Relative divisors Let $X\rightarrow T$ be a fibre bundles with smooth projective fibre $F$ and $X$ and $T$ are also smooth. Let $D$ is relative effective Weil divisor. Suppose $W_1 $ and $W_2$ are relative subvarieties which are isomorphic(by a relative map say $\phi$). Let $L:=\mathcal{O}(D)$. Let $L_t|_{W_{1,t}}\cong L_t|_{W_{2,t}}$
(via $\phi_t$). Is it true that $L|_{W_{1}}\cong L|_{W_{2}}$ (via $\phi$)?
 A: I am posting my comment above as an answer.  Let $T$ be a smooth, projective curve of genus $g\geq 1$.  Let $F$ be $\mathbb{P}^3$.  Let $X$ be the product $\mathbb{P}^3\times T$ with its projection.  Let $L$ be $\text{pr}_{\mathbb{P}^3}^*\mathcal{O}(1)$.  Let $w:W\to T$ be a finite, étale morphism of degree $d>1$ such that $W$ is connected.  (Actually, any finite flat morphism would do, but it is easiest to understand the $w$-fibers when $w$ is étale.)
For any two closed immersions $u_1,u_2:W\hookrightarrow \mathbb{P}^3$, the associated product morphisms $v_i=(u_i,w)$ are closed immersions, $$ v_1,v_2:W\hookrightarrow X.$$  Since $w$ is finite, the pullbacks $v_1^*L$ and $v_2^*L$ have isomorphic restrictions to every $w$-fiber.  Yet typically $[v_1^*L]$ and $[v_2^*L]$ are not equal in $\text{Pic}(W)/w^*\text{Pic}(T)$.  In particular, the congruence class modulo $d$ of $\text{deg}(v_i^*L)$ depends only on the class $[v_i^*L]$.  Thus, if $\text{deg}(v_1^*L)$ is not congruent to $\text{deg}(v_2^*L)$, then $[v_1^*L]$ does not equal $[v_2^*L]$ in $\text{Pic}(W)/w^*\text{Pic}(T)$.  
