Let $X$ be a space and let $\jmath:X\times \mathbb{C}P^1\to X\times \mathbb{C}P^h$ be the obvious inclusion. Then Künneth gives us an isomorphism $\jmath^*:H^2(X\times\mathbb{C}P^h)\to H^2(X\times \mathbb{C}P^1)$ .Now consider maps $$\begin{array}{rcl}s&=&(\mathrm{id}_X,[a_0:a_1]):X\to X\times \mathbb{C}P^1\\\text{and}~~~t&=&(\mathrm{id}_X,[b_0:\dotsb:b_h]):X\to X\times\mathbb{C}P^h\end{array}$$ and assume $a_0(x)=0$ iff $b_0(x)=0$. I want to proof:
$\jmath\circ s$ and $t$ induce the same morphisms $H^2(X\times\mathbb{C}P^h)\to H^2(X)$.
My idea is that we only have to see what the generator $\mu\in H^2(\mathbb{C}P^h)\cong H^2(\mathbb{C}P^1)$ does, and this generator is dual to the submanifold $[0:*:\dotsb:*]$ resp. $[0:*]$. Maybe we can use this.
(In fact, I need the statament in a slightly more complicated way where the occuring bundles are only homologically trivial by Leray–Hirsch, but the problem should be equally hard, I hope.)
I'm not sure this is true. Take $X=\mathbb{C}P^2$ and $h=4$. Then let $s=\Delta$ be the diagonal, and $t$ the composition
$t:\mathbb{C}P^2\xrightarrow{\Delta_3}\mathbb{C}P^2\times \mathbb{C}P^2\times\mathbb{C}P^2\xrightarrow{1\times m}\mathbb{C}P^2\times \mathbb{C}P^4$
where $m:([a_0,a_1],[b_0,b_1])\mapsto [a_0b_0,a_0b_1,a_1b_0,a_1b_1]$ is the Segre map and $\Delta_3$ is the three-fold diagonal. The conditions you ask for hold for these maps (certainly $a_0=0\Leftrightarrow a_0^2=0$), but if $x\in H^2\mathbb{C}P^2$ and $y\in H^2\mathbb{C}P^4$ are generators then
$(j\circ s)^*(1\times y)=\Delta^*(1\times x)=x$
and
$t^*(1\times y)=\Delta_3^*(1\times x\times 1+1\times 1\times x)=2x$.