Extension from subobjects of a generator detects injectivity

To extend a map $A\to I$ across a monomorphism $A\rightarrowtail B$, consider the set of pairs $(A^{\mathrm{\prime }},u^{\mathrm{\prime }})$ with $A\subseteq A^{\mathrm{\prime }}\subseteq B$ a subobject and $u^{\mathrm{\prime }}:A^{\mathrm{\prime }}\to I$ extending the original map. By [L3], the subobjects $A^{\mathrm{\prime }}\subseteq B$ form a set up to equivalence, and local smallness makes the morphisms $A^{\mathrm{\prime }}\to I$ a set, so this collection is a set. Order it by inclusion of subobjects.

If $T$ is a chain of such partial extensions, let $A_{\mathrm{\infty }}$ be the union of the subobjects in that chain inside $B$. Because [L2] gives AB5, this filtered colimit is again a subobject of $B$, and the compatible maps in the chain induce a morphism $u_{\mathrm{\infty }}:A_{\mathrm{\infty }}\to I$ extending the original map. Thus every chain has an upper bound.

By [L5], choose a maximal partial extension $(A_{\max },u_{\max })$.

Suppose $A_{\max }\mathrm{\ne }B$. By [L1], there exists a morphism $\psi :U\to B$ that does not factor through $A_{\max }$. Let $B_0=\mathrm{im}(\psi )$, let $N=A_{\max }\cap B_0$ inside $B$, and let $M={\psi }^{-1}(N)\subseteq U$. The composite $M\to N\to A_{\max }\stackrel{u_{\max }}{\to }I$ extends by hypothesis to a morphism $\chi :U\to I$.

Because $\ker (\psi )\subseteq M$, the map $\chi$ vanishes on $\ker (\psi )$ and therefore factors through $B_0=\mathrm{im}(\psi )$; write the factor map as $u_0:B_0\to I$. The maps $u_{\max }:A_{\max }\to I$ and $u_0:B_0\to I$ agree on $N=A_{\max }\cap B_0$, so they glue to a map $A_{\max }+B_0\to I$ extending $u_{\max }$. Since $\psi$ does not factor through $A_{\max }$, the subobject $A_{\max }+B_0$ is strictly larger than $A_{\max }$, contradicting maximality. Therefore $A_{\max }=B$.

Every map across a monomorphism extends, so $I$ is injective by [L4].