Is the morphism labeled by red rectangle injective?

Question

Is the morphism labeled by red rectangle injective?

Answer 1

No it is not injective in general. For simplicity let's take $\mathcal{F} = \mathcal{O}_X$ and consider schemes instead of general ringed spaces. So we're looking at the equation above the one you boxed. Let $\mathcal{J}$ be a sheaf of ideals on $Y$, with an injective morphism of sheaves $\mathcal{J} \to \mathcal{O}_Y$. If the pulled-back morphism $$ f^* \mathcal{J} \to f^* \mathcal{O}_Y = \mathcal{O}_X $$ was always injective, then $f^*$ would be a left-exact functor and hence exact, but this is true if and only if $f$ is a flat morphism.

Thus to find a counterexample we need to find a non-flat morphism of schemes. The first example I thought of was the inclusion of the origin into the affine line, $\mathrm{Spec}(k[x]/(x)) \hookrightarrow \mathbb{A}^1_k$.