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Hempel, in his 1987 article "Residual Finiteness for 3-Manifolds", shows that if $M$ is a compact Haken 3-manifold, then $\pi_1(M)$ is residually finite. In the proof he starts by reducing the case to '$M$ is closed and irreducible'. Why can he so easily do that?

Thanks in advance!

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    $\begingroup$ The fact that the free prodct of residually finite groups is residually finite was first prved by Gruenberg.. $\endgroup$
    – user6976
    Jan 2, 2020 at 23:03
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    $\begingroup$ K. W. Gruenberg, Residual properties of infinite soluble groups, Proc. London Math. Soc. (3) 7 (1957), 29-62. MR 19, 386. $\endgroup$
    – user6976
    Jan 3, 2020 at 1:30

1 Answer 1

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These are a routine, but very valuable, pair of exercises in the theory of three-manifolds. So if you are trying to learn the material, don't read the following proof sketches until you desperately need some hints!








We assume that $M$ is compact and connected.

Suppose that $M$ is reducible. Then there is a sphere $S$ in $M$ that does not bound a ball on either "side". We deduce that $M$ is either $S^2 \times S^1$ (or perhaps the twisted version) or $M$ is a non-trivial connect sum. Since $\pi_1(S^2 \times S^1) \cong \mathbb{Z}$ the former case can be ignored. In the latter case $M$ is a connect sum $A \, \# \, B$. Thus $\pi_1(M) \cong \pi_1(A) \ast \pi_1(B)$. So the residual finiteness of $M$ reduces to that of $A$ and $B$.

Suppose that $M$ has boundary. Define $D(M)$ to be the double of $M$ across its boundary: that is, take two copies of $M$ and glue them using the "identity" on their boundaries. Thus $\pi_1(M)$ injects into $\pi_1(D(M))$. So the residual finiteness of $M$ reduces to that of $D(M)$.

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