Implication Details

Claim: If a category is Cauchy complete and has kernel pairs and is preadditive, then it has kernels.

Proof: Let f:X→Yf : X \to Y be a morphism. Since ff has a kernel pair, the functor Cop→Set+,T↦{(u,v)∈Hom⁡(T,X)2:f∘u=f∘v}\C^{\op} \to \Set^+, \quad T \mapsto \{(u,v) \in \Hom(T,X)^2 : f \circ u = f \circ v\} is representable. Using w=u−vw = u-v, this functor is isomorphic to T↦{(u,w)∈Hom⁡(T,X)2:f∘w=0}.T \mapsto \{(u,w) \in \Hom(T,X)^2 : f \circ w = 0\}. The functor T↦{w∈Hom⁡(T,X):f∘w=0}T \mapsto \{w \in \Hom(T,X) : f \circ w = 0\} is a retract of this functor. Since the category is Cauchy complete, every retract of a representable functor is representable. Therefore, the kernel of ff exists.