Can an object be a proper direct summand of itself?

Question

Let $\mathcal{A}$ be an additive category, $A,C$ two objects in it. If $A\oplus C\cong A$, is it true that $C=0$?

It seems rather clear, but I am not finding anything on the web and can't really understand how to prove/disprove it.

Answer 1

Consider category of Abelian groups, and let $C=\Bbb Z$ and $A=\Bbb Z\oplus\Bbb Z\oplus\cdots$ (the free Abelian group of countable rank).

Answer 2

Considering the title of your question whether an object can be a direct summand of itself, the answer is yes.

Let $A$ be an abelian group for example. Denoting the trivial subgroup $\{e_A\} = 0$ there is a canonical isomorphism

$$A\oplus 0 \to A,\ (a,0)\mapsto a$$

which implies $$A\oplus 0 \cong A.$$

Answer 3

Assuming that $A \oplus C \cong A$ for some $A$ and $C$ in $\mathcal{A}$, it happens that $C \cong 0$ precisely when the isomorphism $A \oplus C \cong A$ is one of the two couples projection-coprojection exhibiting $A \oplus C$ as a product and a coproduct of $A$ and $C$.

$(\Rightarrow)$ If $C \cong 0$ then one indeed verifies that $p_1 := (A\oplus C\cong A)$ and $p_2:=(A\oplus C \to 0 \cong C)$ are projections and $i_1:=(A \cong A \oplus C)$ and $i_2 :=(C \cong 0 \to A \oplus C)$ are coprojections.

$(\Leftarrow)$ Suppose that the projection $p_1 \colon A \oplus C \to A$ is an isomorphism, i.e. $i_1 \colon A \to A \oplus C$ is not just a section of $p_1$ but also a rectraction, that is, an actual inverse. Let $B$ be an object of $\mathcal{A}$. Being $A \oplus C$ a product, the map: $$F \colon \mathcal{A}(B, A \oplus C)\ni f \mapsto (p_1f,p_2f)\in\mathcal{A}(B, A) \times \mathcal{A}(B, C) $$ is bijective. Let $F^{-1}$ be its inverse and observe that it sends a couple $(g,h)$ to the arrow $i_1g \in \mathcal{A}(B, A \oplus C)$ (for $i_1g$ is the unique arrow $f\in \mathcal{A}(B, A \oplus C)$ such that $p_1f=g$). Now, let us consider $g \in \mathcal{A}(B,A)$, which exists since $\mathcal{A}(B,A)$ is nonempty (for it is an abelian group). Whenever $h_1,h_2 \in \mathcal{A}(B,C)$, it is the case that: $$(g, h_1)=(FF^{-1})(g,h_1)=F(i_1g)=(FF^{-1})(g,h_2)=(g,h_2)$$ which implies that $h_1=h_2$. That means that $\mathcal{A}(B,C)$ is a singleton and $C$ is terminal. Analogously, using that $A \oplus C$ is a coproduct, one gets that $\mathcal{A}(C,B)$ is a singleton as well and this concludes the argument. Observe that there one would use that $i_1p_1$ is an identity, while for this half we have only used that $p_1i_1$ is an identity.