First assume that $x$ is either (1) initial or (2) final. In both cases, it follows that $Mor(x,x)$ is a trivial abelian group containing $\text{id}_x$, thus $\text{id}_x = 0$ in $Mor(x,x)$, which shows that each of (1) and (2) implies (3).

Now assume that $\text{id}_x = 0$ in $Mor(x,x)$. Let $y$ be an arbitrary object of $\mathcal{A}$, $f \in Mor(x,y)$ and $C: Mor(x,x) \times Mor(x,y) \rightarrow Mor(x,y)$ the composition map. Since $C$ is bilinear, it holds that \begin{equation} f = C(0,f) = C(0+0,f) = C(0,f) + C(0,f) = f + f, \end{equation} which yields $f = 0$. Hence $x$ is initial in $\mathcal{A}$. A similar argument for $f \in Mor(y,x)$ can be used to show that $x$ is also final. Thus (3) implies both (1) and (2). $\Box$

]]>Given a pair of objects $x,y$ in a pre-additive category $\mathcal{A}$, the \emph{direct sum} $x \oplus y$ of $x$ and $y$ is the direct product $x \times y$ endowed with the morphisms $i,j,p,q$ as in Lemma \ref{lemma-preadditive-direct-sum} above.

]]>Line 6 of the claim:
Replace
$(A_1)*{\mathfrak m_1}$
by
$(A_1)*{\mathfrak m_i}$

Let $R = \mathbb{Q}[X_n | n \in \mathbb{N}]$ be the polynomial ring in infinitely many variables over $\mathbb{Q}$. Let $I$ be the ideal generated by ${X_n^n | n \in \mathbb{N}}$ and $S = R/I$. Then the ideal $J \subset S$ generated by the image in $S$ of the set ${X_{n + 1} | n \in \mathbb{N}} \subset R$ is locally nilpotent, but not nilpotent. Indeed, since $S$-linear combinations of nilpotents are nilpotent, to prove that $J$ is locally nilpotent it is enough to observe that all its generators are nilpotent (which they obviously are). On the other hand, for each $n \in \mathbb{N}$ it holds that $X_{n+1}^n \neq 0$, so that $J^n \neq 0$. It follows that $J$ is not nilpotent.

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