interior of $S'$ contains two simple loops $a, b$ which intersect transversally in exactly one point. Therefore, one of these loops, say, $a$, is homologically nontrivial in $C$. But $a$ is contained in the 3-ball $B\subset M$ and, hence, is homologically trivial in $M$. (Let me know if you want to see a proof.) Therefore, the inclusion map $C\to M$ is not a homotopy-equivalence. A contradiction. Thus, $i_*$ is injective.

2. I will now prove the surjectivity of $i_*$. Let $C'$ denote $(M - E) \cup S$. Since the inclusion $C\to M$ is $\pi_1$-surjective, so is the inclusion $C'\to M$. By the Seifert - van Kampen theorem, the fundamental group of $M$ is the pushout of the diagram $$ \pi_1(C')\leftarrow \pi_1(S)\rightarrow \pi_1(E). $$ If $i_*$ is not surjective, the homomorphism $\pi_1(C')\to \pi_1(M)$ cannot be surjective either: This is a consequence of the group-theoretic fact that if you have a group amalgam $$G=G_1 \star_{G_3} G_2$$ and $G_3\ne G_2$ then $G_1\ne G$. A contradiction. qed