I am confused by the proof of Lemma 26.4.6:
Lemma 26.4.6: Let $X,Y$ be schemes. Let $I\subset O_Y$ be a quasi-coherent sheaf of ideals. Let $i: Z\to Y$ be the associated closed subscheme. A morphism $f: X\to Y$ factors through $Z$ if and only if the map $f^* I\to f^* O_Y =O_X$ is zero.
The proof of this lemma claims that the composition $I_{f(x)}\to O_{Y,f(x)} \to O_{X,x}$ is zero by assumption. And here is my question:
Does $f^* I\to f^* O_X =O_Y$ be zero implies that $f^{-1} I\to f^{-1} O_X \to O_Y$ is zero?
To be more specific, for $\mathcal{F}\in Mod(O_X)$ and $\mathcal{G}\in Mod(O_Y)$, is there some error in the following proof?
$f^*\mathcal{G} \to \mathcal{F}$ is zero in $Mod(O_X)$
$\Leftrightarrow$ $\mathcal{G} \to f_*\mathcal{F}$ is zero in $Mod(O_Y)$
$\Leftrightarrow$ $\mathcal{G} \to f_*\mathcal{F}$ is zero in $Sh(Y)$
$\Leftrightarrow$ $f^{-1}\mathcal{G} \to \mathcal{F}$ is zero in $Sh(X)$
But this is strange to me since $f^*\mathcal{G} = f^{-1}\mathcal{G} \otimes_{f^{-1}O_Y} O_X$.