Why $h^\sharp - k^\sharp$ factors through $\mathcal{I} = \ker(i^\sharp)$?

54 Views Asked by At

Let $i : Y \longrightarrow T$ be a closed immersion of ideal $\mathcal{I}$. Let $h, k : T \longrightarrow X$ be two morphisms of schemes such that $$ h \circ i = k \circ i. $$ I try to prove that $h^\sharp - k^\sharp$ factors through $\mathcal{I}$.

We have $$ 0 = h_*(i^\sharp) \circ h^\sharp - k_*(i^\sharp) \circ k^\sharp $$ and for an open $V$ in $X$ and $s \in \mathcal{O}_X(V)$, $$ 0 = i^\sharp_{h^{-1}V}(h^\sharp_V(s)) - i^\sharp_{k^{-1}V}(k^\sharp_V(s)). $$ I don't see how to use that.