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](https://arxiv.org/abs/0909.5319). The primitivity is obtained by Thm 2.1 in "Libgober, J. Wood:
[On the topological structure of even-dimensional
complete intersections](https://doi.org/10.2307/1998676)".)



Primitivity of the Lefschetz embedding
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 $H^n(X,\mathbf Z)\_{\operatorname{tf}} = \operatorname{Hom}(H\_n(X,\mathbf Z),\mathbf Z)$.


