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 $\phi$)?



Relative divisors
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 $\text{Pic}(W)/w^\*\text{Pic}(T)$. 


