Let's consider closed simply connected manifold $M^n$ and a $a\in H^k(M)$ and $a\*\in H^{n-k}(M)$ is the dual to $a$. 


Is it true that dual to $a$ is realisable as a immersed sphere or $ a\*=bc $ for some $b,c\in H^\*(M)$ ?


Edit: it is more natural to ask about possibility to decompose dual to $a$ as a product, see example in the answer below.


Edit2: let's assume that $M$ is not decompasable, so there is no $X,Y$ such that $M = X\times Y$



Realisability cohomological class as product or as immersed sphere
I will construct a closed simply-connected $8$-manifold $M$ and an $a\in H^3(M;\Bbb Z)$ such that the Poincare dual $b$ of $a$ is not realizable by a map $S^5\to M$, and a Hom-dual element in $H^5(M;\Bbb Z/2)$ to the $\bmod 2$ reduction of $b$ is not a nontrivial product. 


Let $K$ be the suspension over $\Bbb C P^2$. Then there is an $\alpha\in H^3(K;\Bbb Z/2)$ with $Sq^2(\alpha)\ne 0$, but the $\bmod 2$ cohomology ring of $K$ is trivial. Let $N$ be a regular neighborhood of a PL copy of $K$ in some $\Bbb R^m$. So $N$ is homotopy equivalent to $K$. A loop in $\partial N$ bounds a disk in $N$, which can be pushed off $K$ as long as $5+2\le m-1$. Thus $\partial N$ is simply-connected. Let $M$ be the double of $N$, i.e. $M=\partial (N\times I)$. So $M$ is a closed $m$-manifold, it is simply-connected by Seifert-van Kampen, and the inclusion $N\subset M$ is split by the projection $\phi:M\subset N\times I\to N$. So the cohomology of $N\simeq K$ is a direct summand in the cohomology of $M$. Let $\beta=\phi^\ast(\alpha)$, then $\gamma:=Sq^2\beta\ne 0$, and $\gamma$ is not a nontrivial product. Since the nonzero element of $H^5(S^5;\Bbb Z/2)$ is not in the image of $Sq^2$, there is no map $f:S^5\to M$ such that $f^\*(\gamma)\ne 0$.


Let $b\in H\_5(M;\Bbb Z)$ be such that $\gamma(b)\ne 0$, and let $a\in H^{m-5}(M;\Bbb Z)$ be the Poincare dual of $b$. If $b$ is realized by an immersion, or just a map, $f:S^5\to M$, then $0\ne\gamma\smallfrown f\_\ast[S^5]=f\_\ast(f^\ast(\gamma)\smallfrown[S^5])$, contradicting $f^\*(\gamma)=0$. 


As for $m$, $\Bbb C P^2$ is the mapping cone of the Hopf map $h:S^3\to S^2$. The mapping cylinder of $h$ embeds in $S^2\*S^3=S^6$, so $\Bbb CP^2$ embeds in $\Bbb R^7$ and $K$ in $\Bbb R^8$.


