Nonzero maps from an invertible sheaf to a locally free sheaf are injective

Setup. By [F1] the scheme $X$ is nonempty and irreducible, so by [F2] any two nonempty open subsets of $X$ meet; by [F3] the open sets on which $\mathcal{L}$ is trivial and the open sets on which $\mathcal{M}$ is free form two open covers of $X$, so for a given $x\in X$ we may choose an affine open $U_0\ni x$, intersect it with such a trivialising and such a freeing open set, and apply [F5] to obtain a distinguished open $D(h)\subseteq U_0$ containing $x$ on which both $\mathcal{L}$ and $\mathcal{M}$ are free; $D(h)$ is affine with ring a domain by [F1]. The resulting adapted charts $U=\mathrm{Spec}A$, with $A$ a domain, $\mathcal{L}{\mathrm{\mid }}_U\cong {\mathcal{O}}_U$ and $\mathcal{M}{\mathrm{\mid }}_U\cong {\mathcal{O}}_U^r$, cover $X$.

Chart dictionary. Fix an adapted chart $U=\mathrm{Spec}A$ and trivialisations $\mathcal{L}{\mathrm{\mid }}_U\cong {\mathcal{O}}_U$, $\mathcal{M}{\mathrm{\mid }}_U\cong {\mathcal{O}}_U^r$; these identify ${\mathrm{Hom}}_{{\mathcal{O}}_U}(\mathcal{L}{\mathrm{\mid }}_U,\mathcal{M}{\mathrm{\mid }}_U)$ with ${\mathcal{O}}_U(U{)}^r=A^r$, and $\phi {\mathrm{\mid }}_U$ corresponds to an element $m=(a_1,\dots ,a_r)\in A^r$ acting by multiplication; thus $\phi {\mathrm{\mid }}_U=0$ if and only if $m=0$. If $m\mathrm{\ne }0$, choose $i$ with $a_i\mathrm{\ne }0$ and let $D(f)\subseteq U$ be distinguished; $A$ is a domain and $A\to A_f$ is injective when $f\mathrm{\ne }0$ by [F4], so the image of $a_i$ in the domain $A_f$ is nonzero, and a section $s\in {\mathcal{O}}_U(D(f))=A_f$ with $s\cdot a_i=0$ must be $s=0$; as the distinguished opens cover every open subset of $U$, multiplication by $m$ is injective, that is, $\phi {\mathrm{\mid }}_U$ is injective.

Vanishing propagates. Suppose $\phi {\mathrm{\mid }}_{U_0}=0$ for one adapted chart $U_0$; let $V=\mathrm{Spec}B$ be any adapted chart and let $m_V\in B^s$ correspond to $\phi {\mathrm{\mid }}_V$ as in step 1.2. Since $V\cap U_0$ is nonempty by [F2], [F5] supplies a distinguished open $D(h)\subseteq V\cap U_0$; there $\phi {\mathrm{\mid }}_{D(h)}=0$, and restriction of the element $m_V$ to $D(h)$ is its image in $(B^s{)}_h$ by [F4], so that image is zero; the zero criterion of [F4] gives $h^k{m}_V=0$ in $B^s$ for some $k\ge 0$, and since $B$ is a domain and $D(h)\mathrm{\ne }\mathrm{\varnothing }$ forces $h\mathrm{\ne }0$, this yields $m_V=0$ and hence $\phi {\mathrm{\mid }}_V=0$ by step 1.2. The adapted charts cover $X$, so $\phi =0$ by [F7].

Stalks are injective. Now suppose $\phi \mathrm{\ne }0$. By the contrapositive of step 2.1, $\phi {\mathrm{\mid }}_U\mathrm{\ne }0$ for every adapted chart $U$; by step 1.2 the corresponding element $m$ is nonzero and $\phi {\mathrm{\mid }}_U$ is injective. Every point $x\in X$ lies in an adapted chart $U$, and the stalk map ${\phi }_x$ is the stalk of $\phi {\mathrm{\mid }}_U$ at $x$, hence is injective.

Conclusion. The kernel sheaf $\ker \phi$ is the sheaf of abelian groups with $(\ker \phi {)}_x=\ker ({\phi }_x)$ by [F6]; all these stalks are zero by step 3.1, so every section of $\ker \phi$ over every open set is zero and $\phi$ is injective; the argument made no use of the Axiom of Choice beyond the fixed data recorded in [F8].