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This question should be very basic, which jibes with the fact that I continue to be a beginner in this subject. Here goes: If a functor is not full, or not faithful, can it have a left (or right) adjoint that is full or faithful.

For context, full and faithful describe functors for which the corresponding maps of morphisms are surjective and injective respectively.

It looks like this question might be quite similar.

This question occurred to me when thinking about the group of units functor and its left adjoint, the group ring functor, with $\mathbb Z$ as the ring, over here.

Looking at this answer, it seems the answer is quite straight forward when one of the functors is full and faithful, since then we have an isomorphism of Homs.

Update: that last bit turned out to be false. However, I noticed while reading that if we further require our functor $F:\mathcal C\to\mathcal D $ be essentially surjective, we get an equivalence of categories. These notions are about what one would expect, but I won't define them here.

Arctic Char
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    For one part of the question, the inclusion of abelian groups into groups is full and faithful, but its left adjoint (the abelianization functor) is not faithful. – Daniel Schepler May 21 '21 at 21:51
  • Thanks @DanielSchepler That was easy. Nice example. –  May 21 '21 at 21:53
  • If $G,H$ are abelian groups and $\phi : G \to H$ is a morphism of groups, then the underlying set map $|G| \to |H|$ also induces a morphism of abelian groups. – Daniel Schepler May 21 '21 at 23:08
  • It appears to be a subtle point, but since the functor is the inclusion, fullness follows easily from the fact that we only need to look at morphisms of groups that are in fact abelian. @DanielSchepler correct me –  May 22 '21 at 00:10

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An algebraic example was given in the comments, so here's a topological counterexample.

The forgetful functor $U$ from the category $\mathcal{Top}$ of topological spaces and continuous maps to the category $\mathcal{Set}$ of sets has a left adjoint $D$ which takes a set to the discrete space on that set.

$D$ is fully faithful, since a continuous map between discrete spaces is the same as a function between the underlying sets. More formally, given two sets $X$ and $Y$, every map between $D(X)$ and $D(Y)$ is equal to $D(f)$ for a unique function $f \colon X \to Y$.

However, $U$ is not fully faithful. Given topological spaces $X$ and $Y$, there are (in general) far fewer continuous functions between $X$ and $Y$ than there are functions between $U(X)$ and $U(Y)$.

S.C.
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