Proof. Consider the Yoneda embedding \[ C \, \hookrightarrow \, \Fun (C\catop ,\An ), \qquad Y \, \mapsto \, \Hom _C(-,Y). \] This lands in left exact functors, so by the equivalence \(\Fun ^{\ex }(C\catop ,\Sp ) \iso \Fun ^{\lex }(C\catop ,\An )\) we may uniquely lift it to a (fully faithful) functor \[ C \, \hookrightarrow \, \Fun ^{\ex }(C\catop , \Sp ), \qquad Y \, \mapsto \, \hom _C(-,Y). \] The resulting functor \(\hom _C\colon C\catop \times C \to \Sp \) is exact in the first variable by construction. It remains to show it is also exact in the second variable: for fixed \(X\), we need to show that the functor \(\hom _C(X,-)\colon C \to \Sp \) is exact. It suffices to check this after composing with the functors \((-)_n\colon \Sp \to \An \), which are left exact and jointly conservative. Since we have \(\hom _C(X,Y)_n \simeq \hom _C(X[-n],Y)_0 \simeq \Hom _C(X[-n],Y)\), this follows from the left exactness of \(\Hom _C(X[-n],-)\colon \Sp \to \An \). โก
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