Euler number $\pm 1$ implies the Milnor sphere bundle is a homology seven-sphere

In the Gysin sequence [L1] with $k=0,1,2,3$ and with negative cohomology zero, the only possible intermediate term is $H^3(M_{h,j})\to H^0(S^4)=\mathbb{Z}\stackrel{⌣e}{\to }{H}^4(S^4)=\mathbb{Z}\to H^4(M_{h,j})\to H^1(S^4)=0$; since $e=\epsilon u$ by [L2] and $\epsilon =\pm 1$, the multiplication map is an isomorphism, so $H^3(M_{h,j})=0$ and $H^4(M_{h,j})=0$, while also $H^1(M_{h,j})=H^2(M_{h,j})=0$ and $H^0(M_{h,j})=\mathbb{Z}$.

For $k=5,6$ the sequence reads $0\to H^k(M_{h,j})\to H^{k-3}(S^4)$ with $k-3=2,3$, so $H^5(M_{h,j})=H^6(M_{h,j})=0$.

For $k=7$ the sequence gives $0=H^7(S^4)\to H^7(M_{h,j})\stackrel{\mathrm{\partial }}{\to }{H}^4(S^4)=\mathbb{Z}\stackrel{⌣e}{\to }{H}^8(S^4)=0$, so $\mathrm{\partial }$ is an isomorphism and $H^7(M_{h,j})=\mathbb{Z}$; for $k=8$, $H^4(S^4)\stackrel{⌣e}{\to }{H}^8(S^4)=0$ gives $H^8(M_{h,j})=0$, and all higher degrees vanish.

Combining: $H^0(M_{h,j};\mathbb{Z})=H^7(M_{h,j};\mathbb{Z})=\mathbb{Z}$ and $H^k(M_{h,j};\mathbb{Z})=0$ for $1\le k\le 6$.

Write $H_k(M_{h,j};\mathbb{Z})={\mathbb{Z}}^{r_k}\oplus T_k$ by [L3], [L5]. The UCT exact sequence [L4] and the vanishing of $H^k$ for $1\le k\le 6$ give $r_k=0$ in these degrees and $T_{k-1}=0$. In degree seven, $\mathrm{Ext}(H_6,\mathbb{Z})$ injects into $H^7=\mathbb{Z}$; it is finite, so it is zero and $T_6=0$. The resulting isomorphism $\mathrm{Hom}(H_7,\mathbb{Z})\cong \mathbb{Z}$ gives $r_7=1$. Finally $H^8=0$ from step 3.1 forces $\mathrm{Ext}(H_7,\mathbb{Z})=0$, hence $T_7=0$. Since the bundle is locally path connected, $H^0=\mathbb{Z}$ from step 4.1 forces it to have one component; thus $H_0=\mathbb{Z}$.

Thus $H_0=H_7=\mathbb{Z}$ and $H_1,\dots ,H_6=0$. The closed seven-manifold finiteness supplier [L3] also gives zero groups outside degrees zero through seven, proving the statement in every degree.