If $F$ is a functor such that $F(eq(f,g)) = eq(F(f), F(g))$ and $F(f)$ is an isomophism, then $F$ is faithful 
Suppose that $\mathcal C$ has equalizers of every pairs of morphisms. If $F: \mathcal C  \to \mathcal D$ is a functor such that $F(eq(f,g)) = eq(F(f), F(g))$ and $F(f)$ is an isomophism, then $F$ is faithful.

I am not sure how to approach this question. If $F(f)=F(g)$ then $eq(F(f),F(f)) = F(eq(f,g))$. It seems to me that I must use the fact that $F(f)$ is an isomorphism to "get rid of" the $F$, but I am not sure how to do this, much less gettting a graphical intuition of this.
Any help would be greatly apporeciated.
 A: You may mean this:

Let $F:\scr C\to\scr D$ be a functor between categories.
  If $\scr C$ has equalizers, $F$ preserves them and reflects isomorphisms, then $F$ is faithful.

Let $f,g:X\to Y$ be a pair of parallel arrows in $\scr C$ and assume $F(f)=F(g)$.
Then the equalizer of $F(f)$ and $F(g)$ is an isomorphism.
Since $F$ preserves equalizers and reflects isomorphisms, it follows that the equalizer of $f$ and $g$ is an isomorphisms as well.
Consequently $f=g$, hence $F$ is faithful.
A: Fleshing out Fabio's solution:


*

*The “equaliser” of $x$ and $y$ is the best $e$ with $x ∘ e = y ∘ e$,
in that any other arrow, $d$, with this property uniquely factors through it, $d = e ∘ u$.
Intuitively the equaliser of $f, g$ is the largest sub-part of their
source on which they are identified.

*If $x = y$ then their equaliser is the identity, which is an isomorphism.
Intuitively the largest sub-part of $f$'s source that identifies it with
itself is just the whole source.

*If the equaliser of $x$ and $y$ is an isomorphism, then they are equal.
   x = y
⇐  x ∘ Id = y ∘ Id
⇐  x ∘ e ∘ e⁻¹ = y ∘ e ∘ e ⁻¹
⇐  x ∘ e = y ∘ e
⇐ true, since e equalises x and y


*Hence, by 2 & 3, the equaliser of $x$ and $y$ is an isomorphism precisely when $x = y$.

Now we show

If $F$ preserves equalisers, where the source category has all equalisers, and reflects isomorphisms, then it is faithful; i.e. injective on arrows.

Since the source category has all equalisers, we can let $e$ be the equaliser of $f$ and $g$, we now prove $F f = F g \implies f = g$ as follows:
\begin{align}
 & F f = F g          \\
\iff & F e \text{ iso }   & \text{[$F$ preserves equalisers and (4)]} \\
\implies & e \text{ iso }  & \text{[$F$ reflects isomorphisms]} \\
\iff     & f = g           & \text{[Applying (4) one last time ]}
\end{align}
