Let $H=\textrm{Hilb}^r_P$ be the Hilbert scheme of closed subschemes of $\mathbb P^r$ with fixed Hilbert polynomial $P$. Let $X\subset \mathbb P^r$ be any subscheme. I am trying to characterize $\textrm{Hilb}^X_P$ as a subscheme of $H$. So I take any scheme $T$ and any $T$-family $\eta\in \underline{\textrm{Hilb}}^r_P(T)$, namely a flat morphism $$\eta\,\,\,\,\,:\,\,\,\,\,\mathbb P^r_T\supset \mathfrak X\to T.$$ This defines a $T$-valued point $q:=q_\eta:T\to H$. So I am trying to replace the question mark here below: $$\eta\in \underline{\textrm{Hilb}}^X_P(T)\,\,\,\,\,\iff\,\,\,\,\,?$$
A first thought is that for our family to be a $T$-family with respect to the functor $\underline{\textrm{Hilb}}^X_P$, it is necessary (and enough) that $\mathfrak X\subset X\times_kT$ (closed subscheme). But I found another answer, which I do not understand. It goes as follows (maybe someone can explain it to me). Letting $U\to H$ be the universal family, one has for $m>>0$ these morphisms on $H$: $$H^0(I_X(m))\otimes_kO_H\to H^0(O_{\mathbb P^r}(m))\otimes_kO_H=H^0(O_{\mathbb P^r_H}(m))\to H^0(O_U(m)).$$ If one denotes the composed morphism by $\alpha$, then $$?\,\,\,\,\,\textrm{should be replaced by} \,\,\,\,\, \textrm"q^\ast\alpha=0\textrm".$$
Can anyone explain to me the reason why $\mathfrak X\subset X\times_kT$ is equivalent to $q^\ast\alpha=0$?
Thank you in advance.