A closed three-manifold with H_1 = Z/2 and H_2 = 0 does not embed in S^4

Suppose $M\subset S^4$ is smoothly embedded. Its normal line is oriented by the orientations of $M$ and $S^4$, and has a global positive unit section: in an oriented local line frame the positive unit vector is independent of the frame, so these sections glue. Compactness and [F4] give a product tube $M\times (-ϵ,ϵ)$. By [F2], $H^3(M)=\mathbb{Z}$, since $H_2(M)=0$ and $H_3(M)=\mathbb{Z}$. Thus [F1] gives ${\tilde{H}}_0(S^4\setminus M)=\mathbb{Z}$. The complement is an open manifold and is locally path connected; $H_0$ is free on its path components, so it has exactly two components $U,V$. Every component has nonempty frontier in $M$, since otherwise it is both open and closed in connected $S^4$. Near any frontier point a hypersurface chart has exactly two connected local sides. Each of the two global halves of the product tube is connected because $M$ is connected. Every complementary component meets one of them, by the local side chart at a frontier point. Therefore the two tube halves lie in distinct components, one in $U$ and one in $V$. Their closures $A=\bar{U}$ and $B=\bar{V}$ are compact smooth manifolds with common boundary $M$, are locally contractible, and satisfy $S^4=A\cup B$ and $A\cap B=M$.

Enlarge $A$ and $B$ by a small portion of the opposite tube half to obtain an open cover of $S^4$. The two open sets retract onto $A,B$, and their intersection retracts onto $M$, by moving the collar coordinate linearly to zero on the added halves. Also $U\hookrightarrow A$ and $V\hookrightarrow B$ are homotopy equivalences: a collar map which moves coordinate $s\ge 0$ slightly into $s>0$, and equals $s$ outside a smaller collar, is homotopic to the identity by linear interpolation and supplies homotopy inverses for the interior inclusions. Applying [F3] to the open cover, the segments $H_2(M)\to H_2(A)\oplus H_2(B)\to H_2(S^4)$ and $H_2(S^4)\to H_1(M)\to H_1(A)\oplus H_1(B)\to H_1(S^4)$ show $H_2(A)=H_2(B)=0$ and $H_1(A)\oplus H_1(B)\cong \mathbb{Z}\mathrm{/}2$. Hence one of $H_1(A),H_1(B)$ is zero and the other is $\mathbb{Z}\mathrm{/}2$.

Since $S^4\setminus A=V$, applying [F1] to $A$ and then the interior equivalence in step 2.1 gives $H_1(B)\cong H_1(V)\cong H^2(A)$. By [F2] and $H_2(A)=0$, this equals ${\mathrm{Ext}}_{\mathbb{Z}}^1(H_1(A),\mathbb{Z})$. The latter is zero if $H_1(A)=0$ and is $\mathbb{Z}\mathrm{/}2$ if $H_1(A)=\mathbb{Z}\mathrm{/}2$: for the second calculation use the free resolution $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$, whose dual has cokernel $\mathbb{Z}\mathrm{/}2$. Thus $H_1(A)$ and $H_1(B)$ are simultaneously zero or simultaneously $\mathbb{Z}\mathrm{/}2$, contradicting step 2.1. No smooth embedding in $S^4$ exists. Composing a putative embedding in ${\mathbb{R}}^4$ with inverse stereographic projection would give one in $S^4$, proving the last assertion.