= Solution
Let $D_\alpha,D_\beta$ be the disjoint complete meridian-disk system of $H_1$, and let $D_\gamma,D_\delta$ be that of $H_2$. The identification $\phi(\alpha)=\gamma$ joins $D_\alpha$ and $D_\gamma$ into an embedded sphere $S$ in $M$. Each disk is nonseparating in its <handlebody>: cutting along it leaves a genus-one <handlebody>, and cutting along the other then gives a ball. After cutting $M$ along $S$, the two cut <handlebodies> are connected and are still joined along the complement of the common boundary circle. Hence $S$ is nonseparating.
Cap the two new sphere boundaries with balls. The resulting <closed manifold> $N$ inherits a genus-one <Heegaard splitting>: its two sides are the <solid tori> obtained by cutting the original <handlebodies> along their common disks. Their meridians are the surviving curves $\beta$ and $\delta$. The hypothesis that $\phi(\beta)$ avoids $\gamma$ makes it descend to the compressed torus, and its single intersection with $\delta$ remains a single intersection there.
A genus-one splitting whose two meridians intersect once is $S^3$. To see this directly, choose the target torus basis $(\mu,\lambda)$ with $\mu$ its meridian. The other meridian is $a\mu\pm\lambda$. A meridional <Dehn twist>, extending across the target <solid torus>, removes the integer $a$, leaving the longitude. Gluing a second <solid torus> with meridian equal to that longitude is the standard splitting of $S^3\subset\mathbb C^2$ into the neighborhoods of its two coordinate circles. Therefore $N\cong S^3$.
By the <common nonseparating disks in a Heegaard splitting> construction and part (b), reversing the cut and caps gives
$$
\boxed{M\cong(S^1\times S^2)\mathbin\#S^3\cong S^1\times S^2.}
$$
The product metric on $S^2\times\mathbb R$ is homogeneous, and quotienting by an integer translation in the second factor gives $S^2\times S^1$. Thus $M$ carries this <Thurston geometry> and is \b[a <geometrizable three-manifold>].
There is a wording qualification: if “<Heegaard genus> two” means the minimum defined in part (a), the disk data contradict it. The product splits as $S^1\times D^2$ on each hemisphere of $S^2$, giving a genus-one splitting, while genus zero gives only $S^3$, so $g_H(S^1\times S^2)=1$. The genuine construction is a genus-two splitting of a genus-one manifold; the geometrizability conclusion remains valid. The proof used the complete disk systems, the common-sphere cut-and-cap construction, and the explicitly justified genus-one meridian-intersection criterion.
Back to article page
