# Is the morphism labeled by red rectangle injective?

Is the morphism labeled by red rectangle injective?

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, $$\mathrm{Spec}(k[x]/(x)) \hookrightarrow \mathbb{A}^1_k$$.