A rational homology ball bounded by L(4,1)

Choose the filtration ${\mathbb{C}\mathbb{P}}^0\subset P\subset {\mathbb{C}\mathbb{P}}^2$ with $P=\{y=0\}\cong {\mathbb{C}\mathbb{P}}^1$ and ${\mathbb{C}\mathbb{P}}^0=[1:0:0]$. Its successive complements are $\mathbb{C}$ and ${\mathbb{C}}^2$, so the cellular boundary maps vanish for degree reasons and $H_2({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})=\mathbb{Z}[P]$, $H_4({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})=\mathbb{Z}[{\mathbb{C}\mathbb{P}}^2]$; UCT gives free cohomology, and put $h=\mathrm{P}\mathrm{D}[P]$. The projective maps $T_t([x:y:z])=[x:y+tx:z]$ ($0\le t\le 1$) give a homotopy from the inclusion of $P$ to that of $P^{\mathrm{\prime }}=\{y=x\}$. The two lines meet transversely at $[0:0:1]$ with positive sign, since a complex-linear isomorphism between their tangent complex lines and the ambient complex plane has positive real determinant. By [F3], $[P{]}^2=1$. On the charts $s\mathrm{\ne }0$ and $t\mathrm{\ne }0$, the conic map $[s:t]\mapsto [s^2:st:t^2]$ has target coordinates $(u,u^2)$ with $u=t\mathrm{/}s$ and $(v^2,v)$ with $v=s\mathrm{/}t$, respectively; these recover the source coordinate and have injective derivatives. Its image is exactly $xz-y^2=0$: if $x\mathrm{\ne }0$ the point is $[1:y\mathrm{/}x:(y\mathrm{/}x{)}^2]$, and if $x=0$ then $y=0$ and the point is $[0:0:1]$. The same chart descriptions show the zero set is a smooth complex-oriented sphere and the map is an embedding. The line $P$ meets it at the two simple roots $st=0$, both transversely and positively in the complex orientations. Thus [F3] gives $[Q]\cdot [P]=2$; because $[P]$ generates $H_2$, this implies $[Q]=2[P]$, hence $[Q]\cdot [Q]=4$.

The normal Euler number of $Q$ is therefore $4$ by [F4]. Its unit normal circle bundle is the boundary of its tubular disk bundle. Compute its genus-one gluing directly from the conic charts, without assuming an Euler-to-surgery identification. In the chart $x=1$ write the surface as $(y,z)=(u,u^2)$ and use normal coordinate $n=z-y^2$. In the chart $z=1$ use normal coordinate $n^{\mathrm{\prime }}=x-y^2$. Their change is $n^{\mathrm{\prime }}=(z-y^2)\mathrm{/}{z}^2$; differentiating in the normal direction on the conic gives $n^{\mathrm{\prime }}=u^{-4}n$ in the normal line bundle. On the equator $u=e^{i\phi }$ its unit normal circle transition is therefore ${\theta }_N={\theta }_S-4\phi$. Each closed hemisphere has its actual displayed complex normal frame, so its unit circle bundle is $D^2\times S^1$, a solid torus. On their common torus let $m$ be the equator loop at fixed ${\theta }_S$, and $l$ the fiber loop. The southern solid-torus meridian is $m$, and the northern meridian is $m+4l$, since holding ${\theta }_N$ fixed gives that exact slope. Identify the southern solid torus with the standard unknot exterior by $l={\mu }_U$, $m={\lambda }_U$: its fiber is the core circle, while its base disk is the exterior meridian disk. The second filling slope is therefore $4{\mu }_U+{\lambda }_U$, with the global boundary orientation determined by the product identifications. If those identifications reverse the desired global orientation, reverse the orientation of the total space; the statement permits exactly that choice. Thus [F5] identifies the total circle bundle with $L(4,1)$ up to orientation, by an actual meridian calculation. The complement has the opposite boundary orientation to its tubular neighborhood in the complex-oriented projective plane; choose the orientation of $B_2$ as allowed in the Statement so its boundary agrees with the declared lens-space convention.

By [F2], the relative cohomology groups of $({\mathbb{C}\mathbb{P}}^2,B_2)$ are $\mathbb{Z}$ in degrees $2$ and $4$ and zero otherwise, since $Q\cong S^2$. The relative-to-absolute map in degree two sends the Thom generator to $\mathrm{P}\mathrm{D}[Q]$: the normal-Thom sign is $(-1{)}^{2\cdot 2}=+1$ by [F2]. By step 1.1, $[Q]=2[P]$, so this map is multiplication by $2$ under $h=\mathrm{P}\mathrm{D}[P]$. For degree four, let $a$ be the positive generator of $H^2(Q;\mathbb{Z})$, with $⟨a,[Q]⟩=1$. The Thom image of $a$ evaluates on $[{\mathbb{C}\mathbb{P}}^2]$ as $⟨a,[Q]⟩$: excision reduces the evaluation to the normal disk fibers, and the normalized Thom class evaluates $+1$ on each such fiber by [F2]. Thus the degree-four relative-to-absolute map sends a generator to a generator and is an isomorphism.

By [F6], the cohomology long exact sequence of the pair gives $H^1(B_2;\mathbb{Z})=0$, $H^2(B_2;\mathbb{Z})\cong \mathbb{Z}\mathrm{/}2$, and $H^j(B_2;\mathbb{Z})=0$ for every $j\ge 3$: in degree one the map $\mathbb{Z}\to \mathbb{Z}$ is injective multiplication by $2$, in degree two its cokernel is $\mathbb{Z}\mathrm{/}2$ and the next relative group is zero, and in degree four the relative-to-absolute map is an isomorphism. In degree zero, the relative group vanishes, so $H^0({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})\to H^0(B_2;\mathbb{Z})$ is an isomorphism; hence $B_2$ is connected. Repeat the relative calculation over $\mathbb{Q}$: the degree-two map is multiplication by $2$ and the degree-four map is the identity, both isomorphisms, so the pair sequence gives $H^j(B_2;\mathbb{Q})=0$ for $j>0$ and $H^0(B_2;\mathbb{Q})=\mathbb{Q}$. The UCT over $\mathbb{Q}$ then gives the rational homology of a point. Together with step 2.1 this proves that $B_2$ is a rational homology ball bounded by $L(4,1)$.