Let $E\times \mathbb{P}^1$ be the product of an elliptic curve $E$ and a rational curve. Let $\eta\in\mathrm{Pic}^0(E)$ be torsion point such that $2\eta=0$ and $\delta=\mathrm{pr}_1^*\eta+\mathrm{pr}_2^*\mathcal{O}(3)$, take $D=\sum_{i=1}^6\mathrm{pr}_2^*q_i\in|2\delta|$ the six distinct fibres of $\mathrm{pr}_2$. Let $r:T\longrightarrow E\times\mathbb{P}^1$ be the double cover ramified along $D$. we have $K_T\sim r^*(K_{E\times\mathbb{P}^1}+\delta')$. Let $a\geq 2$ be an integer, and $A\in\mathrm{Pic}(E)$. By the consruction, $\phi:T\longrightarrow \mathbb{P}^1$, where $\psi=\mathrm{pr_2}\circ r$, is an elliptic fibration, and $C_i=(\psi^*q_i)_{\mathrm{red}}$ are the multipul fibres with multiplicity 2. Take $\delta=\phi^*A+C_1-C_2$, where $\phi=\mathrm{pr_1}\circ r$. My question is why the linear system $|K_T+\delta|$ is composed with a pencil?

By construction, we can see $K_T+\delta\sim \phi^*(A+\eta)+\psi^*\mathcal{O}(1)+C_1-C_2\sim \phi^*(A+\eta)+C_1+C_2$, since $\psi^*\mathcal{O}(1)\sim 2C_2$. It's enough to show that $C_1+C_2$ is the fixed part of $|K_T+\delta|$. But the intersection number of $(K_T+\delta).(C_1+C_2)=2a$ can not expain this. I don't know any other method to test a divisor is the fixed part of a linear system.