2. Degree-one maps which induce surface-group injective homomorphisms

In this section, we will prove results about degree-one maps which induce surface-group injective homomorphisms on $\pi_1$ .

A group $\Gamma$ is a surface group if it is isomorphic to the fundamental group of a closed orientable surface. Let $\varphi\,{:}\, G\to H$ be a group homomorphism. We say $\varphi$ is surface-group injective, if the restriction of $\varphi$ to every surface subgroup of G is injective.

Proof. Let $\Gamma$ be a surface subgroup of $G_1*G_2$ . By the Kurosh Subgroup Theorem [ Reference Hempel8 , theorem 8·3], $\Gamma$ is the free product of a free group and conjugates of subgroups of $G_i$ , $i=1,2$ . Since $\Gamma$ is not a nontrivial free product, it is conjugated to a subgroup of $G_i$ for some i. Since $\varphi|_{G_i}$ is surface-group injective, $\varphi$ is also injective on $\Gamma$ .

The importance of the concept of surface-group injective maps is illustrated by the next lemma.

Proof. Let $\iota\,{:}\, R\to X$ be the inclusion map. Since $f(R)=S$ is a sphere,

\begin{align*}\iota_*(\pi_1(R))\subset\ker f_*.\end{align*}

If R is not a sphere, since $f_*$ is surface-group injective, R must be compressible. Let R $^{\prime}$ be the surface obtained by compressing R, then R can be obtained from R $^{\prime}$ by adding a tube. Let $R_1^{\prime}$ be a component of R $^{\prime}$ , let $\iota'\,{:}\, R_1^{\prime}\to X$ be the inclusion map, then

\begin{align*}\iota^{\prime}_*(\pi_1(R_1^{\prime}))=\iota_*(\pi_1(R_1^{\prime}\cap R))\subset\iota_*(\pi_1(R))\subset\ker f_*.\end{align*}

If R $^{\prime}$ does not consist of spheres, since $f_*$ is surface-group injective, R $^{\prime}$ must be compressible. So R $^{\prime}$ can be obtained from another surface R $^{\prime\prime}$ by adding a tube.

Continue with the above process, we conclude that R can be obtained from some spheres by adding tubes. We can rearrange the order of the tubes, so that some tubes connecting different spheres are added first to get a single sphere E, then R is obtained by adding other tubes to E.

Let $y_1,y_2\in Y$ be two points separated by S. Since f is surjective, both $f^{-1}(y_1)$ and $f^{-1}(y_2)$ are non-empty. These two sets are clearly separated by R, so R is separating. The process of compressing a surface does not change the homology class of the surface, hence E is also separating.

In the rest of this section, let Y be a 3–manifold which has no $S^1\times S^2$ summand, $S_1,S_2,\dots,S_n$ be a collection of disjoint 2–spheres in Y satisfying the following conditions: $Y\setminus (\cup_{i=1}^n S_i)$ has $n+1$ components whose closures are $\check Y_1,\check Y_2,\dots,\check Y_n,\check Y_{n+1}$ , where $\check Y_{n+1}$ is $S^3$ with n open balls removed, and a closed irreducible manifold $Y_i\ne S^3$ can be obtained from $\check Y_i$ by capping off $\partial\check Y_i=S_i$ with a ball $B_i$ , $1\le i\le n$ . Then

\begin{align*}Y=\#_{i=1}^nY_i.\end{align*}

When Y is irreducible, it is understood that $n=0$ .

Proof. We induct on n. When $n=0$ , there is nothing to prove. So we assume $n>0$ and the result is proved for $n-1$ .

Using [ Reference Rong and Wang17 , theorem 1·1], we may assume $R_1=f^{-1}(S_1)$ is a connected surface. Let $\check Y_0=\overline{Y\setminus\check Y_1}$ , and let $Y_0$ be obtained by capping off $\partial \check Y_0$ with a ball $B_0$ . Let $U_i=f^{-1}(\check Y_i)$ , $i=0,1$ . Then $\partial U_1=\partial U_0=R_1$ .

By Lemma 2·2, there exists a separating 2–sphere $E_1\subset X$ , so that $R_1$ is obtained by adding tubes to $E_1$ . Now $E_1$ splits X into two parts $\check X_1,\check X_0$ , so that $\check X_i$ can be obtained from $U_i$ by adding 1–handles and digging tunnels, $i=0,1$ . Let $X_i$ be the closed manifold obtained by capping off $\partial\check X_i$ with a ball, $i=0,1$ .

We claim that each map $f|_{U_i}\,{:}\,U_i\to \check Y_i$ can be extended to a degree-one map $f_i\,{:}\, X_i\to Y_i$ . In fact, the manifold $X_i$ can be obtained from $U_i$ by gluing a 3–manifold $V_i$ which is obtained from $B^3$ by digging tunnels and adding 1–handles. Since $B_i$ is a ball which is contractible, we can extend $f|_{U_i}\,{:}\,U_i\to \check Y_i$ to a map $f_i\,{:}\, X_i\to Y_i$ by sending $V_i$ to $B_i$ . The degree of $f_i$ is 1 since the degree of $f|_{U_i}$ is 1.

Since $\deg f_i=1$ , after a homotopy supported in $V_i$ , we may assume there exists a ball $B^{\prime}_i\subset\mathrm{int}(B_i)$ , such that $f_i$ sends $B^{\star}_i=f_i^{-1}(B^{\prime}_i)$ homeomorphically onto $B_i^{\prime}$ . Now we can glue $X_i\setminus\mathrm{int}(B_i^{\star})$ , $i=0,1$ , together along their boundary, to get back

\begin{align*}X=(X_0\setminus\mathrm{int}(B_0^{\star}))\cup_{S^2} (X_1\setminus\mathrm{int}(B_1^{\star})),\end{align*}

and define a map

\begin{align*}f_0\#f_1\,{:}\,X\to Y=(Y_0\setminus\mathrm{int}(B_0^{\prime}))\cup_{S^2} (Y_1\setminus\mathrm{int}(B_1^{\prime}))\end{align*}

by gluing the restrictions of $f_0,f_1$ . We rename

\begin{align*}S_1=\partial (Y_0\setminus\mathrm{int}(B_0^{\prime})), E_1=\partial (X_0\setminus\mathrm{int}(B_0^{\star})),\end{align*}

then $E_1=(f_0\#f_1)^{-1}(S_1)$ .

We claim that $(f_0\#f_1)_*=f_*$ . Let $D\subset E_1$ be a disk such that all tubes in $S_1$ are added to the interior of D, and let $D^c=\overline{E_1\setminus D}$ . Let $V\subset X$ be the handlebody obtained by adding the 1–handles bounded by the tubes to $\nu(D)$ . Since $\pi_1(V)$ is a quotient of $\pi_1(\partial V)$ , the map $X\setminus V\to X$ induces a surjective map on $\pi_1$ . To prove $(f_0\#f_1)_*=f_*$ , we only need to prove that $(f_0\#f_1)_*(\alpha)=f_*(\alpha)$ when $\alpha$ is a homotopy class represented by a loop in $X\setminus V$ . We observe that $f_0\#f_1=f$ on $X\setminus(V\cup\nu(D^c))$ , and $\pi_1(X\setminus V)$ is the free product of the $\pi_1$ of the two components of $X\setminus(V\cup\nu(D^c))$ , so $(f_0\#f_1)_*(\alpha)=f_*(\alpha)$ .

By the induction hypothesis, there exists a map $g_0\,{:}\,X_0\to Y_0$ , such that $g_1^{-1}(S_i)$ is a sphere for $2\le i\le n$ , and $(g_1)_*=(f_0)_*$ . We can define a map $g=g_0\#f_1$ in a similar way as $f_0\#f_1$ , then $g^{-1}(S_i)$ is a sphere for $1\le i\le n$ , and

\begin{align*}g_*=(g_0\#f_1)_*=(f_0\#f_1)_*=f_*.\end{align*}

This finishes the induction step.

3. Zero surgery on a null-homotopic knot

The aim of this section is to prove the following proposition.

We first prove a general result about surgery on null-homotopic knots.

Proof. The cobordism V deformation retracts to the space V $^{\prime}$ obtained from Y by adding a 2–cell $e^2$ along K. Since K is null-homotopic, the identity map on Y can be extended over $e^2$ . Hence we have a retraction $V'\to Y$ , which implies the existence of the retraction $p\,{:}\,V\to Y$ . The degree of the restriction $p|_{Y_m(K)}\,{:}\,Y_m(K)\to Y$ is 1 since it induces an isomorphism on $H_3$ .

The existence of the above degree-one map is a well-known result. See [ Reference Boileau and Wang3 , Proposition 3·2] and [ Reference Gadgil7 ].

In the rest of this section, let V be the 2–handle cobordism from Y to $X=Y_0(K)$ , $p\,{:}\,V\to Y$ be the retraction in Lemma 3·2, and $f=p|_{Y_0(K)}$ . Moreover, we assume $Y_0(K)=Z\#(S^1\times S^2)$ . Let $\check Z$ be the submanifold of $Y_0(K)$ which is Z with a ball removed. Since $Y_0(K)=Z\#(S^1\times S^2)$ , we can add a 3–handle to V to get a cobordism $W\,{:}\,Y\to Z$ .

Proof. Consider the following commutative diagram, where all maps except p are inclusions:

Then $f=p\circ\iota_0$ .

Since the 2–handle in V is added along the null-homotopic knot K, the inclusion $\iota_Y\,{:}\,Y\to V$ induces an isomorphism on $\pi_1$ . Since $p\circ\iota_Y=\mathrm{id}_Y$ , $p_*$ is an isomorphism.

Since W is obtained from V by adding a 3–handle, the inclusion $V\subset W$ induces an isomorphism on $\pi_1$ . We have the commutative diagram

The manifold W (after being turned up-side-down) can be obtained from $Z\times I$ by adding a 1–handle and a 2–handle, and the 2–handle cobordism is exactly V being turned up-side-down. By [ Reference Daemi, Lidman, Vela–Vick and Wong4 , Proposition 2·1], $(\iota_Z)_*$ is injective, so $(\iota_{\check Z})_*$ is also injective.

Now consider the commutative diagram

The restriction of $f_*$ to $\pi_1(\check Z)$ is just $p_*\circ(\iota_{\check Z})_*$ , which is injective since $(\iota_{\check Z})_*$ is injective and $p_*$ is an isomorphism.

Proof. This follows from Lemmas 2·1 and 3·3.

Proof of Proposition 3·1. We will use the notations in Section 2. Since $f_*$ is surface-group injective, we can apply Proposition 2·3 to get a degree-one map $g\,{:}\,X=Y_0(K)\to Y$ so that $g_*=f_*$ and $g^{-1}(S_i)=E_i$ is a separating sphere whenever $1\le i\le n$ .

Since X has an $S^1\times S^2$ summand, by the uniqueness part of the Kneser–Milnor theorem, one component of $X\setminus(\cup_{i=1}^n E_i)$ has an $S^1\times S^2$ summand.

If the $S^1\times S^2$ summand is in $g^{-1}(\check Y_{n+1})$ , then $g_*(S^1\times\{point\})$ is null-homotopic. It follows that

\begin{align*}g_*(\pi_1(X)=g_*(\pi_1(Z\#(S^1\times S^2)))=g_*(\pi_1(Z)).\end{align*}

Since $\deg g=1$ , $g_*$ is surjective. So $g_*|_{\pi_1(Z)}$ is surjective. Our result follows from Lemma 3·3 since $g_*=f_*$ .

If $\check X_i=g^{-1}(\check Y_i)$ has an $S^1\times S^2$ summand for some i satisfying $1\le i\le n$ , without loss of generality, we may assume $i=1$ . The map $g|_{\check X_1}$ extends to a map $g_1\,{:}\,X_1\to Y_1$ . Suppose that $X_1=Z_1\#(S^1\times S^2)$ , let $P\subset X_1$ be $\{\mathrm{point}\}\times S^2$ . Then $X_{1}\setminus \nu^{\circ}(P)$ is homeomorphic to $Z_1$ with two open balls removed. Since $\pi_2(Y_1)=0$ , $(g_1)|_P$ is null-homotopic in $Y_1$ . We can then extend $g_1|_{X_1\setminus \nu^{\circ}(P)}$ to a map $h_1\,{:}\, Z_1\to Y_1$ . The new map $h_1$ is again a degree-one map, so $(h_1)_*$ is surjective.

Using Lemma 3·3, we see that $g_*$ is injective on $\pi_1(X_0\#Z_1)$ . (Recall that $X_0$ is obtained from X by replacing $\check X_1$ with a ball.) In particular, $(h_1)_*$ is injective, so

\begin{align*}\pi_1(Z_1)\cong\pi_1(Y_1).\end{align*}

We also get that $g_*$ is injective on $\pi_1(X_0)$ . Since $g|_{\check X_0}\,{:}\,\check X_0\to\check Y_0$ is a degree-one proper map, $(g|_{\check X_0})*$ is surjective. So

\begin{align*}\pi_1(X_0)\cong \pi_1(Y_0).\end{align*}

Since $Z\cong X_0\#Z_1$ , $Y=Y_0\#Y_1$ , we have

\begin{align*}\pi_1(Z)\cong\pi_1(X_0)*\pi_1(Z_1)\cong\pi_1(Y_0)*\pi_1(Y_1)\cong\pi_1(Y).\end{align*}

4. Proof of the main theorem

In this section, we will prove Theorem 1·1 and Corollary 1·2.

Proof of Theorem 1·1. when $b_1(Y)>0$ Without loss of generality, we may assume $M=Y\setminus\nu^{\circ}(K)$ is irreducible. Since $b_1(Y)>0$ , there exists a closed, oriented, connected surface S in the interior of M, so that S is taut in M. Notice that for the $\infty$ slope on K, the core of the surgery solid torus, which is K, is null-homotopic. Using [ Reference Lackenby11 , theorem A.21], which is a stronger version of the main result in [ Reference Gabai5 ], we conclude that each 2–sphere in $Y_0(K)$ bounds a rational homology ball. Hence $Y_0(K)$ does not have an $S^1\times S^2$ summand.

Proposition 4·1. Let $Y_1,Y_2$ be two closed, oriented, connected 3–manifolds. If $\pi_1(Y_1)\cong\pi_2(Y_2)$ , then

\begin{align*}\mathrm{rank}\widehat{HF}(Y_1)=\mathrm{rank}\widehat{HF}(Y_2).\end{align*}

Proof. This is a well-known consequence of the Geometrisation Theorem. As in [ Reference Aschenbrenner, Friedl and Wilton2 , theorem 2·1·3], if $\pi_1(Y_1)\cong\pi_2(Y_2)$ , then there is a one-to-one correspondence between the summands of $Y_1$ and the summands of $Y_2$ , such that any pair of summands in the correspondence consists of either homeomorphic manifolds, (the homemorphism does not necessarily preserve the orientation), or lens spaces with the same $H_1$ . Our result then follows from basic properties of Heegaard Floer homology [ Reference Ozsváth and Szabó15, Reference Ozsváth and Szabó16 ].

Proof of Theorem 1·1. when $b_1(Y)=0$ Without loss of generality, we may assume $Y\setminus K$ is irreducible. Assume that $Y_0(K)=Z\#(S^1\times S^2)$ . By Proposition 3·1, $\pi_1(Y)\cong\pi_1(Z)$ . Using Proposition 4·1, we get $\mathrm{rank}\widehat{HF}(Y)=\mathrm{rank}\widehat{HF}(Z)$ . Our conclusion follows from [ Reference Hom and Lidman9 , theorem 1·1].

Proof of Corollary 1·2. The first statement follows from Theorem 1·1. We only need to prove the second statement. In this case, $Y_0(K)$ is a torus bundle over $S^1$ . By Lemma 3·2, there exists a degree-one map $f\,{:}\,Y_0(K)\to Y$ . Then $f_*$ is surjective. We claim that Y is $S^1\times S^2,\mathbb RP^3\#\mathbb RP^3$ or a spherical manifold. Then our conclusion follows from [ Reference Ai and Ni1 ].

If Y is reducible, then either $Y=S^1\times S^2$ or Y is a nontrivial connected sum. If $Y=S^1\times S^2$ , we are done. Now we assume Y is a nontrivial connected sum, so $\pi_1(Y)\cong A*B$ with A, B nontrivial. Let T be a fiber of $Y_0(K)$ , then $f_*(\pi_1(T))$ is an abelian normal subgroup of $\pi_1(Y)=f_*(\pi_1(Y_0(K)))$ . By the Kurosh Subgroup Theorem [ Reference Hempel8 , theorem 8·3], $f_*(\pi_1(T))$ is also a free product of free groups and conjugates of subgroups of A, B. Since $f_*(\pi_1(T))$ is abelian, it must be either a subgroup of $\mathbb Z$ or the conjugate of a subgroup of A or B. Since $f_*(\pi_1(T))$ is a normal subgroup of $\pi_1(Y)\cong A*B$ , the latter case cannot happen.

Now $f_*(\pi_1(T))$ is a subgroup of $\mathbb Z$ . Since $f_*$ is surjective, $\pi_1(Y)/f_*(\pi_1(T))$ is a cyclic group. If $f_*(\pi_1(T))=\{1\}$ , then $\pi_1(Y)$ is cyclic, a contradiction to $\pi_1(Y)\cong A*B$ . So $f_*(\pi_1(T))\cong\mathbb Z$ .

If $\pi_1(Y)/f_*(\pi_1(T))\cong\mathbb Z$ , then $\pi_1(Y)$ contains $\mathbb Z^2$ as a finite index subgroup, which is not possible since $\mathbb Z^2$ is not the fundamental group of any closed 3–manifold. If $\pi_1(Y)/f_*(\pi_1(T))$ is finite, then $\pi_1(Y)$ contains $\mathbb Z$ as a finite index subgroup. It follows that Y is finitely covered by $S^1\times S^2$ , thus it must be $S^1\times S^2$ or $\mathbb RP^3\#\mathbb RP^3$ .

Now we consider the case Y is irreducible. If Y is a spherical manifold, we are done. If Y is irreducible and not a spherical manifold, then $f_*$ is injective by [ Reference Wang18 , theorem 4]. So $f_*$ is an isomorphism, a contradiction to the fact that $b_1(Y_0(K))>b_1(Y)$ .