Projective epimorphisms onto the simples generate every finite-length object

The assertion to be proved by induction on the natural number $p$ is: for every object $Y$ of $\mathcal{C}$ of length $p$ there are integers $m_i\ge 0$ and an epimorphism ${⨁}_i{Q}_i^{m_i}\twoheadrightarrow Y$. At $p=0$ a composition series of $Y$ has no factors, so $Y=0$ by [F1]; taking all $m_i=0$, the empty biproduct is the zero object and the identity $0\to Y$ is an epimorphism by [F5], so the assertion holds at $p=0$.

Assume the assertion at the natural number $p$: for every object $Y$ of $\mathcal{C}$ with $\mathrm{\ell }(Y)=p$ there are integers $m_i\ge 0$ and an epimorphism ${⨁}_i{Q}_i^{m_i}\twoheadrightarrow Y$.

Suppose $X$ has length $p+1$ and let $0=X_0<\cdots <X_{p+1}=X$ be a composition series of $X$; then the last factor $S=X_{p+1}\mathrm{/}{X}_p$ is simple by [F1], hence $S\cong S_i$ for some $i$ by [F8], and $\mathrm{\ell }(X_p)=p$ by [F1].

Successor step. Let $X$ have length $p+1$, with composition series, last simple factor $S\cong S_i$ and $Y:=X_p$ of length $p$ as in step 1.3. By the induction hypothesis of step 1.2 applied to $Y$ there are integers $m_j\ge 0$ and an epimorphism $e:W\twoheadrightarrow Y$, where $W:={⨁}_j{Q}_j^{m_j}$ is a finite biproduct of the chosen projectives. The quotient $\pi :X\twoheadrightarrow X\mathrm{/}{X}_p=S$ of [F2] is an epimorphism, and composing the chosen epimorphism ${\phi }_i:Q_i\twoheadrightarrow S_i$ with an isomorphism $S_i\cong S$ gives an epimorphism $\phi :Q_i\twoheadrightarrow S$, so by the lifting property [F4] of the projective $Q_i$ there is $\psi :Q_i\to X$ with $\pi \psi =\phi$. Let $u:W\oplus Q_i\to X$ be the morphism with components the composite $W\stackrel{e}{\to }Y\rightarrowtail X$ and $\psi$, which exists and is unique by the biproduct property [F6].

The morphism $u$ of step 2.1 is an epimorphism. Let $c:X\to C$ be a morphism with $cu=0$. Composing with the first biproduct injection gives $c\circ u\circ {\mathrm{i}\mathrm{n}\mathrm{j}}_W=c\circ (\text{inclusion}\circ e)=0$, and $e$ is epic, so $c\circ \text{inclusion}=0$ for the inclusion $X_p\rightarrowtail X$; since $\pi$ is a cokernel of that inclusion by [F2], the universal property [F3] gives $c=\bar{c}\circ \pi$ for some $\bar{c}:S\to C$. Then $0=cu$ composed with the second biproduct injection gives $\bar{c}\circ \pi \circ \psi =\bar{c}\circ \phi =0$, and $\phi$ is epic, so $\bar{c}=0$ and therefore $c=0$. Since $\mathcal{C}$ is preadditive [F7], a morphism $u$ with the property that every $c$ satisfying $cu=0$ is zero is an epimorphism: from $gu=hu$ one gets $(g-h)u=0$, hence $g-h=0$.

The source of the epimorphism $u$ of step 3.1 is a finite biproduct of the chosen objects $Q_1,\dots ,Q_n$: regrouping its summands by index, $W\oplus Q_i\cong {⨁}_{i=1}^n{Q}_i^{m_i^{\mathrm{\prime }}}$ for integers $m_i^{\mathrm{\prime }}\ge 0$ by the associativity and commutativity of biproducts [F6]. Hence the successor case holds: $X$ of length $p+1$ admits an epimorphism ${⨁}_i{Q}_i^{m_i^{\mathrm{\prime }}}\twoheadrightarrow X$.

Step 1.1 is the base case and steps 1.3, 2.1, 3.1 and 4.1 pass from $p$ to $p+1$ using the induction hypothesis of step 1.2, so by induction on $p$ every object $X$ of finite length admits an epimorphism ${⨁}_i{Q}_i^{m_i}\twoheadrightarrow X$ for suitable integers $m_i\ge 0$. For the final clause put $P={⨁}_i{Q}_i$ and $m={\sum }_i{m}_i$; by [F6] the power $P^m$ is the biproduct ${⨁}_i{Q}_i^m$, whose subfamily of summands $Q_i^{m_i}$ has ${⨁}_i{Q}_i^{m_i}$ as a biproduct, and the corresponding projection $P^m\twoheadrightarrow {⨁}_i{Q}_i^{m_i}$ is a split epimorphism, hence epic by [F5]; composing it with an epimorphism onto $X$ gives an epimorphism $P^m\twoheadrightarrow X$ by [F5]. The proof selects only finite data (a composition series of the object at hand, finitely many summand indices and biproduct structure maps, and the $n$ supplied epimorphisms ${\phi }_i$), so no choice principle is used.