Lemma 23.1.2. Let \(p\colon E \to C\) be a functor. Consider the functor \[ Q\colon \Ar (E) \longrightarrow \Ar (C) \times _{s,C,p} E, \qquad (\phi \colon e \to e') \longmapsto (p\phi ,e). \] An arrow \(\phi \colon e\to e'\) satisfying \(Q(\phi )=(f,e)\) is \(p\)-cocartesian if and only if, equipped with the identity map of \((f,e)\), it is a left adjoint object to \((f,e)\) under \(Q\).
Proof. Set \(B:=\Ar (C)\times _{s,C,p}E\). Since \(Q(\phi )=(f,e)\), the identity of \((f,e)\) exhibits \(\phi \) as a left adjoint object under \(Q\) if and only if, for every arrow \(\psi \colon a\to b\) in \(E\), the induced map \[ Q_*\colon \Hom _{\Ar (E)}(\phi ,\psi )\longrightarrow \Hom _B((f,e),Q(\psi )) \] is an equivalence. By Axiom D, Chapterexercise 1.4, its source and target admit canonical descriptions \begin {align*} \Hom _{\Ar (E)}(\phi ,\psi ) &\iso \Hom _E(e,a)\times _{\Hom _E(e,b)}\Hom _E(e',b), \\ \Hom _B((f,e),Q(\psi )) &\iso \Hom _E(e,a)\times _{\Hom _C(pe,pb)}\Hom _C(pe',pb). \end {align*}
Here the maps out of \(\Hom _E(e,a)\) are given by postcomposition with \(\psi \) in the first line and by postcomposition with \(p(\psi )\) after applying \(p\) in the second line. Under these identifications, \(Q_*\) is induced by \(p\colon \Hom _E(e',b)\to \Hom _C(pe',pb)\) on the second factor.
If \(\phi \) is \(p\)-cocartesian, then for every \(b\) the defining pullback square identifies \[ \Hom _E(e',b) \longrightarrow \Hom _E(e,b)\times _{\Hom _C(pe,pb)}\Hom _C(pe',pb) \] as an equivalence. Pulling this equivalence back along the postcomposition map \(\psi \circ -\colon \Hom _E(e,a)\to \Hom _E(e,b)\) shows that \(Q_*\) is an equivalence for every \(\psi \). Thus \(\phi \) is a left adjoint object under \(Q\).
Conversely, suppose that \(\phi \) is a left adjoint object under \(Q\). Taking \(\psi =\id _b\) identifies \(Q_*\) with the preceding comparison map. It is therefore an equivalence for every object \(b\) of \(E\), which says precisely that the squares defining \(p\)-cocartesianness of \(\phi \) are pullback squares. โก
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