For a closed embedding of compact complex manifolds
$$
\iota : Y \hookrightarrow X
$$
and any $\alpha \in H^\*(X,\mathbb Q)$, we have trivially:
$$
\iota^\*(\alpha)=0\quad \Rightarrow \quad\iota\_\*\iota^\*(\alpha)=0.
$$
My question is: does the converse hold? So does $\iota\_\*\iota^\*(\alpha)=0$ imply that $\iota^\*(\alpha)=0$?
If not: can you give a counterexample?



Kernel of projection formula
No, it is not true.


Consider a Hopf surface $X=(\mathbb C^2\backslash \{0\})/((z\_1,z\_2)\sim (2z\_1,2z\_2))\simeq S^3\times S^1$. It contains a "coordinate" elliptic curve $Y=(\mathbb C\backslash \{0\})/(z\_1\sim 2z\_1)\simeq S^1\times S^1$.


Let us use Poincare-dual language to operate $i^\*$ and $i\_\*$. Take $\alpha\in H^1(X)$ to be the class dual to the cycle $S^3\times pt$, then the class
$$
i^\*(\alpha)\in H^1(Y)
$$
is dual to the cycle $S^1\times pt\subset S^1\times S^1$ and nonzero. Direct image of this cycle in $H\_1(X)$ is zero, hence
$$
i\_\*i^\*(\alpha)\in H^3(X)
$$
also vanishes.


