For any morphism of presheaves, if all maps of sections are injective, then all stalk maps are injective.

Question

This is a remark in Vakil's notes of Algebraic Geometry.

For any morphism of presheaves, if all maps of sections are injective, then all stalk maps are injective.

Before this remark, I just showed that this works for sheaves. I think the same proof should work for presheaves. But I am not sure whether it used the identity axiom somehow.

Here is my proof.

Suppose $\phi_p(f_{1p},U)= \phi_p(f_{2p},U)=(g_p,U)$ for some $p\in U$ and $(f_{1p},U)\ne(f_{2p},U)$. Then $f_{1p}\ne f_{2p}$ in any neighborhood of $p$. Suppose they are induced from $\phi(U_{p1})(f_1)=\text{res}_{U,U_{p1}}g$ and $\phi(U_{p2})(f_2)=\text{res}_{U,U_{p2}}g$ for some $U_{p1}, U_{p2}$ containing $p$. Then there exist some $U_p$, such that $\phi(U_p)(\text{res}_{U_{p1},U_p}f_1)=\phi(U_p)(\text{res}_{U_{p2},U_p}f_2)$, By the injectivity, $\text{res}_{U_{p1},U_p}f_1=\text{res}_{U_{p2},U_p}f_2$. This is a contradiction.

Could someone check my proof? Thank you very much!

Answer 1

Yes the proof is absolutely right. For a fact , injectivity is an easier thing to deal with. The main part where sheaves and presheaves create a difference is the issue of surjectivity and thats where sheafification comes into play.