Karoubi envelope

Remarks

The condition in clause 2 is equivalent to $f \circ e = f$ and $e' \circ f = f$. The functor $\mathcal{C} \to \mathrm{Kar}(\mathcal{C})$ with $A \mapsto (A, \id_A)$ is fully faithful. Every idempotent of $\mathrm{Kar}(\mathcal{C})$ splits. For example, an idempotent $e$ of $(A, \id_A)$ is the composite of the retraction $e \colon (A, \id_A) \to (A, e)$ and the section $e \colon (A, e) \to (A, \id_A)$. Other names are idempotent completion and Cauchy completion.

Example: let $\mathcal{C}$ have one object $*$ and morphisms $1$ and $e$, with $e \circ e = e$. Then $\mathrm{Kar}(\mathcal{C})$ has the objects $(*, 1)$ and $(*, e)$. The morphisms $(*, e) \to (*, e)$ are the $f$ with $e f e = f$, and only $f = e$ qualifies, so $(*, e)$ has only its identity $e$.

Non-example: with the same $\mathcal{C}$, the morphism $1 \colon (*, e) \to (*, e)$ is not a morphism of $\mathrm{Kar}(\mathcal{C})$. Indeed $e \circ 1 \circ e = e \ne 1$, so clause 2 fails.