Do Adjoint Functors preserve mono-sources? A functor $F: \mathfrak{C} \to \mathfrak{D}$ is called adjoint if for any $D \in \mathfrak{D}$ there exist an object $C \in \mathfrak{C}$ and a morphism $f: D \to F(C)$ such that whenever we consider $C' \in \mathfrak{C}$ and $f: D \to F(C')$ we can find a unique morphism $\hat f: C \to C'$ in $\mathfrak{C}$ such that $f'=F(\hat f) \circ f$.
Now, I want to show that $F$ preserves mono-sources, namely if $\mathcal{S}=(m_i: A \to A_i)_{i \in I}$ is a mono-source ($\mathcal{S} \circ r=\mathcal{S} \circ s \implies r=s$), then $F(\mathcal{S})=(F(m_i): F(A) \to F(A_i))_{i \in I}$ is a mono-source.
Here there is my attempt of proof:
I consider a mono-source $\mathcal{S}=(m_i: A \to A_i)_{i \in I}$ in $\mathfrak{C}$. Let $r,s: B \to F(A)$ be such that $F(\mathcal{S}) \circ r=F(\mathcal{S}) \circ s$. As $F$ is adjoint, there exist an object $C \in \mathfrak{C}$ and a morphism $f: B \to F(C)$ as in the definition. By our choice of $r$ and $s$, there exist $\hat r,\hat s: C \to A$ such tht $r=F(\hat r) \circ f$ and $s=F(\hat s) \circ f$.
At this point, we get the equalities
$$
F(\mathcal{S}) \circ r=F(\mathcal{S}) \circ F(\hat r) f=F(\mathcal{S} \circ \hat r) \circ f
$$
and
$$
F(\mathcal{S}) \circ r=F(\mathcal{S}) \circ F(\hat s) f=F(\mathcal{S} \circ \hat s) \circ f,
$$
whence the equality $F(\mathcal{S} \circ \hat r) \circ f=F(\mathcal{S} \circ \hat s) \circ f$. I would like to conclude that
$$
\mathcal{S} \circ \hat r=\mathcal{S} \circ \hat s
$$
but I have no clue.
In this way, using the assumption that $\mathcal{S}$ is a mono-source in $\mathfrak{C}$, then I would get $\hat r=\hat s$ and hence $r=F(\hat r) \circ f=F(\hat s) \circ f$.
 A: As I said in my comment, an easy translation of the property of being an "adjoint" for a functor $F$ is that for every object D of its codomain, the comma category $(D/F)$ has an initial object.
Note that this condition can be very useful: each functor with that property is final, which is good to know sometimes.
The two notions (finality and being an adjoint) are related, but I can't think of a sharp theorem right now (I reserve myself the right to come back at this later). For the moment observe that every right adjoint functor is final. I'd try to ask myself if an "adjoint" functor is a weak form of being a right adjoint, or something like that.
So far, this is just context. Coming to your question instead, you're in the following situation: let's consider a "mono-source" $\cal S$ (modern terminology: a "jointly monic" family ${\cal S} = \{s : A_s\to B_s\}$ of arrows). You want to show that the family $F{\cal S} = \{Fs : FA_s\to FB_s\}$ is also jointly monic.
To this end, consider two arrows $r,t : E\to FA_s$ such that $Fs\circ r = Fs\circ t$; since $F$ is adjoint, there exist unique arrows $r',t'$ such that $r=Fr'\circ z, t=Ft'\circ z$ for the initial object $z : E\to FU$ in the comma category $(E/F)$. So you're left with the equality
$$
F(s\circ r')\circ z = F(s\circ t')\circ z
$$ valid for every $s\in\cal S$. Now, this common composition is a certain map $E\to FB_s$, which must be $Fu\circ z$ for a unique map $u$. So, $s\circ r'=s\circ t'$. This concludes.
