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.


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.


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.

  • $\begingroup$ 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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