A functor $F \colon C\to D$ is conservative if it is “isomorphism-reflecting”, i.e. if $g \colon a\to b$ is a morphism in $C$ such that $F(g)$ is an isomorphism in $D$, then $g$ is an isomorphism in $C$.
Sometimes (e.g. in the Elephant) conservative functors are assumed to be faithful as well. If $C$ has, and $F$ preserves, equalizers, then conservativity implies faithfulness.
See conservative morphism for a generalization to an arbitrary 2-category.
Every fully faithful functor, and more generally any pseudomonic functor, is a conservative functor.
Conversely, every faithful conservative functor is pseudomonic.
But further would-be converses of Exp. fail: not every conservative functor is full or faithful:
An example of a functor that is conservative but not full is the inclusion of the groupoid core $Core(\mathcal{C}) \longrightarrow \mathcal{C}$ of a category $\mathcal{C}$ that is itself not a groupoid, into that category.
An example of a functor that is conservative but not faithful is the unique functor from any groupoid with two distinct isomorphisms $f, g : x \to y$ to the terminal groupoid. \end{example}
Every monadic functor is a conservative functor (see also at monadicity theorem):
For a $T$-algebra homomorphism given by an invertible morphism $f \colon A \to B$, the inverse $f^{-1} \colon B \to A$ is easily seen to also be a $T$-algebra homomorphism.
A functor $F \colon C \to D$ is conservative if and only if $F(f)$ being an identity morphism implies that $f$ is an isomorphism.
Let $C$ be a category with pullbacks. Given any morphism $f \colon X \longrightarrow Y$ in $C$ write
for the functor of pullback along $f$ between slice categories (“base change”). If strong epimorphisms in $\mathcal{C}$ are preserved by pullback, then the following are equivalent:
$f$ is a strong epimorphism;
$f^\ast$ is conservative.
(e.g. Johnstone, lemma A.1.3.2)
When $C$ and $D$ are pretoposes, a pretopos morphism $F \colon C \to D$ is conservative if and only if for every object $c \in C$, the induced map between subobject lattices $F^{(c)} : \operatorname{Sub}(c) \to \operatorname{Sub}(F(c))$ is injective.
A conservative functor $F \colon C \to D$ reflects all limits and colimits that it preserves and which exist in the source category.
We discuss the case of limits (the argument for colimits is formally dual):
Let $K \colon J \to C$ be a diagram in $C$ whose limit $\lim K$ exists and such that $\lim (F\circ K) \,\simeq\, F (\lim K)$. Now if $const_c \to K$ is a cone in $C$ that is sent to a limiting cone $F const_c$ in $D$, then by the universal property of the limit in $D$ the morphism $F( c \to \lim K)$ is an isomorphism in $D$, hence must have been an isomorphism in $C$ (by the assumption that $F$ is conservative), hence $const_c$ must have been a limiting cone in $C$.
Geun Bin Im, Gregory Maxwell Kelly, Some remarks on conservative functors with left adjoints, J. Korean Math. Soc. 23 (1986), no. 1, 19–33, MR87i:18002b, pdf; On classes of morphisms closed under limits, J. Korean Math. Soc. 23 (1986), no. 1, 1–18, Adjoint-triangle theorems for conservative functors, Bull. Austral. Math. Soc. 36 (1987), no. 1, 133–136, MR88k:18005, doi
For an example of a conservative, but not faithful, functor $f: A\to Set$ having a left adjoint see Example 2.4 in:
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