For any nonempty set $X$, the endofunctor $- \times X$ of $\mathbf{Set}$, which is left adjoint to $Hom_{\mathbf{Set}}(X, -)$, is faithful and reflects isomorphisms.
More generally, given any object $X$ of a cocomplete category $C$, the functor $Hom_{C}(X, -)$ has a left adjoint $F$ given by copowers of $X$.
In this context, is it true that the following three statements are equivalent?
- The left adjoint $F$ is faithful.
- The left adjoint $F$ is conservative (reflects isomorphisms).
- The unique morphism from the initial object $0$ to $X$ is not an epimorphism.
Here's a full proof (that incidentally dismisses my "counterexample" from the comments, as you said)
i) Suppose $F$ is conservative, let's prove that it's faithful. Let $f,g: A\to B$ be set maps. Then $F(\mathrm{coeq}(f,g)) = \mathrm{coeq}(F(f),F(g))$ as $F$ is a left adjoint and so preserves colimits. In particular, if $F(f)=F(g)$, then $\mathrm{coeq}(F(f),F(g))$ is an isomorphism, and so by conservativity, so is $\mathrm{coeq}(f,g)$. It follows that $f=g$.
ii) Suppose $F$ is faithful and let's prove that $0\to X$ is not an epimorphism. Let $f,g: 1\to 2$ be the two distinct maps of sets, and $h:0\to 1$ the unique map. then $f\circ h = g\circ h$.
Thus $F(f)\circ F(h) = F(g)\circ F(h)$. $F(h) : 0\to X$ is the unique map, so if it were an epimorphism, we'd have $F(f)=F(g)$, so by faithfulness $f=g$, a contradiction. So $0\to X$ is not an epimorphism.
iii) Suppose $0\to X$ is not an epimorphism. Let's prove that $F$ is conservative. We do this in 2 steps : first we prove that $F$ reflects epimorphisms, then that it reflects monomorphisms. As any isomorphism is a bimorphism, it follows that if $F(f)$ is an isomorphism, $f$ is a bimorphism, so an isomorphism ($\mathbf{Set}$ is balanced). So if we prove this we will be done.
So let $f:A\to B$ be such that $F(f)$ is an epimorphism. Let $b\in B$ and assume that there is no $a$ such that $f(a) = b$.
Let moreover $p,q: X\to Y$ be two distinct arrows agreeing on $0\to X$ (they exist by assumption).
Then we get $$\coprod_{a\in A}X \coprod 0\to \coprod_{c\in B\setminus\{b\}}X \coprod X_b \rightrightarrows Y$$
Where the two arrows agree on $\coprod_{c\in B\setminus\{b\}}X $ (say fixed to $p$ on each coordinate) and are respectively $p,q$ on $X_b$
The two maps we get agree on the $\coprod_A X$ factor and on the $0$ factor, so they agree on everything. But the middle map is essentially $F(A)\to F(B)$ which is assumed to be an epimorphism, so that the two end-maps $\coprod_{c\in B\setminus\{b\}}X \coprod X_b \rightrightarrows Y$ must be equal. Looking at the $X_b$ coordinate, we find that $p=q$, which is absurd.
Therefore $f$ is surjective, so $F$ reflects epimorphisms.
Somewhat surprisingly, I will use this to prove that it reflects monomorphisms.
Indeed consider the unit $\eta$ of the adjunction : $\eta_A: A\to \hom(X, \coprod_A X)$ which of course sends $a$ to the "inclusion" $i_a: X\to \coprod_AX $. The claim is that this is injective.
If we prove this, then assuming $F(f)$ is mono, as $G=\hom(X,-)$ preserves monos (it is a right adjoint) we get that $GF(f)$ is mono and so we have a commutative square
$\require{AMScd}\begin{CD}A @>{f}>> B \\ @V{\eta_A}VV @V{\eta_B}VV \\ GF(A) @>{GF(f)}>> GF(B)\end{CD}$ with $GF(f)\circ \eta_A$ injective, so that $f$ must be too.
To prove that $\eta_A$ is injective, assume $i_a = i_b$, and let $C=A/\sim$ where $a\sim b$ and nothing else. We have a projection map $A\to C$ of course but it has a section that sends $\{a,b\}$ to (say) $a$. Call this $s:C\to A$. Then the claim is that $F(s)$ is epi.
Let $m,n : F(A)\to Z$ be two maps that agree on $F(s)$. Then the thing is that any $i_e: X\to \coprod_A X$, $e\in A$ factors through $F(C)$. For $e\neq b$ it is clear cause $e$ is in the image of $C$, and for $e=b$ it follows from $i_a = i_b$, so it suffices to factor $i_a$, and that we know how to do.
Therefore, $m\circ i_e = n\circ i_e$ for any $e\in A$, so $m=n$ (universal property of coproduct). So $F(s)$ is an epimorphism, and therefore as $F$ reflects epimorphisms, so is $s$. But this implies that $b$ is in the image so $a=b$.
Therefore $\eta_A$ is indeed injective, and we are done.