[proofplan] Surgery in dimension three cuts only along embedded two-spheres and caps by three-balls, so it changes the topology only by separating connected-sum factors and discarding components. We prove that every component appearing after any finite sequence of surgeries has fundamental group isomorphic to a free factor of $\pi_1(M)$. Since $\pi_1(M)=1$, every surviving component is simply connected. A closed two-sided incompressible torus or closed two-sided incompressible higher-genus surface would inject a surface group different from the identity group into the fundamental group of its ambient component, which is impossible. [/proofplan] [step:Track the fundamental group through one spherical surgery] Let $N$ be a connected closed orientable three-manifold, and suppose a surgery cuts $N$ along a finite disjoint family of embedded two-spheres $S_1,\dots,S_k \subset N$ and caps every resulting spherical boundary component with a three-ball. Let $N_1,\dots,N_m$ denote the connected closed components obtained after this operation. For any path-[connected space](/page/Connected%20Space) $X$ with a chosen basepoint, let $\pi_1(X)$ denote its fundamental group; in the free-factor statements below, changing basepoints inside a connected component only conjugates the identified subgroup and does not change whether it is a free factor. We claim that each group $\pi_1(N_j)$ is isomorphic to a free factor of $\pi_1(N)$. It is enough to treat one embedded two-sphere $S \subset N$, because the finite case follows by induction over $S_1,\dots,S_k$. The group-theoretic point used in the induction is that a free factor of a free factor is a free factor: if $G \cong H * K$ and $H \cong L * Q$, then associativity of free products gives $G \cong L * Q * K$, so $L$ is a free factor of $G$. For each use of van Kampen below, replace the closed pieces by open collar neighbourhoods of those pieces inside the glued manifold. The collars have the same homotopy type as the corresponding pieces, their intersections deformation retract onto the common sphere or disk attachments, and the hypotheses of the [Seifert-van Kampen Theorem](/theorems/1905) are then satisfied. If $S$ is separating, write $N \setminus S = A^\circ \sqcup B^\circ$, where $A$ and $B$ are the compact manifolds with boundary $\partial A = S = \partial B$. Let $D_A$ and $D_B$ be three-balls, and define the closed manifolds $A^+$ and $B^+$ by \begin{align*} A^+ &:= A \cup_S D_A. \end{align*} Also set \begin{align*} B^+ &:= B \cup_S D_B. \end{align*} Since $S$ and each three-ball are path-connected and simply connected, the inclusion $A \hookrightarrow A^+$ induces an isomorphism $\pi_1(A) \cong \pi_1(A^+)$, and the inclusion $B \hookrightarrow B^+$ induces an isomorphism $\pi_1(B) \cong \pi_1(B^+)$. Applying the [Seifert-van Kampen Theorem](/theorems/1905) to the decomposition $N=A\cup_S B$, whose intersection $S$ is path-connected and has fundamental group equal to the identity group, gives \begin{align*} \pi_1(N) \cong \pi_1(A^+) * \pi_1(B^+). \end{align*} Thus $\pi_1(A^+)$ and $\pi_1(B^+)$ are free factors of $\pi_1(N)$. If $S$ is nonseparating, cutting along $S$ gives one compact connected manifold $A$ with two spherical boundary components $S_-$ and $S_+$. Let $D_-$ and $D_+$ be three-balls glued to $S_-$ and $S_+$, and set \begin{align*} A^+ := A \cup_{S_-} D_- \cup_{S_+} D_+. \end{align*} The original manifold $N$ is recovered from $A$ by identifying the two boundary spheres $S_-$ and $S_+$ by the gluing diffeomorphism determined by the cut. Equivalently, after capping both boundary components, this reconstruction identifies $N$ with the connected sum \begin{align*} A^+ \# (S^2 \times S^1). \end{align*} Indeed, removing the interiors of the two capping balls from $A^+$ returns $A$, while the product region $S^2 \times [0,1]$ with its two end spheres identified supplies the missing neck and contributes the $S^2 \times S^1$ summand. Applying the [Seifert-van Kampen Theorem](/theorems/1905) to the connected-sum decomposition along a separating two-sphere, whose fundamental group is the identity group, gives \begin{align*} \pi_1(N) \cong \pi_1(A^+) * \pi_1(S^2 \times S^1) \cong \pi_1(A^+) * \mathbb Z. \end{align*} Therefore $\pi_1(A^+)$ is again a free factor of $\pi_1(N)$. [guided] We isolate the only topological operation used in three-dimensional surgery: cutting along two-spheres and filling the resulting spherical boundary components by three-balls. Let $N$ be a connected closed orientable three-manifold and let $S \subset N$ be one embedded two-sphere. We must understand how $\pi_1(N)$ compares with the fundamental groups of the capped pieces. Basepoints cause no ambiguity in the conclusion: we choose the basepoint in a connected collar of the common boundary sphere when applying van Kampen, and changing the basepoint inside a connected component only conjugates the subgroup, while being a free factor is unchanged by this identification. First suppose $S$ separates $N$. Then $N \setminus S$ has two components. Let $A$ and $B$ be the compact manifolds obtained by including the common boundary sphere $S$, so $\partial A = S = \partial B$. Let $D_A$ and $D_B$ be three-balls, and define closed manifolds $A^+$ and $B^+$ by \begin{align*} A^+ &:= A \cup_S D_A. \end{align*} Also set \begin{align*} B^+ &:= B \cup_S D_B. \end{align*} Why do the capped manifolds have the same fundamental groups as the uncapped pieces? The gluing set $S$ is path-connected and simply connected, and each three-ball $D_A$ and $D_B$ is path-connected and simply connected. Applying the [Seifert-van Kampen Theorem](/theorems/1905) to $A^+=A\cup_S D_A$ gives $\pi_1(A^+)\cong \pi_1(A)$, and applying it to $B^+=B\cup_S D_B$ gives $\pi_1(B^+)\cong \pi_1(B)$. Now apply the [Seifert-van Kampen Theorem](/theorems/1905) to the decomposition $N=A\cup_S B$. The hypotheses are satisfied because $A$ and $B$ are path-connected after choosing the components cut out by the separating sphere, and the intersection $S$ is path-connected. The amalgamating subgroup is $\pi_1(S)=1$, since $S$ is a two-sphere. Therefore the amalgamated free product reduces to an ordinary free product: \begin{align*} \pi_1(N) \cong \pi_1(A) * \pi_1(B) \cong \pi_1(A^+) * \pi_1(B^+). \end{align*} Thus each capped component has fundamental group a free factor of $\pi_1(N)$. Now suppose $S$ is nonseparating. Cutting $N$ along $S$ produces one connected compact manifold $A$ with two boundary spheres $S_-$ and $S_+$. Let $D_-$ and $D_+$ be three-balls glued to these two boundary spheres, and define \begin{align*} A^+ := A \cup_{S_-} D_- \cup_{S_+} D_+. \end{align*} The capping balls again do not change the fundamental group, by the same van Kampen argument using the simply connected gluing spheres and simply connected three-balls. How is $N$ reconstructed from $A^+$? Cutting $N$ along the nonseparating sphere produces $A$ with two boundary spheres $S_-$ and $S_+$; recovering $N$ means identifying $S_-$ with $S_+$ by the gluing diffeomorphism determined by the original cut. Since $A^+$ was obtained from $A$ by filling those boundary spheres with the balls $D_-$ and $D_+$, the same reconstruction can be described after capping as a connected sum \begin{align*} N \cong A^+ \# (S^2 \times S^1). \end{align*} The reason is that the region which records the identification of the two boundary spheres is a product $S^2\times[0,1]$ with its two end spheres identified, namely $S^2\times S^1$; removing one ball from this product and one ball from $A^+$ and gluing along the resulting boundary sphere gives exactly the reglued manifold. Now apply the [Seifert-van Kampen Theorem](/theorems/1905) to the connected-sum decomposition. The separating sphere used in the connected sum is path-connected and has fundamental group equal to the identity group, so the amalgamated free product over that subgroup is an ordinary free product: \begin{align*} \pi_1(N) \cong \pi_1(A^+) * \pi_1(S^2 \times S^1) \cong \pi_1(A^+) * \mathbb Z. \end{align*} Thus $\pi_1(A^+)$ is a free factor of $\pi_1(N)$ in the nonseparating case as well. Repeating this argument for finitely many disjoint surgery spheres proves the same assertion for an entire surgery time. [/guided] [/step] [step:Induct over the surgery history to show every surviving component is simply connected] Let $I \subset [0,\infty)$ denote the time interval on which the Ricci flow with surgery is defined, let $t \in I$ be a finite time, and let $C$ be a connected component of $M_t$. By the local-finiteness clause in the definition of Ricci flow with surgery, the set of surgery times in every compact time interval is finite; applying this to $[0,t]$, there are only finitely many surgery times before time $t$. Hence $C$ is obtained from $M$ by a finite sequence of the following operations: evolve by diffeomorphism between surgery times, cut along embedded two-spheres, cap by three-balls, and discard some connected components. Evolution by the Ricci flow does not change the diffeomorphism type of a time-slice component between surgeries. By the previous step, at each surgery time the fundamental group of each newly produced component is a free factor of the fundamental group of the component immediately before surgery. Therefore, by induction over the surgery times leading to $C$, the group $\pi_1(C)$ is isomorphic to a free factor of $\pi_1(M)$. Since $\pi_1(M)=1$, the only free factor of $\pi_1(M)$ is the identity group. Hence \begin{align*} \pi_1(C)=1. \end{align*} [/step] [step:Exclude incompressible tori and higher-genus surfaces] Assume, for contradiction, that some component $C$ of some finite time slice $M_t$ contains a closed two-sided incompressible surface \begin{align*} \Sigma \subset C \end{align*} of genus $g \ge 1$. Let \begin{align*} i:\Sigma \to C \end{align*} be the inclusion map. By the definition of an [incompressible surface](/page/Incompressible%20Surface), the [induced homomorphism on fundamental groups](/theorems/1879) \begin{align*} i_*:\pi_1(\Sigma) \to \pi_1(C) \end{align*} is injective. If $g=1$, then $\Sigma$ is a torus and \begin{align*} \pi_1(\Sigma) \cong \mathbb Z^2. \end{align*} If $g \ge 2$, then $\pi_1(\Sigma)$ is the closed orientable surface group of genus $g$, hence is different from the identity group. In both cases $\pi_1(\Sigma)$ is different from the identity group. Since $i_*$ is injective, $\pi_1(C)$ contains a subgroup different from the identity group. This contradicts the conclusion of the previous step that $\pi_1(C)=1$. Therefore no component of any finite time slice contains a closed two-sided incompressible surface of genus at least one. In particular, no connected component produced by the Ricci flow with surgery contains a closed two-sided incompressible torus or a closed two-sided incompressible surface of genus at least two. [/step]