= Solution
For a properly embedded oriented <topological surface> $F=\coprod F_j$, put
$$
c(F)=\chi_-(F)=\sum_j\max\{0,-\chi(F_j)\},\qquad \|x\|_T=\min_{[F]=x}c(F).
$$
This defines the <Thurston norm> on integral <relative homology> classes in $H_2(Y,\partial Y;\mathbb Z)$. Homogeneity extends it to rational classes, and continuity extends it to real classes. It is a <seminorm>: nonzero classes carried by spheres, disks, <annuli> or <tori> can have zero value.
\b[Not every integral class has a connected embedded representative.] In $Y=S^1\times S^2$, the class $2[\{*\}\times S^2]$ is a counterexample. A connected embedded oriented surface either separates, in which case it is null-homologous, or has connected complement. In the latter case join its two local sides through its complement to obtain a loop intersecting it exactly once. Its <homology class> is consequently a <primitive lattice element>. The proposed double has all intersection numbers even, so cannot have such a representative. Two disjoint parallel spheres do represent it.
For the <knot exterior> in an <integral homology sphere>, $H_2(Y,\partial Y;\mathbb Z)=\mathbb Z$ is generated by a <Seifert surface>. Its <Thurston norm> is
$$
\|y\|_T=\max\{0,2g(K)-1\},
$$
where $g(K)$ is the <Seifert genus> in the ambient <integral homology sphere>. To justify using a one-boundary-component <Seifert surface>, simplify a minimizing representative's boundary to parallel essential longitudes. The algebraic sum of these longitudes is one. Join oppositely oriented pairs by boundary annuli and push inward; this preserves <Euler characteristic> and does not increase $\chi_-$. Discard closed components, which represent zero because the ambient <integral homology sphere> has $H_2=0$. The component retaining the single boundary is a <Seifert surface>. This proves the formula, including the disk case.
When $\|y\|_T>0$, $g(K)\geq1$. In zero <Dehn surgery> the longitude bounds a meridional disk in the filling <solid torus>. Cap a minimizing <Seifert surface> with that disk. The resulting closed surface $\widehat F$ has the same <genus> and represents a generator of $H_2(Z;\mathbb Z)$: its intersection with the filling core is one. Therefore
$$
\boxed{\|z\|_T\leq\chi_-(\widehat F)=2g(K)-2=\|y\|_T-1.}
$$
The case $g(K)=1$ gives a zero-cost <torus>, rather than a negative value.
Here is a rigorous family of strict examples with $Z=S^1\times S^2$. Use the <excellent knot representative theorem> to choose a <knot> $J$ representing the generator of $\pi_1(Z)$ with an <excellent three-manifold> as exterior $E_0$. Inside $V=\nu J$, choose a <winding number of a satellite pattern> one pattern $P$ whose two-boundary-component exterior $E_1=V\setminus\operatorname{int}\nu P$ is also an <excellent three-manifold>. Both choices are available because neither ambient manifold has a spherical boundary component. Put $\Gamma=P(J)$ and $E=Z\setminus\operatorname{int}\nu\Gamma=E_0\cup_{\partial V}E_1$.
The interface is an <incompressible surface>. On a <Thurston norm> minimizing surface in $E$ with boundary the meridian $\mu_\Gamma$, arrange all interface intersections to be essential. The <winding number of a satellite pattern> one condition forces the oriented boundary on that interface to have net class $\mu_J$. The piece in $E_0$ has odd meridional boundary sum and costs at least one: a zero-cost representative would require an essential disk or <annulus>, forbidden by excellence. The piece in $E_1$ has opposite net meridians on its two boundary <tori>. It costs at least two: zero-cost components are boundary-parallel <annuli> or <tori> and carry no such boundary class, and its total meridional boundary count is even, so its nonzero negative <Euler characteristic> has even magnitude. Cutting along the interface adds the costs, since the essential pieces have no disk or sphere components. Thus
$$
\|[\mu_\Gamma\text{-bounding surface}]\|_T\geq1+2=3.
$$
Here the notation denotes the relative class whose boundary is $\mu_\Gamma$, not the meridian curve itself. This is the usual <Thurston norm gluing along an incompressible torus> argument.
Because $\Gamma$ represents the generator of $H_1(Z)$, $H_1(E)=\mathbb Z$, its meridian is null-homologous in $E$, and a longitudinal curve $\lambda_\Gamma$ generates $H_1(E)$. Fill $E$ along $\lambda_\Gamma$ to obtain $X$. The <Mayer–Vietoris sequence> gives $H_1(X)=0$, so $X$ is an <integral homology sphere>. Let $K$ be the filling core. Its preferred longitude is $\mu_\Gamma$; zero <Dehn surgery> on $K$ consequently recovers $Z$. For this example $\|y\|_T\geq3$, whereas the product sphere generates $H_2(Z)$ with zero cost. Hence
$$
\boxed{0=\|z\|_T<\|y\|_T-1.}
$$
The construction needs a nontrivial winding-one pattern; merely tying a local <knot> into the product core would not provide this lower bound.
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