A rational homology four-ball has square boundary torsion order

By [F1] the integral homology groups of $W$ and $M$ are finitely generated. By [F5], the rational homology hypothesis makes $H_i(W;\mathbb{Z})\otimes \mathbb{Q}=0$ for $i>0$; finite generation and the decomposition of [F3] therefore make those integral groups finite. Apply [F3] also with coefficient module $\mathbb{Q}$: the Hom terms vanish in positive degrees and the Ext terms for finite cyclic groups vanish because multiplication by their orders is onto on $\mathbb{Q}$. Thus $H^i(W;\mathbb{Q})=0$ for $i>0$ and $H^0(W;\mathbb{Q})=\mathbb{Q}$. Duality [F2] gives $H_i(W,M;\mathbb{Q})=0$ for $i\mathrm{\ne }4$, and $H_4(W,M;\mathbb{Q})=\mathbb{Q}$. The pair sequence now gives $H_3(M;\mathbb{Q})=\mathbb{Q}$, $H_1(M;\mathbb{Q})=H_2(M;\mathbb{Q})=0$, and $H_0(M;\mathbb{Q})=\mathbb{Q}$. Hence $M$ is a rational homology sphere and $H_1(M)$ is finite. Integral duality on $M$, followed by [F3], gives $H_2(M)\cong H^1(M)=0$: the Ext term for $H_0(M)=\mathbb{Z}$ and the Hom term for finite $H_1(M)$ both vanish. Likewise $H_3(W,M)\cong H^1(W)=0$. These are exactly the vanishings needed in the order calculation.

Put $A:=H_2(W)$ and $B:=H_1(W)$; both are finite, because $W$ is rationally a point and the groups are finitely generated by [F1]. Duality [F2] and the cohomology universal coefficient sequence [F3] give $H_2(W,M)\cong H^2(W)\cong \mathrm{Ext}(B,\mathbb{Z})$ and $H_1(W,M)\cong H^3(W)\cong \mathrm{Ext}(A,\mathbb{Z})$, the Hom terms vanishing because $A,B$ are finite. For a finite abelian group $F$ decomposed into cyclic summands, the sequence $0\to \mathbb{Z}\stackrel{n}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}n\to 0$ computes $\mathrm{Ext}(\mathbb{Z}\mathrm{/}n,\mathbb{Z})=\mathbb{Z}\mathrm{/}n$, so $\mathrm{\mid }\mathrm{Ext}(F,\mathbb{Z})\mathrm{\mid }=\mathrm{\mid }F\mathrm{\mid }$.

The long exact sequence of the pair $(W,M)$ in low degrees, using $H_2(M)=0$ and the isomorphism $H_0(M)\to H_0(W)$ of connected spaces, reads $0\to A\to \mathrm{Ext}(B,\mathbb{Z})\to H_1(M)\to B\to \mathrm{Ext}(A,\mathbb{Z})\to 0$. Split it into the short exact sequences $0\to A\to \mathrm{Ext}(B,\mathbb{Z})\to K\to 0$ and $0\to K\to H_1(M)\to B^{\mathrm{\prime }}\to 0$, where $K$ is the image of $\mathrm{Ext}(B,\mathbb{Z})$ and $B^{\mathrm{\prime }}$ the image of $H_1(M)$; then $\mathrm{\mid }K\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }A\mathrm{\mid }$ and $\mathrm{\mid }{H}_1(M)\mathrm{\mid }=\mathrm{\mid }K\mathrm{\mid }\cdot \mathrm{\mid }{B}^{\mathrm{\prime }}\mathrm{\mid }$. The third short exact sequence $0\to B^{\mathrm{\prime }}\to B\to \mathrm{Ext}(A,\mathbb{Z})\to 0$, the last map being surjective by exactness at the final term, gives $\mathrm{\mid }{B}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }\mathrm{Ext}(A,\mathbb{Z})\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }A\mathrm{\mid }$ by step 2.1. Hence $\mathrm{\mid }{H}_1(M)\mathrm{\mid }=(\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }A\mathrm{\mid }{)}^2$. The injection $A\hookrightarrow \mathrm{Ext}(B,\mathbb{Z})$ makes $\mathrm{\mid }A\mathrm{\mid }$ divide $\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }\mathrm{Ext}(B,\mathbb{Z})\mathrm{\mid }$, so $\mathrm{\mid }B\mathrm{\mid }\mathrm{/}\mathrm{\mid }A\mathrm{\mid }$ is a positive integer and $\mathrm{\mid }{H}_1(M)\mathrm{\mid }$ is a square. Choice enters only through [F4] and the cited inputs.