Let $F\colon \mathcal{C}\to \mathcal{D}$ be a functor.

  1. Characterisations. The following conditions are equivalent:
    1. The functor $F$ is conservative.
    2. For each $f\in \textup{Mor}\webleft (\mathcal{C}\webright )$, the morphism $F\webleft (f\webright )$ is an isomorphism in $\mathcal{D}$ iff $f$ is an isomorphism in $\mathcal{C}$.
  2. Interaction With Fully Faithfulness. Every fully faithful functor is conservative.
  3. Interaction With Precomposition. The following conditions are equivalent:
    1. For each $\mathcal{X}\in \text{Obj}\webleft (\mathsf{Cats}\webright )$, the precomposition functor
      \[ F^{*} \colon \mathsf{Fun}\webleft (\mathcal{D},\mathcal{X}\webright ) \to \mathsf{Fun}\webleft (\mathcal{C},\mathcal{X}\webright ) \]

      is conservative.

    2. The equivalent conditions of Item 5 of Proposition 9.6.1.1.2 are satisfied.

Item 1: Characterisations
This follows from Item 1 of Proposition 9.5.1.1.6.
Item 2: Interaction With Fully Faithfulness
Let $F\colon \mathcal{C}\to \mathcal{D}$ be a fully faithful functor, let $f\colon A\to B$ be a morphism of $\mathcal{C}$, and suppose that $F_{f}$ is an isomorphism. We have

\begin{align*} F\webleft (\text{id}_{B}\webright ) & = \text{id}_{F\webleft (B\webright )}\\ & = F\webleft (f\webright )\circ F\webleft (f\webright )^{-1}\\ & = F\webleft (f\circ f^{-1}\webright ). \end{align*}

Similarly, $F\webleft (\text{id}_{A}\webright )=F\webleft (f^{-1}\circ f\webright )$. But since $F$ is fully faithful, we must have

\begin{align*} f\circ f^{-1} & = \text{id}_{B},\\ f^{-1}\circ f & = \text{id}_{A}, \end{align*}

showing $f$ to be an isomorphism. Thus $F$ is conservative.


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