Projective covers in O are indecomposable and unique

By [F1] write $P=P_1\oplus \cdots \oplus P_n$ with each $P_i$ indecomposable. If every composite ${\pi }_i:P_i\hookrightarrow P\stackrel{\pi }{\to }L$ were zero, then $\pi ={\sum }_i{\pi }_i=0$, contradicting that $\pi$ is an epimorphism onto the nonzero object $L$; so some ${\pi }_j\mathrm{\ne }0$, and ${\pi }_j$ is an epimorphism because $L$ is simple and $0\mathrm{\ne }\mathrm{im}{\pi }_j\subseteq L$.

A direct summand of a projective is projective: if $P=P_j\oplus Q$, $q:E\twoheadrightarrow M$ is an epimorphism and $f:P_j\to M$ is a morphism, extend $f$ by zero on $Q$ to $F:P\to M$; by [F2] there is a lift $\tilde{F}:P\to E$ with $q\tilde{F}=F$, and its restriction to $P_j$ is a lift of $f$. Hence $P_j$ is projective.

Any two projective covers $(P,\pi )$ and $(P^{\mathrm{\prime }},{\pi }^{\mathrm{\prime }})$ of the same object $L$ are isomorphic: by [F2] applied to ${\pi }^{\mathrm{\prime }}$ there is $f:P\to P^{\mathrm{\prime }}$ with ${\pi }^{\mathrm{\prime }}f=\pi$, and applied to $\pi$ there is $g:P^{\mathrm{\prime }}\to P$ with $\pi g={\pi }^{\mathrm{\prime }}$. Then ${\pi }^{\mathrm{\prime }}(fg-{\mathrm{id}}_{P^{\mathrm{\prime }}})={\pi }^{\mathrm{\prime }}fg-{\pi }^{\mathrm{\prime }}={\pi }^{\mathrm{\prime }}-{\pi }^{\mathrm{\prime }}=0$, so $\mathrm{im}(fg-{\mathrm{id}}_{P^{\mathrm{\prime }}})\subseteq \ker {\pi }^{\mathrm{\prime }}$; from ${\mathrm{id}}_{P^{\mathrm{\prime }}}=fg-(fg-{\mathrm{id}}_{P^{\mathrm{\prime }}})$ it follows that $P^{\mathrm{\prime }}=\mathrm{im}(fg)+\ker {\pi }^{\mathrm{\prime }}$, and since $\ker {\pi }^{\mathrm{\prime }}$ is superfluous by [F3] we get $\mathrm{im}(fg)=P^{\mathrm{\prime }}$, so $fg$ is an epimorphism; symmetrically $gf$ is an epimorphism, and finite length makes each of these epimorphic endomorphisms injective: $\mathrm{\ell }(P)=\mathrm{\ell }(\ker (gf))+\mathrm{\ell }(P)$ forces $\mathrm{\ell }(\ker (gf))=0$, and similarly for $fg$. Thus $\ker f\subseteq \ker (gf)=0$ makes $f$ a monomorphism and an epimorphism, hence an isomorphism.

The epimorphism ${\pi }_j:P_j\twoheadrightarrow L$ of step 1.1 is essential in the sense of [F3]: if $Q\subseteq P_j$ is a proper subobject with $Q+\ker {\pi }_j=P_j$, then $Q$ and $\ker {\pi }_j$ are proper subobjects of the indecomposable projective $P_j$ whose sum is all of $P_j$, contradicting the proper-sum property of [F1] (note $\ker {\pi }_j\mathrm{\ne }{P}_j$ because $L\mathrm{\ne }0$). Hence $(P_j,{\pi }_j)$ is a projective cover of $L$.

A projective cover is indecomposable: if $P=P_1\oplus P_2$ with $P_i\mathrm{\ne }0$ and $\pi :P\twoheadrightarrow L$ essential, then not both components $\pi {\mathrm{\mid }}_{P_i}$ vanish, so some component is nonzero; a nonzero map to the simple object $L$ is an epimorphism, so $\pi (P_i)=L$, whence $P=P_i+\ker \pi$, and essentiality forces $P_i=P$, contradicting $P_2\mathrm{\ne }0$. Hence the endomorphism ring of a projective cover is local by [F1]. Moreover, if an indecomposable projective $P$ has head $L$, meaning its unique simple quotient is $L$, then the canonical epimorphism onto $P\mathrm{/}J(P)$ is essential because the unique maximal proper subobject $J(P)$ of [F1] contains every proper subobject; so such a $P$ is a projective cover of $L$, and step 1.3 makes any two of them isomorphic. Thus for each simple $L$ there is at most one isomorphism class of indecomposable projectives with head $L$, written $P(L)$ when it exists.