Let $Y \to B$ be the universal family of smooth quadric surfaces in $\mathbb {CP}^3$, and let $T=\{(l,q) \mid l \subset Y_q \text{ where $l$ is a line}\}$ parametrize the lines contained in these $Y_q$'s. Then we get the pull-back family $Y_T \to T$. On Voisin's Hodge Theory and Complex Algebraic Geometry, Vol 2, page 218 it says that it is obvious that
The cohomology class of the tautological divisor $$D=\{(x,l,q)\mid x \in l \subset Y_q \text{ where $l$ is a line}\}$$ in $H^2(Y_T,\mathbb Q)$ does not come from $H^2(T\times \mathbb {CP}^3, \mathbb Q)$.
But I did not see why this is clear?
Thanks!
A short answer: First, there is a commutative diagram $\require{AMScd}$ \begin{CD}\tag{1}\label{1} T\times \mathbb P^3 @<\supseteq<i<Y_T@>>>Y\\ @. @VfVV@VgVV\\ @. T@>\pi>>B \end{CD} where $g$ is the universal family of smooth quadric surfaces; $\pi$ is a $\mathbb P^1\coprod \mathbb P^1$-fiber bundle (one can think this as homotopically a double cover), since every smooth quadric surface has two disjoint rulings of lines, and $f:Y_T\to T$ is the base change of $g$ along $\pi$.
A quick observation is the restriction of the hyperplane class $H\in H^2(\mathbb P^3)$ via $$H^2(\mathbb P^3)\xrightarrow{\cup H^0(T)} H^2(T\times \mathbb P^3)\xrightarrow{i^*} H^2(Y_T)$$ is represented by the divisor $$E:=\{(x,l,q)|l,l'\subseteq Y_q,\ x\in l\cup l',\ \text{where}\ l'\ \text{is the line such that}\ l\cup l'\ \text{is a hyperplane section}\},$$
as the set-theoretical intersection with hyperplane in $\mathbb P^3$. $E$ is an "invariant class" in the sense that its image in $Y$ via the forgetful map $(x,l,q)\mapsto (x,q)$ is the sum $[L_1]+[L_2]$ (more precisely this class should live in $H^0(B,R^2g_*\mathbb Q)$), which is invariant under the monodromy action, where $L_1$ and $L_2$ are lines in two different rulings, and the monodromy action is by permutation of two rulings. On the other hand, the image of tautological class $D$ is the class $[L_1]$ which is not invariant under monodromy action. In particular $E$ and $D$ are not the same class.
To see $D\notin \text{Im}(i^*)$, especially not from restriction of $H^2(T)$, one need Leray filtration, which need some spectral sequence theory to be discussed below.
A significance of the non-sujectivity of $i^*$ shows that the sufficient ampleness assumption in Nori's connectivity theorem [note by Nagel, p.20] is necessary. (In particular, $Y_q$ is not sufficiently ample in $\mathbb P^3$.)
Here is a longer answer: One can compute cohomology of $Y_T$ and $T\times \mathbb P^3$ rather explicitly. Before we get into the calculation, let's get some geometric intuition of what's going on here. Consider a degeneration of smooth quadric surfaces into a nodal quadric surface
$$\mathbb P^3\times \Delta\supseteq \{t(x^2+y^2+z^2+w^2)+(y^2+z^2+z^2)=0\ ,\ t\in \Delta\},\tag{2}\label{2}$$
it is an exercise to show that the generator of $\pi_1(\Delta^*,t_0)$ permutes the set of two rulings. In other words, $\Delta^*\hookrightarrow B$ is an embedding of holomorphic punctured disk such that its completion $\Delta$ intersect the boundary divisor transversely at a smooth point, then the restriction of $g$ to the punctured family $g_{|\Delta^*}$ has the property that the local system $R^2(g_{|\Delta^*})_*\mathbb Q$ is determined by the $\mathbb Z_2$-action $(1,0)\mapsto (0,1),\ (0,1)\mapsto (1,0)$ on $\mathbb Q^2$.
Although there are more singular quadric surfaces (e.g., union of two planes), the global monodromy of lines is generated by the monodromy on $(\ref{2})$. Essentially, one can restrict the universal family to $\Delta^*$ in the diagram $(\ref{1})$.
With this preparation, we claim:
Proof. The proof will be based on three steps. In the first two steps, we will compute $H^2(T\times \mathbb P^3)$ and $H^2(Y_T)$. In the third step, we will reduce to the discussion on restriction of hyperplane class.
Step 1. By Kunneth formula and connectivity of $T$, one has $$H^2(T\times \mathbb P^3)\cong H^2(T)\otimes H^0(\mathbb P^3)\oplus H^0(T)\otimes H^2(\mathbb P^3)\cong H^2(T)\oplus \mathbb Q.$$
(Edition remark: one can further compute $H^2(T)$ using the fiber bundle structure on $T$, but is unnecessary for our problem.)
Step 2. $\dim H^2(Y_T)=\dim H^2(T\times \mathbb P^3)+1.$ In particular, $i^*$ cannot be surjective.
Note that $Y_T$ is a quadric surface bundle over $T$ arising as pullback of $g$, we can use Leray spectral sequence on $f:Y_T\to T$. Working over $\mathbb Q$-coefficient, Deligne's theorem [Voisin, Thm 4.15] says that the spectral sequence degenerates at $E_2$-page, with $E_2^{p,q}=H^p(B,R^q\pi_*\mathbb Q)$, and for each $k$, $H^k(Y_T)$ is made up of $H^p(T,R^qf_*\mathbb Q)$ with $p+q=k$.
In order to compute $H^2(Y_T)$, let's compute the three groups with $p+q=2$ case by case:
(1) $q=0$, $R^0f_*\mathbb Q=f_*\mathbb Q$, so $H^2(T,f_*\mathbb Q)=H^2(T,\mathbb Q)$, since $f$ is proper and fiber is connected;
(2) $q=1$, $R^1f_*\mathbb Q=0$ because the quadric surface has no $H^1$, so $H^1(T,R^1f_*\mathbb Q)=0$;
(3) $q=2$, $R^2f_*\mathbb Q\cong \mathbb Q^2$ is a constant local system, this is because $R^2g_*\mathbb Q$ and $\pi_*\mathbb Q$ are isomorphic as local systems, so base change along $\pi$ eliminate the monodromy. It follows that $H^0(T,R^2f_*\mathbb Q)=\mathbb Q^2$.
Finally, $\dim H^2(Y_T)=\sum_{p+q=2}\dim H^p(T,R^qf_*\mathbb Q)=\dim H^2(T)+2=\dim H^2(T\times \mathbb P^3)+1$.
Step 3. $D$ is not in the image of $i^*$.
By functoriality of Leray spectral sequence, we have the following comparison diagram: $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} 0 & \ra{} & \big (H^2(T)=\big )H^2(T,H^0(\mathbb P^3)) & \ra{} & H^2(T\times \mathbb P^3) & \ra{} & H^0(T,H^2(\mathbb P^3))(=\mathbb Q) & \ra{} & 0 \\ & & \da{\cong} & & \da{i^*} & & \da{\bar{i}^*} & & \\ 0 & \ras{} & \big (H^2(T)=\big )H^2(T,f_*\mathbb Q) & \ras{} & H^2(Y_T) & \ras{} & H^0(T,R^2f_*\mathbb Q)(=\mathbb Q^2) & \ras{} & 0 \\ \end{array} $$
where the first row is the Leray filtration of the projection $T\times \mathbb P^3\to T$ which is just the Kunneth formula. The generator of $H^0(T,H^2(\mathbb P^3))=H^2(\mathbb P^3)$ is the hyperplane class $H$. Now from the diagram, we can see that the restriction $\bar{i}^*$ is not surjective. This map is identified with
$$\bar{i}^*:\mathbb Q\to \mathbb Q^2,\ 1\mapsto (1,1),$$
whose image is the cohomology class of divisor $E$. On the other hand, the image of the tautological cycle $D$ in $\mathbb Q^2$ is $(1,0)$ or $(0,1)$ depending on the choice of the basis. So $D$ is not in the image of restriction. $\tag*{$\blacksquare$}$