On the <principal open subscheme> $D(a)\subseteq\operatorname{Spec}A$, the induced <morphism of sheaves> is the <ring homomorphism>
$$
A_a\longrightarrow B_{\alpha(a)},\qquad \frac{b}{a^n}\longmapsto\frac{\alpha(b)}{\alpha(a)^n}.
$$
If $\alpha$ is <injective>, this map is injective by <exactness of localization>: it is the localization of the injective $A$-module map $A\to B$. Explicitly, if the displayed fraction vanishes, some power of $\alpha(a)$ annihilates $\alpha(b)$; injectivity gives $a^mb=0$, so the original fraction also vanishes. The principal opens form an <open basis>, so the <kernel sheaf> is zero on a basis and hence zero everywhere. Conversely, an injective <morphism of sheaves> induces an injection on <global sections>. On the whole <affine scheme> these sections are $A\to B$, exactly $\alpha$. Thus
$$
\boxed{\alpha\text{ is injective}\ \Longleftrightarrow\ \phi\text{ is injective}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 113 / 1 / i / Solution

{title2=$\ker\phi$}

For a <morphism of sheaves> of abelian groups or modules, the kernel sheaf has sections $\ker(\mathcal F(U)\to\mathcal G(U))$. These already satisfy the <sheaf gluing axiom>. Its <stalk> at $x$ is the kernel of $\mathcal F_x\to\mathcal G_x$, since filtered colimits of modules preserve exactness. A sheaf morphism is injective exactly when its kernel sheaf is zero.

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 Incoming links: Kernel sheaf