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Let $M$ be a 3-manifold with boundary $\partial M$. Suppose $\partial M$ contains a sphere or a projective plane, which is contractable in $M$. Show that $M$ is also contractable.

The above statement is shown in the proof of Sphere Theorem, so it should not be used to show the above argument.

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As Stallings writes in his book Group theory and three-dimensional manifolds, where this proof of the Sphere Theorem originates, "It is easy to see that, if $\partial M$ contains a $2$-sphere which is contractible in $M$, then $M$ is itself contractible." I gave a talk about his proof recently, and this was the easiest way I could see it:

(Let $M$ denote a compact connected $3$-manifold in what follows.)

Claim. If there is a nullhomotopic $S^2\subset \partial M$, then $\pi_1(M)=1$.

Proof. If not, take copies $M_1,M_2$ of $M$ and join them along the $S^2$ boundary to form $M'$. By van Kampen, $\pi_1(M')=\pi_1(M_1)*\pi_1(M_2)$, which is infinite since we are assuming $\pi_1(M)\neq 1$, so the universal cover $\widetilde{M'}$ is non-compact. Lift $S^2$ to a sphere $\Sigma\subset \widetilde{M'}$, which separates $\widetilde{M'}$ since $S^2$ separates $M'$. Call the pieces $N_1,N_2$, which are both noncompact by construction. By excision with $\Sigma$ being thought of as the boundary of a $3$-ball, $H_3(N_i,\Sigma)=H_3(N_i)=0$, so the LES gives that $H_2(\Sigma)\to H_2(N_i)$ is injective. Mayer Vietoris gives $$0\to H_2(\Sigma)\to H_2(N_1)\oplus H_2(N_2)\to H_2(\widetilde{M'})$$ so the kernel of the map to $H_2(\widetilde{M'})$ is generated by $([\Sigma],[\Sigma])$, hence $([\Sigma],0)$ is not in the kernel, and so $H_2(\Sigma)\to H_2(\widetilde{M'})$ has non-trivial image. But $S^2$ being nullhomotopic in $M\subset M'$ implies $\Sigma$ is nullhomotopic in $\widetilde{M'}$ by lifting the nullhomotopy, implying $[\Sigma]$ is zero in $H_2(\widetilde{M'})$! Therefore, $M$ must have been simply connected.

Claim. If $M$ is simply connected, then $M$ is orientable.

Proof. If $M$ is non-orientable, then $H^1(M;\mathbb{Z}/2\mathbb{Z})$ is non-trivial since there is an oriented connected double-cover.

Claim. If $M$ is simply connected with a nullhomologous boundary component $\Sigma\subset \partial M$, then $H_2(M)=0$.

Proof. Let $[M]\in C_3(M)$ denote the fundamental class, and recall that $\partial[M]$ is a sum of fundamental classes of each boundary component. If $\Sigma$ is a boundary of some $3$-chain $A$, then $A$ is non-trivial in $H_3(M;\partial M)=\mathbb{Z}$, so $A$ and $[M]$ must be homologous, hence there is only a single boundary component. By Poincare duality, $H_2(M)=H^1(M,\Sigma)$, and the LES gives $\widetilde{H}^0(\Sigma)\to H^1(M,\Sigma)\to \widetilde{H}^1(M)$, and since $\Sigma$ is connected and $M$ is simply connected, we get $H_2(M)=0$.

Claim. If $M$ has a nullhomotopic $S^2\subset \partial M$, then $M$ is contractible.

Proof. We have that $M$ is simply connected with $H_2(M)=0$. By Hurewicz, $\pi_2(M)=0$. (Note that this is all we need for the Sphere Theorem, since $\pi_2(M)\neq 0$ is a hypothesis.) Since $M$ has boundary, $H_3(M)=0$. Again by Hurewicz, $\pi_n(M)=0$ for $n\geq 3$. Since $M$ has a CW structure, the map $*\to M$ is a homotopy equivalence by Whitehead's theorem, so $M$ is contractible.

Claim. If $M$ has an $\mathbb{R}\mathrm{P}^2\subset\partial M$ that as a map $S^2\to M$ is nullhomotopic, then $\pi_2(M)=0$.

Proof. Let $P$ denote this $\mathbb{R}\mathrm{P}^2$ in the boundary. If it were the case that $\pi_1(P)\to \pi_1(M)$ has trivial image, then the Loop Theorem gives an embedded disk $D$ in $M$ with $\partial D$ being nontrial in $\pi_1(P)$. The normal bundle of $D$ is trivial since $H^1(D;\mathbb{Z}/2\mathbb{Z})=0$, so $D$ orients the normal bundle for $\partial D$ in $P$, which is a contradiction. So, the oriented double cover of $M$ has $P$ lift to an $S^2$ in the boundary. By hypothesis, $P$ is nullhomologous in $M$, so this $S^2$ would be nullhomologous in the oriented double cover, and the preceding claims imply that the oriented double cover is contractible, hence $\pi_2(M)=0$.

Hence, if $\pi_2(M)\neq 0$:

  • If there is an $S^2$ in the boundary, it is not nullhomotopic.

  • If there is a projective plane in the boundary, its map $S^2\to M$ is not nullhomotopic.

That is, such boundary components satisfy the conclusion of the Sphere Theorem, when pushed into $M$ slightly so they become properly embedded $2$-sided surfaces.

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First I'll prove that given the condition that a $S^2$ boundary contracts inside $M$ implies $M$ is simply connected.

If $M$ has a sphere in the boundary, then glue two copies of $M$ together with respect to boundary sphere. If $M$ is not simply connected then $\pi_1(M_1\cup M_2)$ is a free group. Then consider two non-trivial loops $a_1\in M_1$ and $a_2\in M_2$. Let $\gamma$ be $a_1*a_2$. Since $S^2$ is contractible in $M$, that implies we can do some homotopy such that $\gamma$ doesnot intersect $S^2$. Which implies $\gamma$ either lies in $M_1$ or $M_2$. And this contracdicts the free property of $\pi_1(M_1\cup M_2)$. Thus $M$ is simply connected.

If $M$ is simply connected with $S^2$ boundary, then Hurewicz theorem (since all its homology groups are zero by Poincare Dulality) implies that it is contractible.

BTW, $M$ cannot have an non-orientable boundary componenet such as $\mathbb RP^2$. Then $M$ will not be contractible, since any simply connected manifold is orientable. And boundary of any orientable manifold is orientable.

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  • $\begingroup$ very nice reasoning, but it doesn't work in case $M$ has another boundary components but one $S^2$ (in this case $M$ can be simply connected but not contractible) $\endgroup$ – Andrey Ryabichev Nov 8 '17 at 19:51
  • $\begingroup$ How is it that $S^2$ being nullhomologous means there is a homotopy of $\gamma$ such that $\gamma$ does not intersect $S^2$? $\endgroup$ – Kyle Miller Feb 24 at 3:50
  • $\begingroup$ @KyleMiller nullhomotopic is stronger than nullhomologous $\endgroup$ – Anubhav Mukherjee Feb 24 at 5:49
  • $\begingroup$ @AnubhavMukherjee Sorry, I meant to type "nullhomotopic," or rather "contractible." I don't see how you can homotope $\gamma$ to avoid the $S^2$. Is there some result that the nullhomotopy of $S^2$ can be through isotopies? $\endgroup$ – Kyle Miller Feb 24 at 8:15
  • $\begingroup$ @KyleMiller S^2 is homotopic to a point. And we can assume that S^2 and the loop $\gamma$ intersect transversally but then so is the loop and the point. So we can assume they are infact disjoint. $\endgroup$ – Anubhav Mukherjee Feb 24 at 16:06

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