Primitivity of the Lefschetz embedding Let $X$ be a smooth projective variety of dimension $2n+1$, let $i\colon Y\subset X$ be an ample hypersurface, by Lefschetz hyperplane theorem, the pullback $i^*\colon H^{2n}(X,\mathbb{Z})\to H^{2n}(Y,\mathbb{Z})$ is an injection.
When $X$ is a product of two projective spaces or more general (e.g., projective bundle over projective space), is it necessarily true that $i^*\colon H^{2n}(X,\mathbb{Z})\to (H^{2n}(Y,\mathbb{Z})/H^{2n}(Y,\mathbb{Z})_{\mathrm{tors}})$ is always a primitive embedding of lattices (i.e., the cokernel of $i^*$ is torsion free)?
(The question rose when trying to calculate lattice of primitive middle cohomology of $Y$, and hopefully it is primitive embedding. For $X$ being projective space, there is a reference Thm 2.3 The primitive cohomology lattice of a complete intersection. The primitivity is obtained by Thm 2.1 in "Libgober, J. Wood:
On  the  topological  structure  of  even-dimensional
complete intersections".)
 A: This is actually a general consequence of the Lefschetz hyperplane theorem and the universal coefficient theorem.
For simplicity, write $A_{\operatorname{tf}} = A/A_{\operatorname{tors}}$. Note that this association is functorial, additive, and preserves injections and surjections, but is not exact in the middle! It follows from the universal coefficient theorem that $H^k(X,\mathbf Z)_{\operatorname{tf}} = \operatorname{Hom}(H_k(X,\mathbf Z)_{\operatorname{tf}},\mathbf Z)$.
Lemma. Let $X$ be a smooth projective complex manifold of (complex) dimension $n+1$, and let $Y \subseteq X$ be a smooth hypersurface. Then the image of $H^n(X,\mathbf Z)_{\operatorname{tf}} \hookrightarrow H^n(Y,\mathbf Z)_{\operatorname{tf}}$ is saturated.
Proof. By the Lefschetz hyperplane theorem, the map $H_n(Y,\mathbf Z) \to H_n(X,\mathbf Z)$ is surjective, hence the same goes for $H_n(Y,\mathbf Z)_{\operatorname{tf}} \to H_n(X,\mathbf Z)_{\operatorname{tf}}$. Since $H_n(X,\mathbf Z)_{\operatorname{tf}}$ is finite free, we may choose a splitting, realising $H_n(X,\mathbf Z)_{\operatorname{tf}}$ as a summand of $H_n(Y,\mathbf Z)_{\operatorname{tf}}$, and hence $H^n(X,\mathbf Z)_{\operatorname{tf}}$ as a summand of the finite free module $H^n(Y,\mathbf Z)_{\operatorname{tf}}$. In particular, it is saturated. $\square$
Alternatively, if you don't like making these choices, use the pairings between $H_n$ and $H^n$ together with the push-pull formula
$$(i^* \psi)(\beta) = \psi(i_* \beta)$$
for $\psi \in H^n(X,\mathbf Z)$ and $\beta \in H_n(Y,\mathbf Z)$. If $n\phi = i^* \psi$ for some $n > 1$ and some $\phi \in H^n(Y,\mathbf Z)_{\operatorname{tf}}$, then surjectivity of $i_*$ shows that $\psi(\alpha)$ is divisible by $n$ for all $\alpha \in H_n(X,\mathbf Z)$, thus $\psi$ is divisible by $n$ in $H^n(X,\mathbf Z)_{\operatorname{tf}} = \operatorname{Hom}(H_n(X,\mathbf Z),\mathbf Z)$.
